Chapter 7
Continuity
There are many processes and eects that depends on certain set of variables in such a way
that a small change in these variables acts as small change in the process. Changes of this
type are commonly called continuous. As an example we can consider the growth of a living
organism or dilation of heated wire. It is common to perform a computation of value of a
function in an approximate argument. Doing this we assume that the values of the function
obtained by such a manner are close enough to the exact values. However, there are function that do not obey such a nice property. For example, one cannot approximate values of
the Dirichlet function in an irrational number by its values in an (arbitrary close) rational
number. Therefore we are naturally forced to distinguish between two classes of functions
with respect to the above mentioned property of continuity. We shall introduce the notion
of continuous function (mapping). The notions of "small change" and so on will be replaced
by exact notion of a neighborhood.
In the following chapters we will draw our attention mainly to the real-valued functions
of one or more real variables, however, this chapter will consider also the mappings dened
on a subset of Rm with values in Rk . Let us mention that the proofs of the theorems considering such the mappings are essentially not dierent from the case of real-valued functions.
Besides this, we shall prove that the problem of continuity of such a mapping is equivalent
with the problem of continuity of the set of k real-valued functions each of them depending
on m real variables. In spite of this fact we recommend to the beginners to formulate all the
notions and prove relevant theorems in the case of real-valued function of one real variable.
On the other hand, the experienced readers could try to generalize some notions and/or
results to the cases of the mappings dened on a subset of a topological space and with
values in a Hausdor topological space.
Through this chapter the symbol Rs , for s ∈ N means, as usually, Hausdor topological
55
7.1 Continuity of a mapping at a point
56
space equipped with the metric topology generated by the Euclidean metric ds
Ì
ds (x, y) =
s
X
(xj − yj )2
j=1
for x = (x1 , x2 m . . . , xs ) ∈ Rs , y = (y1 , y2 , . . . , ys ) ∈ Rs .
7.1 Continuity of a mapping at a point
Denition 7.1.1. Let f : (M ⊂ Rm ) → Rk . We say that the mapping f is continuous at a
point a ∈ M , if for any neighborhood O(f (a)) of the point f (a) exists such a neighborhood
O(a) of the point a that for all x in M ∩ O(a) we have f (x) ∈ O(f (a)). We write down
lim f (x) = f (a).
x→a
(we read: the limit of f in a equals f (a))
In logical symbolic we write down:
lim f (x) = f (a) := ∀O(f (a))∃O(a)∀x ∈ M ∪ O(a) : f (x) ∈ O(f (a)).
x→a
The set M ∩ O(a) is the neighborhood of the point a in a relative topology Tr on M (see
section 5.2). Thus the set M ⊂ Rm can be understood in denition 7.1.1 as an topological
subspace (M, T ) of the topological space (Rm , Tdm ).
Let us write down another and often used denition of continuity of a mapping f : (M ⊂
Rm ) → Rk in a point a:
Denition 7.1.2. We say that the mapping f : (M ⊂ Rm ) → Rk is continuous at a point
a ∈ M if the following holds
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ M : (dm (x, a) < δ ⇒ dk (f (x), f (a)) < ²).
The equivalency of the denitions 7.1.1 and 7.1.2 follows from the fact that any neighborhood of a point in Rs contains some open ball centered at this point. Simultaneously,
this ball is a neighborhood of its own center (see section 5.2).
In the case k = 1 we have the notion of continuity of a real-valued function of many real
variables. If m = k = 1 we have simplest case: continuity of a real-valued function of one
real variable. Let us write denition 7.1.2 in explicit forms for these two important cases:
k = 1 the mapping f : (M ⊂ Rm ) → R is said to be continuous at a point a ∈ M if:
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ M : (dm (x, a) < δ ⇒ |f (x) − f (a)| < ²).
56
7.1 Continuity of a mapping at a point
57
k = m = 1 the mapping f : (M ⊂ R) → R is said to be continuous at a point a ∈ M if:
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ M : (|x − a| < δ ⇒ |f (x) − f (a)| < ²).
The geometrical interpretation of the notion of continuity of a real-valued function of one
real variable in a point is given in g. 7.1.
y
f (a) + f (x)
f (a)
f (a) − a−δ
O
x
a+δ
x
Figure 7.1:
Let us examine the notion of continuity in a point in more detail. Let us consider the situation when a is an isolated point of the set M ⊂ Rm , where the mapping f is dened. Then
the mapping f is continuous at a. In fact, a is isolated point in M implies that there exists
such a neighborhood O0 (a) of a that O0 (a) ∩ M = {a}. Therefore for any x ∈ M ∩ O0 (a) we
have f (x) = f (a) ∈ O(f (a)) for arbitrary neighborhood of the point f (a), this simply means
that f is continuous at a. This example shows us that the essential part of the denition
of continuity of a mapping in a point a lies in the nontrivial case when the point a is limit
point of the denition domain of the mapping f .
It also follows the above mentioned that if a mapping is not continuous at given point
then this point is limit point belonging to the denition domain of considered mapping.
Moreover, denition 7.1.1 implies: the mapping f : (M ⊂ Rm ) → Rk is continuous at
a point a ∈ M if and only if the restriction f |O(a)∩M , where O(a) is a neighborhood of the
point a, is continuous at a.
57
7.1 Continuity of a mapping at a point
58
THEOREM 7.1.1. Let f : (M ⊂ Rm ) → Rk and let a ∈ M be limit point of the set M .
The mapping f is continuous at a if and only if
lim f (x) = f (a).
x→a
(7.1)
Proof. a) Let the mapping f be continuous at a point a. Then it holds:
∀O(f (a))∃O(a)∀x ∈ M ∩ O(a) : f (x) ∈ O(f (a)).
(7.2)
∀O(f (a))∃O(a)∀x ∈ M ∩ Ô(a) : f (x) ∈ O(f (a)),
(7.3)
This implies that
thus (7.1) holds.
b) Let
lim f (a) = f (a).
x→a
Then (7.3) holds. Since f (a) ∈ O(f (a)) for any neighborhood O(f (a)) of the point f (a),
then (7.2) holds also, this means f is continuous at a.
Example 7.1.1. Let k ∈ R. Show that the constant function
f : x 7→ k,
x∈R
is continuous at any a ∈ R.
Solution: Any real number a is limit point of the set R and
lim f (x) = k = f (a).
x→a
Therefore, with respect to theorem 7.1.1, this function is continuous at a.
Example 7.1.2. Show that the function
a)
f : x 7→ x + [x],
x ∈ M = {0, 1} ∪ [2, 5]
is continuous at 1.
b)
g : x 7→
√
x,
is continuous at any non-negative number a.
58
x ∈ R+
0
7.1 Continuity of a mapping at a point
59
Solution: a) The continuity of the considered function in the point 1 follows from the fact
that 1 is isolated point of the denition domain of the function.
b) Any non-negative number a is a limit point of the set D(g) = R+
0 and
√
lim g(x) = a = g(a)
x→a
(as shown in example 11.6.1). With respect to the theorem 7.1.1, the function g is continuous
at a.
Example 7.1.3. Let
¨
f : x 7→
2x for x ∈ R \ {1}
0 for
x = 1.
Show that f is not continuous at 1.
Solution: It is obvious that 1 is limit point of the set R = D(f ) and
lim f (x) = 2 6= f (1) = 0.
x→1
Thus, theorem 7.1.1 implies that f is not continuous at the point 1.
Example 7.1.4. Show that the function sin is continuous at any real a.
Solution: Let ² > 0 and 0 < δ ≤ ². Then for any x such that |x − a| < δ holds:
| sin(x) − sin(a)| =

x − a x + a‹
2 cos
sin
2
2
≤
x − a 2 sin
2
≤2
|x − a|
=
2
|x − a| < δ ≤ ².
And this means that f is continuous at a.
Example 7.1.5. Show that the function
f : (x1 , x2 ) 7→
È
|x1 x2 |,
(x1 , x2 ) ∈ R2
is continuous at a = (0, 0).
Solution: For any (x1 , x2 ) ∈ R2 the inequality
|x1 x2 | ≤
x21 + x22
2
holds. Therefore
È
1 È
1
| |x1 x2 | − 0| ≤ √ x21 + x22 = √ d2 (x, a), x = (x1 , x2 ).
2
2
√
Let ² be any positive number and 0 < δ ≤ 2². Then for any x ∈ R2 such that d2 (x, a) < δ
the dierence |f (x) − f (0)| obeys the following inequality
δ
1
|f (x) − f (0)| ≤ √ d2 (x, a) ≤ √ ≤ ².
2
2
And this shows that f is continuous at (0, 0).
59
7.1 Continuity of a mapping at a point
60
Example 7.1.6. Show that each of the following functions: expa , loga , a polynomial (in one
as well as in many variables) and a rational function (in one as well as in many variables) is
continuous at any point of its denition domain.
Solution: The statement follows from the theorem 7.1.1 and from the relevant examples
discussed in previous chapter (chapter 6). Namely, see examples 6.1.13, 6.1.14, 6.3.9, 6.3.12,
6.7.5, 6.7.6 to verify that each of listed functions is continuous at any point of its denition
domain.
THEOREM 7.1.2 (On continuity of composition). Let g : (E ⊂ Rs ) → Rk and f :
(M ⊂ Rm ) → E and let a ∈ M . If f is continuous at a and g is continuous at f (a) then
the composition g ◦ f is continuous at a. (This theorem is, in fact, reformulation of the
implication 2 of theorem 6.9.4)
THEOREM 7.1.3. Let f : (M ⊂ Rm ) → R is continuous at a ∈ M . Then the following
two statements hold:
1. there exists such a neighborhood O0 (a) of the point a that f is bounded in O0 (a) ∩ M
2. if f (a) > 0 (< 0) then there exists such an neighborhood O0 (a) of the point a that f
is positive (negative) in O0 (a) ∩ M
THEOREM 7.1.4. Let f and g be a pair of real-valued functions dened on a subset M
of the Euclidean space Rm . Then their continuity in a ∈ M implies continuity in a of each
function of the following list: |f |, λf (where λ ∈ R), f + g , f g . If g(a) 6= 0 then the function
f /g is continuous at a also.
The proofs of the theorems 7.1.2-7.1.4 can be done using the proofs of the theorems on
limits from the section 6.3. One has to realize that the punctured neighborhoods in R are
to be replaced by the neighborhoods of the points in Rm . As an example, we will show the
proof of the statement 1 of the theorem 7.1.3.
Proof. (Statement 1 of the theorem 7.1.3) The assumption that f is continuous at a implies
that for ² = 1 exists such an neighborhood O0 (a) of the point a that
∀x ∈ M ∩ O0 (a) : |f (x) − f (a)| < 1.
This can be easily rewritten into the form
f (a) − 1 < f (x) < f (a) + 1.
The system of the last two inequalities directly tells us that f is bounded in M ∩ O0 (a).
With help of the proof of the theorem 6.2.1 one can easily prove the following theorem
60
7.1 Continuity of a mapping at a point
61
THEOREM 7.1.5. The mapping f : (M ⊂ Rm ) → Rk is continuous at a ∈ M if and only
if for any sequence {xn }n∈N ⊂ M such that limn→∞ xn = a we have
lim f (xn ) = f (a).
n→∞
Example 7.1.7. Show that function cos is continuous at any real number.
Solution: Let us dene the translation in the real axis
f : x 7→
π
+ x,
2
x∈R
and consider the sine function g : x 7→ sin(x), x ∈ R. Then obviously

(g ◦ f )(x) = sin
‹
π
+ x = cos(x),
2
x ∈ R.
This means the identity g ◦ f = cos holds. Let a be arbitrary real number. The continuity of
f in a (as already shown in example 6) and continuity of the sine function in entire real axis
imply (with respect to the theorem 7.1.2) that their composition g ◦ f = cos is continuous
at a, too.
Example 7.1.8. Let α ∈ R. Show that the power function
x 7→ xα ,
x ∈ R+
is continuous at any positive x0 .
Solution: We will use example 6 in which we have already shown that the exponential
function and the logarithmic function are continuous at their denition domains. Let us
dene g : y 7→ exp(y), y ∈ R and f : x 7→ α ln(x), x ∈ R+ . Then our power function can
be expressed as the following composition of f and g :
(g ◦ f )(x) = xa ,
x ∈ R+ .
Then the continuity of ln function in x0 as well as the continuity of exp function in ln(x0 )
imply that their composition g ◦ f is also continuous at x0 .
Example 7.1.9. Let a process go on in such a way that in any time t the speed of creation
of given substance is proportional to the total rate (volume, quantity) of this substance in
this time t. Find the dependence of the total rate of the quantity on time in such a process.
Solution: Let q0 be the total rate of given substance at the time t = 0. Let us divide the
interval [0, t] into n-equal subintervals
t
0,
,
n
–
™
t 2t
(n − 1)t nt
,
,...,
,
.
n n
n
n
Let the number n be large enough to be able to consider the speed of creation of the
substance in each of the subintervals is constant. With respect to our assumption this
61
7.1 Continuity of a mapping at a point
62
speed is proportional to the total rate of the substance. Let this constant be denoted as
k, (k > 0). Then the total rates of the substance in time moments
t 2t
nt
,
,...,
=t
n n
n
are given by
‚
Œ
t
kt
q1 = q0 + kq0 = q0 1 +
n
n
‚
Œ
t
kt 2
q2 = q1 + kq1 = q0 1 +
n
n
... ... ...
‚
Œ
t
kt n
qn = qn−1 + kqn−1 = q0 1 +
.
n
n
The setting of the problem implies that the process of the substance creation continuously
depends on time t. In order to obtain an exact expression for the total rate of the substance
at any given time within considered interval [0, t] one needs to perform the limit n → ∞.
This will result in
Œ
‚
kt n
.
q(t) = lim q0 1 +
n→∞
n
Now, let us compute the limit in question:
‚
kt
1+
n
lim
n→∞
Let
kt
g1 : x 7→ (1 + x) x ,
and
f1 : n 7→
kt
,
n
Œn
.
x ∈ R+ ,
n ∈ N.
Then
lim f1 (n) = 0 (f1 (n) 6= 0, n ∈ N),
n→∞
and
‚
kt
(g1 ◦ f1 )(n) = 1 +
n
Œn
,
n ∈ N.
The theorem 2.6.3 implies that if g1 has a limit at point 0 then
‚
lim
n→∞
kt
1+
n
Œn
= lim g(x).
x→0
Actually, function g1 has a limit in 0, let us show this. Let g = exp and
(
f : x 7→
62
kt ln(1+x)
for x ∈ R+
x
kt
for x = 0.
7.1 Continuity of a mapping at a point
63
Then (g ◦ f )(x) = g1 (x) for all x ∈ R+ . Since
lim+
x→0
ln(1 + x)
=1
x
(example 4.6.5), we have
lim f (x) = kt = f (0).
x→0
Thus, the theorem 7.1.1 tells us the function f is continuous at 0. Then, the continuity
of the function g = exp in kt together with the theorem 7.1.2 guarantee continuity of the
composition g ◦ f in 0. The point 0 is limit point of the denition domain of g ◦ f and
therefore theorem 7.1.1 implies
lim (g ◦ f )(x) = (g ◦ f )(0) = ekt ,
x→0
and
‚
kt
lim g1 (x) = lim (g ◦ f )(x) = e = lim
x→0
n→∞
x→0
kt
1+
n
Œn
.
So, we can conclude
q(t) = q0 ekt .
(7.4)
This shows that at given setup the total rate of the substance q in time t is given by the
formula (7.4).
The exponential dependence like (7.4) arises in many situations. Exponential growth
(k > 0) or decay (k < 0) describes such important processes as radioactive decay, bacteria
reproduction and so on. This shows great importance of the e-number in many applications
of mathematics.
Example 7.1.10. Show that the function
(x1 , x2 ) 7→ cos(x21 + x22 ),
(x1 , x2 ) ∈ R2
is continuous at any (a1 , a2 ) ∈ R2 .
Solution: Let us denote
f : (x1 , x2 ) 7→ x21 + x22 ,
(x1 , x2 ) ∈ R2 ,
and
g = cos .
Then their composition is: (g ◦ f ) = cos(x21 + x22 ) for (x1 , x2 ) ∈ R2 . The continuity of
this composition in any (a1 , a2 ) ∈ R2 is guaranteed by the theorem 7.1.2 because from the
previous examples we know that both f and g are continuous at any value of their arguments.
Example 7.1.11. Let b = (b1 , b2 , . . . , bm ) ∈ Rm . Show that the function
Ì
x 7→ dm (x, b) =
m
X
(xj − bj )2 ,
j=1
63
x = (x1 , x2 , . . . , xm ) ∈ Rm
7.1 Continuity of a mapping at a point
64
is continuous at any point a = (a1 , a2 , . . . , am ) ∈ Rm .
Solution: Let us denote
f : x = (x1 , x2 , . . . , xm ) 7→
m
X
(xj − bj )2 ,
x ∈ Rm
j=1
and
g : y 7→
√
y,
y ∈ R+
0.
The function f is continuous at a as was shown in example 6 and
lim f (x) = f (a).
x→a
Further,
lim g(y) =
y→f (a)
È
f (a) = g(f (a)).
and theorem 7.1.1 tells us that g is continuous at the point f (a). And nally, theorem 7.1.2
guarantees the continuity in a of the composition g ◦ f : Rm → R, (g ◦ f )(x) = dm (x, b) in
a.
Example 7.1.12. Show that each of the functions: tan, cot, sinha , cosha , tanha and cotha
is continuous at any point of its denition domain.
Solution: We will show that the function
tan(x) : x 7→
sin(x)
,
cos(x)
x∈M =
[ 
−
k∈Z
‹
π
π
+ kπ, + kπ
2
2
is continuous at any number x0 from M .
The functions sin cos are continuous at x0 (this was shown in examples 4 and 7 respectively)
and cos(x0 ) 6= 0, therefore theorem 7.1.4 implies that their fraction - the function tan - is
continuous at x0 .
Now we show that the function
cosha : x 7→
ax + a−x
,
2
x∈R
is continuous at any real x0 .
This statement follows the following: the function expa is continuous at x0 (example 6) and
1
are
for any x0 ∈ R: expa (x0 ) > 0. In fact, this means that the functions 12 expa and 12 exp
a
continuous at x0 and then also their linear combination
cosha =
1
1 1
expa +
2
2 expa
is, due to theorem 7.1.4, continuous at x0 .
In other cases, the continuity in the point of denition domain of given functions can be
proven analogically.
64
7.1 Continuity of a mapping at a point
65
Let f : (M ⊂ Rm → Rk ). Let us denote by fi (x) the i-th component of the value f (x)
for x ∈ M . Then f (x) = (f1 (x), f2 (x), . . . , fk (x)) for x ∈ M . All the functions fi are dened
on M :
fi : (M ⊂ Rm ) → R, (i = 1, 2, . . . , k).
Within this notation it is natural to understand the mapping f as k -tuple of functions dened
over M . This is usually represented by
f = (f1 , f2 , . . . , fk ).
The following theorem will allow for reducing the problem of continuity of the mapping f in
given point a ∈ M to the problem of continuity in a of the system of k real-valued functions
dened over M .
THEOREM 7.1.6. Let f1 , f2 , . . . , fk be real-valued functions dened on M ⊂ Rm . Then
the mapping f : M → Rk dened by
f (x) = (f1 (x), f2 (x), . . . , fk (x)),
x ∈ M,
i.e. f = (f1 , f2 , . . . , fk ) is continuous at a ∈ M if and only if any of the functions f1 , f2 , . . . , fk
is continuous at a.
Proof.
a) Let the mapping f be continuous at a. For all ² > 0 exists such δ > 0 that for x ∈ M
such that dm (x, a) < δ the inequality dk (f (x), f (a)) < ² holds. Making use of the inequality:
Ì
|f (x) − f (a)| ≤
k
X
(fj (x) − fj (a))2 = dk (f (x), f (a)) (i = 1, 2, . . . , k)
j=1
we derive that the following statement is true:
∀x ∈ M : (dm (x, a) < δ ⇒ |fi (x) − fi (a)| < ²),
(i = 1, 2, . . . , k).
This states exactly the continuity in the point a of each of the functions f1 , f2 , . . . , fk .
b) Let us suppose the functions f1 , f2 , . . . , fk be continuous at a ∈ M . This means that the
following statement is true:
²
∀i ∈ {1, 2, . . . , k}∀² > 0∃δi > 0 ∀x ∈ M : (dm (x, a) < δi ) ⇒ |fi (x) − fi (a)| < √ .
k
Now, let 0 < δ ≤ min{δ1 , δ2 , . . . , δk }. Then for x ∈ M such that dm (x, a) < δ we have
Ì
dk (f (x), f (a)) =
k
X
s
(fj (x) − fj (a))2 <
j=1
k
²2
= ².
k
And the last inequality shows that the mapping f is continuous at the point a.
65
7.1 Continuity of a mapping at a point
66
This theorem together with theorem 7.1.4 imply the following
Corollary 7.1.1. Let f1 , f2 , . . . , fk , g1 , g2 , . . . , gk be real-valued functions dened on M ⊂
Rm and let they be continuous at a point a ∈ M . Then the mappings: λf (λ ∈ R) and f + g
dened in the following way:
(λf )(x) = (λf1 (x), λf2 (x), . . . , λfk (x)),
x ∈ M,
(f + g)(x) = (f1 (x) + g1 (x), f2 (x) + g2 (x), . . . , fk (x) + gk (x))
are continuous at a. Moreover, the function (f, g) : M → R dened as
(f, g)(x) =
k
X
fj (x)gj (x),
x∈M
j=1
is also continuous at a.
Example 7.1.13. Show that the mapping f : R → R2 given by
f (t) = (cos(t), sin(t)),
t∈R
is continuous at any real t0 .
Solution: We can write the mapping in question as a pair f = (cos, sin) where the components of this mapping, i.e. the functions cos and sin are continuous at any real number.
Then theorem 7.1.6 implies that f is continuous at t0 .
Example 7.1.14. Show that the mapping
È
È
f : (x1 , x2 ) → ( |x1 x2 |, cos(x21 + x22 ), x21 + x22 ),
(x1 , x2 ) ∈ R2
is continuous at (0, 0).
Solution: Let us denote:
f1 : (x1 , x2 ) 7→
È
|x1 x2 |,
(x1 , x2 ) ∈ R2 ,
f2 : (x1 , x2 ) 7→ cos(x21 + x22 ),
f3 : (x1 , x2 ) 7→
È
x21 + x22 ,
(x1 , x2 ) ∈ R2 ,
(x1 , x2 ) ∈ R2 .
We have already shown (see examples 5, 10 and 11) that all the three functions f1 , f2 , f3
are continuous at (0, 0), thus, with respect to theorem 7.1.6, the mapping f is continuous at
(0, 0).
66
7.2 Criteria of continuity of a mapping at a point. Points of
discontinuity
67
Problems
1. Show that the following functions: x 7→ |x|, x ∈ R; x 7→ ln(1 + x), x ∈ (−1, 0] are
continuous at the point 0. Further, show that the function: x 7→ sgn(x), x ∈ R is not
continuous at the point 0.
2. Show that the function
(
x 7→
sin(x)
x
1
for x ∈ R \ {0}
for
x=0
is continuous at 0.
3. Show that the functions
x 7→
√
2
x cos(x) − x sin (x), x ∈
R+
0;
π‹
x 7→ e cos x −
, x∈R
3
2x

are continuous at any point of their denition domains.
4. Show that the function
(x1 , x2 ) 7→
È
|x1 x2 | cos(x21 + x22 ) − x2 sin(x1 x2 ), (x1 , x2 ) ∈ R2
ic continuous at any point of R2 .
5. Show that the mapping
t 7→ (et , e−t cos(t), ln(t)),
t ∈ R+
is continuous at any positive real number.
6. Show that the mapping
È
(x1 , x2 ) 7→ (x1 − x2 , ex1 x2 , sin( x21 + x22 ), 1),
(x1 , x2 ) ∈ R2
is continuous at any point of R2 .
7.2 Criteria of continuity of a mapping at a point. Points
of discontinuity
Let ∅ 6= S ⊂ M ⊂ Rm and let f : M → Rk . If the mapping
f |S : x 7→ f (x), x ∈ S
is continuous at a point a ∈ S , i.e. if
∀O(f (a))∃O(a)∀x ∈ S ∩ O(a) : f (x) ∈ O(f (a))
67
7.2 Criteria of continuity of a mapping at a point. Points of
discontinuity
68
holds, or equivalently, if the statement:
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ S : (dm (x, a) < δ ⇒ dk (f (x), f (a)) < ²)
is true, then we will note down this by
lim f (x) = f (a).
S3x→ai
In the case m = k = 1, i.e. in the case of real-valued function of one real variable, we say
that f : (M ⊂ R) → R is continuous at a point a ∈ M from the right (left) if the function
f |Ma+ (M |Ma− ), where
Ma+ = {x ∈ M ; x ≥ a} (Ma− = {x ∈ M ; x ≤ a})
is continuous at the point a, i.e. if
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ Ma+ : (|x − a| < δ ⇒ |f (x) − f (a)| < ²)
(∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ Ma− : (|x − a| < δ ⇒ |f (x) − f (a)| < ²))
holds. This statement can be rewritten into the form
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ M : (a ≤ x < a + δ ⇒ |f (x) − f (a)| < ²)
∀² ∈ R+ ∃δ ∈ R+ ∀x ∈ M : (a − δ < x ≤ a ⇒ |f (x) − f (a)| < ²)
and write down accordingly with the notation in the section 6.9 as
lim f (x) = f (a)
( lim− f (x) = f (a)).
x→a+ i
x→a i
Now we will introduce conditions that will allow for deciding whether a mapping is or is
not continuous at a given point.
THEOREM 7.2.1 (Necessary condition of continuity in a point). Let ∅ 6= S ⊂ M ⊂ Rm .
If the mapping f : M → Rk is continuous at a point a ∈ S . then the restriction of this
mapping f |S is continuous at a, i.e.
lim f (x) = f (a)
x→ai
⇒
lim f (x) = f (a).
S3x→ai
THEOREM 7.2.2 (Necessary and sucient condition of continuity in a point). Let f :
(M ⊂ Rm ) → Rk and let S1 and S2 be such subsets in M that S1 ∪ S2 = M . Let a ∈ S1 as
well as a ∈ S2 . The mapping f is continuous at a if and only if both of its restrictions f |S1
and f |S2 are continuous at a.
68
7.2 Criteria of continuity of a mapping at a point. Points of
discontinuity
69
Corollary 7.2.1. Let f : (M ⊂ R) → R and let a ∈ M . The function f is continuous at a
if and only if f is continuous at a from the left as well as from the right.
The theorems and the corollary can be proven in a way analogical to that one used to
prove the theorems from the section 6.4 (on the other hand, they follows from the theorems
1.6.10 and 2.6.10).
Example 7.2.1. Show that the function
¨
f : x 7→
2
for x ∈ {0, −1, −2}
2
1 + x + ln(x) for
x ∈ [1, 9)
is continuous at 1.
Solution: Let M := D(f ). Then M1− = {1, 0, −1, −2}, M1+ = [1, 9). The equalities:
lim f (x) =
x→1− i
lim
M1− 3x→1i
2 = 2 = f (1)
and
lim f (x) =
x→1+ i
lim
(1 + x2 + ln(x)) = 2 = f (1)
M1+ 3x→1i
imply (with respect to theorem 7.2.2) that f is continuous at 1.
Example 7.2.2. Show that the function
¨
f : (x1 , x2 ) 7→
x1
for (x1 , x2 ) ∈ {(y1 , y2 ) ∈ R2 ; y2 = y1 > 0}
x1 + x2 for (x1 , x2 ) ∈ {(y1 , y2 ∈ R2 ; y1 ≤ 0, y2 ≤ 0)}
is continuous at (0, 0).
Solution: Let S1 = {(x1 , x2 ) ∈ R2 ; x2 = x1 > 0} and S2 = {(x1 , x2 ) ∈ R2 ; x1 ≤ 0, x2 ≤ 0)}.
Let f1 : (x1 , x2 ) 7→ x1 , (x1 , x2 ) ∈ R2 , f2 : (x1 , x2 ) 7→ x1 + x2 , (x1 , x2 ) ∈ R2 . The functions
f1 , f2 are continuous at (0, 0) and therefore their restrictions f1 |S1 and f2 |S2 are continuous
at (0, 0), too. This means that the functions f |S1 and f |S2 are continuous at (0, 0) because
f |S1 = f1 |S1 and f |S2 = f2 |S2 . Obviously, S1 ∪S2 = D(f ). Therefore theorem 7.2.2 guarantees
continuity of f in the point (0, 0).
Points of discontinuity
Let f : (M ⊂ Rm ) → Rk . A point a ∈ M , in which the mapping f is not continuous, is
called point of discontinuity of the mapping f .
If a mapping f : (M ⊂ Rm ) → Rk is not continuous at a ∈ M then
∃O(f (a))∀O(a)∃x ∈ M ∩ O(a) : f (x) ∈
/ O(f (a))
holds. Furthermore, the point of discontinuity is necessarily the limit point of the set M .
So, we can speak about the limit of the mapping f in the point a. With respect to this and
69
7.2 Criteria of continuity of a mapping at a point. Points of
discontinuity
70
theorem 7.1.1 we obtain:
A mapping f : (M ⊂ Rm ) → Rk is not continuous at a point a ∈ M if and only if the point
a is limit point of the set M and the mapping f has no limit in the point a or if f has a
limit in the point a then this limit is dierent from f (a).
Example 7.2.3. Show that the function
¨
f : (x1 , x2 ) 7→
1
for (x1 , x2 ) ∈ S1 = {(x1 , x2 ) ∈ R2 ; x21 + x22 > 1}
1 − x21 − x22 for (x1 , x2 ) ∈ R2 \ S1
is not continuous at any point of the circle x21 + x22 = 1.
Solution: Let S2 = R2 \ S1 and let a = (a1 , a2 ) be a point of the circle x21 + x22 = 1. The
point a is limit point of both sets S1 and S2 and therefore we can compute the limits:
lim
S1 3(x1 ,x2 )→(a1 ,a2 )
f (x1 , x2 ) = 1,
and
lim
S2 3(x1 ,x2 )→(a1 ,a2 )
f (x1 , x2 ) =
lim
(1 − x21 − x22 ) = 0.
S2 3(x1 ,x2 )→(a1 ,a2 )
We see that f has no limit at a and we conclude f is not continuous at a.
In the case of real-valued functions we distinguish between two kinds of points of discontinuity.
Let f : (M ⊂ R)
mbbR. A point a ∈ M is called point of discontinuity of the rst kind of the function f if :
a) point a is limit point of the sets Ma+ and Ma− and the function f has a limit from the
left in a as well as it has a limit from the right in a, at least one of these limits is
dierent from f (a), or
b) the point a is limit point of the set Ma+ (Ma− ) and the function f has a limit from the
right (left) in a and this limit diers from f (a).
Point of discontinuity of a real-valued function of one real variable that is not of the rst
kind is called point of discontinuity of the second kind.
Example 7.2.4. Let
¨ 1
g : x 7→
for x = m
∈ [0, 1] ∩ Q, where
n
0 for x ∈ [0, 1] \ Q.
n
m
n
is in the simple form
This function is commonly called Riemann function. Show that this function is continuous
at any irrational number in the interval [0, 1] and that it has point of discontinuity of the
rst kind in any rational number in the interval [0, 1].
70
7.2 Criteria of continuity of a mapping at a point. Points of
discontinuity
71
Solution: Let a be any number in [0, 1]. Let ² be arbitrary positive number. There is only
a nite number of fractions m
in simple form such that n ≤ 1/². This implies that there is
n
a truncated δ -neighborhood of a that contains rational numbers (in simple form) with the
denominator greater than 1/epsilon. Thus, for x ∈ [0, 1], 0 < |x − a| < δ we have |g(x)| < ².
This means that in any point a from [0, 1] the Riemann function g has a limit and this limit
is given by
lim g(x) = 0.
x→a
Now, if a is an irrational number then g(a) = 0, thus g is continuous at a. On the other
hand, if a is a rational number, then g(a) 6= 0 and the Riemann function is not continuous
at a. Existence of the limit in a means that this point of discontinuity is of the rst kind.
Example 7.2.5. Show that the so-called Dirichlet function :
¨
χ : x 7→
1 for x ∈ Q
0 for x ∈ R \ Q
has point of discontinuity of the second kind in any real number.
Solution: Example 1.6.4 tells us that the χ-function has no limit in a real number and
therefore it has in any given real number the point of discontinuity of the second kind.
Example 7.2.6. Let
8
>
<
f : x 7→ >
:
sin(1/x) for x ∈ R+
0
for x = 0
.
−1
for x ∈ (−5, 0)
Show that the point 0 is the point of discontinuity of the second kind of the function f .
Solution: The function f evidently has no limit from the right in the point 0, therefore the
point 0 is the point of discontinuity of the second kind if f . Let us note that f has a limit
from the left in 0 - it is simply −1.
Example 7.2.7. Let f : [0, 1] → R be dened by:
¨ 1
f (x) =
for x ∈ (0, 1]
x
−1 for x = 0.
Show that 0 is the point of discontinuity of the second kind of this function.
Solution: Obviously, there is no limit of the function f in the point 0, and this implies that
the point 0 is the point of discontinuity of the second kind of the function f .
Problems
1. Show that the function x 7→ (−1)[x] , x ∈ R is continuous at any point of the set R \ Z
and that it has point of discontinuity of the rst kind in any integer.
71
7.2 Criteria of continuity of a mapping at a point. Points of
discontinuity
2. Show that the function given by:
¨
x 7→
72
x
for x ∈ Q
−x for x ∈ R \ Q
is continuous at the point 0 and that it has point of discontinuity of the second kind
in any other number.
3. Prove that the functions:
x 7→ sgn(x),
and
( sin(x)
x 7→
x∈R
for x ∈ R \ {0}
for x = 0
|x|
33
are continuous at any real number except 0 and that they have point of discontinuity
of the rst kind in 0.
4. Prove that the functions:
(
(x1 , x2 ) 7→
and
x1 x2
x21 +x22
, (x1 , x2 ) ∈ R2 \ {(0, 0)}
, (x1 , x2 ) = (0, 0)
x21 x2
x41 +x42
, (x1 , x2 ) ∈ R2 \ {(0, 0)}
, (x1 , x2 ) = (0, 0)
0
8
<
(x1 , x2 ) 7→ :
0
are discontinuous at the point (0, 0) and continuous at any other point of the Euclidean
plane.
5. Show that the function
(x1 , x2 ) 7→
¨
1 , (x1 , x2 ) ∈ S = {(y1 , y2 ) ∈ R2 ; y1 y2 = 0}
0 , (x1 , x2 ) ∈ R2 \ S
is discontinuous at any point of the set S and is continuous at any point of the set
R2 \ S .
6. Prove that the mapping
x 7→ (x + 1, sgn(x)),
x∈R
is not continuous at the point 0.
7. Let
(
f1 : (x1 , x2 ) 7→
x1 x2
x21 +x22
0
, (x1 , x2 ) ∈ R2 \ {(0, 0)}
, (x1 , x2 ) = (0, 0)
and
f2 : (x1 , x2 ) 7→ x1 x2 ,
(x1 , x2 ) ∈ R2
and let f : R2 → R2 be given by f = (f1 , f2 ). Prove that the mapping f is not
continuous at the point (0, 0).
72
7.3 Continuity of a mapping on a set
73
7.3 Continuity of a mapping on a set
Denition 7.3.1. We say that the mapping f : (M ⊂ Rm ) → Rk is continuous on M if f
is continuous at any point of M .
The set of all continuous mappings from M to Rk is denoted by C(M ; Rk ) or briey by
C(M ).
The examples 7.1.4, 7.1.6, 7.1.7, 7.1.8, 7.1.12 show us that the functions sin, cos, expa , loga ,
tan, cot, sinha , cosha , tanha , cotha , a power function, a polynomial and a rational function
of one or many real variables are continuous on their denition domains.
Moreover, example 4.6.7 and theorem 7.1.1 imply that the functions
(x1 , x2 , . . . , xm ) ∈ R,
(x1 , x2 , . . . , xm ) 7→ xi ,
(i = 1, 2, . . . , m)
are continuous on Rm .
The function
x 7→ dm (x, b),
x ∈ Rm ,
where b is a point of Rm , is continuous on Rm as shown in example 11.7.1.
The following theorem is a direct consequence of the theorem 7.1.6:
THEOREM 7.3.1. Let f1 , f2 , . . . , fk be real-valued functions dened on a set M ⊂ Rm .
The mapping f : (M ⊂ Rm ) → Rk given by f = (f1 , f2 , . . . , fk ) is continuous on M if and
only if each of the functions f1 , f2 , . . . , fk is continuous on M .
This theorem and example 7.1.13 imply that the mapping f : R → R2 , f = (cos, sin) is
continuous on R, i.e. f ∈ C(R, R2 ).
Example 7.3.1. Let a = (a1 , a2 , . . . , am ), b = (b1 , b2 , . . . , bm ) be two points of Rm . Let the
mapping ϕ : [0, 1] → Rm be dened by the prescription: ϕ(t) = bt + a(1 − t) for t ∈ [0, 1],
i.e. ϕ = (ϕ1 , ϕ2 , . . . , ϕm ) where
ϕi : t 7→ bi t + ai (1 − t),
t ∈ [0, 1] (i = 1, 2, . . . , m).
Show that the mapping ϕ ∈ C([0, 1]; Rm ).
Solution: The functions ϕ1 , ϕ2 , . . . , ϕm are continuous on [0, 1], consequently (with respect
to theorem 7.3.1) the mapping ϕ is continuous on [0, 1], i.e. ϕ ∈ C([0, 1]; Rm ).
Example 7.3.2. Let aij , i = 1, 2, . . . , k; j = 1, 2, . . . , m be given real numbers. Prove that
the linear mapping l : Rm → Rk dened by
„
l(x1 , x2 , . . . , xm ) =
m
X
j=1
73
a1j xj ,
m
X
j=1
a2j xj , . . . ,
m
X
j=1
Ž
akj xj
7.3 Continuity of a mapping on a set
74
is continuous on Rm .
Solution: The polynomials in m-variables:
(x1 , x2 , . . . , xm ) 7→
m
X
a1j xj ,
(i = 1, 2, . . . , k)
j=1
are continuous on Rm , thus, with respect to theorem 7.3.1, the mapping l is continuous on
Rm , i.e. l ∈ C(Rm ; Rk ).
Problems
1. Show that the function x 7→ ln(x +
√
1 + x2 ), x ∈ R is continuous on R.
2. Prove that the function x 7→ |x| cos(x) + e−x sin(x) + 3x − 1, x ∈ R is continuous on
R.
3. Show that the function
√
(x1 , x2 , x3 ) 7→ e
x21 +x22 +x23
+ x3 sin(x1 x2 ),
(x1 , x2 , x3 ) ∈ R3
is continuous on R3 .
4. Prove that the function
(x1 , x2 , x3 ) 7→
(
x1 x2 x3
x21 +x22 +x23
0
, (x1 , x2 , x3 ) ∈ R3 \ {(0, 0, 0)}
, (x1 , x2 , x3 ) = (0, 0, 0)
is continuous on R3 .
5. Let M ⊂ Rm and f, g ∈ C(M ; R). Show that then the following functions:
x 7→ max{f (x), g(x)}; x 7→ min{f (x), g(x)}
are also from C(M ; R).
6. Let f ∈ C(R; R). Show that the mapping
t 7→ (t, f (t)),
t∈R
is continuous on R.
7. Show that any mapping Z × Z → R5 is continuous on Z × Z.
8. Show that the mapping
È
(x1 , x2 ) 7→ (x21 − x22 , |x1 x2 |, cos(x21 + x22 ), 3),
(x1 , x2 ) ∈ R2
belongs to C(R2 , R4 ).
9. Let M ⊂ Rm and let (B(M ), dB ) be a metric space dened in example 5.2.1. Let
BC(M ; R = B(M ) ∩ C(M ; R). Prove that (BC(M ; R), dB ) is complete metric space.
(Hint: use of example 6.8.9.)
74
7.4 Uniformly continuous mappings
75
7.4 Uniformly continuous mappings
Let a mapping f : (M ⊂ Rm ) → Rk be continuous on M . This means f is continuous at
each point of M :
∀a ∈ M ∀² > 0∃δ > 0∀x ∈ M : (dm (x, a) < δ ⇒ dk (f (x), f (a)) < ²).
In this denition the number δ may depend upon choice of the point a except its dependence on ². (δ may vary from point to point without change in ²). However, there are
such mappings for which δ does not depend on a ∈ M . These mappings form an important
subset of all continuous mappings and are called: uniformly continuous. Let us write down
the denition:
Denition 7.4.1. We say that a mapping f : (M ⊂ Rm ) → Rk is uniformly continuous on
M if
∀² > 0∃δ > 0∀x ∈ M, y ∈ M : (dm (x, y) < δ ⇒ dk (f (x), f (y)) < ²).
holds.
In the special case m = k = 1 we obtain: A real-valued function of one real variable
f : (M ⊂ R) → R is uniformly continuous on M if
∀² > 0∃δ > 0∀x ∈ M, y ∈ M : (|x − y| < δ ⇒ |f (x) − f (y)| < ²).
THEOREM 7.4.1. If a mapping f : (M ⊂ Rm ) → Rk is uniformly continuous on M then
it is continuous on M .
Proof. Let ² be positive and δ > 0 be such that for any pair x, y ∈ M obeying dm (x, y) < δ
we have: dk (f (x), f (y)) < ². Let a be a point of the set M . Then for all x ∈ M such that
dM (x, a)δ we have dk (f (x), f (a)) < ² and this means continuity of f in a. As a was an
arbitrary point of M we have proved that f is continuous on M .
If a mapping f : (M ⊂ Rm ) → Rk is continuous, it has not to be uniformly continuous.
This is shown in the following example.
Example 7.4.1. Let
1
, x ∈ (0, 1).
x
Show that this continuous function on (0, 1) is not uniformly continuous on (0, 1).
Solution: Let ² > 0. Let δ be arbitrary positive number. Let us consider x1 ∈ (0, 1) ∩ (0, δ)
and dene
x1
x2 =
.
1 + (2 + ²)x1
x 7→
75
7.4 Uniformly continuous mappings
76
Then x2 ∈ (0, δ), x2 < x1 < δ and |x1 − x2 | < δ . However,
1
1
− = 2 + ² > ².
x1 x2
The last inequality, since δ was considered as arbitrary positive number, veries that our
function is not uniformly continuous on (0, 1).
Example 7.4.2. Let a, b be real numbers such that a < b. Prove then that the function
f : x 7→ x2 , x ∈ [a, b] is uniformly continuous on [a, b].
Solution: Let ² > 0. Let x, y ∈ [a, b] be such that |x − y| < δ with some positive δ . Then:
|f (x) − f (y)| = |x2 − y 2 | = |x − y||x + y| ≤ δ|x + y| ≤ 2δ(|a| + |b|).
So, if we choose (for given ² > 0) as:
δ<
²
2(|a| + |b|)
we obtain: |f (x) − f (y)| < ². We have shown f is uniformly continuous on [a, b].
Example 7.4.3. Show that the function
f : x 7→ x2 ,
x ∈ [0, ∞)
is not uniformly continuous on [0, ∞).
√
Solution: Let us x ² > 0. Let δ be arbitrary positive number. Then: for x = n and
√
y = n + 3², with n > 9²2 /δ 2 , we obtain
|f (x) − f (y)| = 3² > ².
At the same time
|x − y| =
√
n + 3² −
√
n= √
3²
δ
3²
√
< √ < < δ.
2 n
2
n + n + 3²
This implies that f is not uniformly continuous on [0, ∞).
Example 7.4.4. Prove that the function
f : (x, y) 7→ 1 + x + y,
(x, y) ∈ R2
is uniformly continuous on R2 .
Solution: Let ² be arbitrary positive number. Let
A = (x, y) ∈ R2 , B = (x0 , y 0 ) ∈ R2 .
76
7.4 Uniformly continuous mappings
77
Then
|f (A) − f (B)| = |1 + x + y − 1 − x0 − y 0 | ≤ |x − x0 | + |y − y 0 | ≤
√
2d2 (A, B).
√
So we see that if we choose 0 < δ ≤ ²/ 2, we obtain, for any pair A, B such that d2 (A, B) < δ ,
that
|f (A) − f (B)| < ².
And this demonstrates that the linear function f is uniformly continuous on R2 .
THEOREM 7.4.2. Let f1 , f2 , . . . , fk be real-valued functions dened on M ⊂ Rm . The
mapping f : M → Rk dened by
f (x) = (f1 (x), f2 (x), . . . , fk (x)),
x∈M
is uniformly continuous on M if and only if each of the functions f1 , f2 , . . . , fk is uniformly
continuous on M .
Proof. This theorem can be easily proved in the same way as theorem 7.1.6.
Theorem 7.4.2 implies that uniform continuity of the mapping f = (f1 , f2 , . . . , fk ) : (M ⊂
R ) → Rk is equivalent with the uniform continuity on the set M of k -tuple of real-valued
functions f1 , f2 , . . . , fk .
m
Problems
1. Prove that the functions
x 7→
√
x,
x ∈ R+
0;
x 7→ x +
√
1 + x2 ,
x∈R
are uniformly continuous on their denition domains.
2. Find the functions f : (M ⊂ R) → R, g : (M ⊂ R) → R such that: they are
uniformly continuous on M and their product f g is not uniformly continuous on M .
3. Prove that the functions
x 7→ sin
1
,
x
x ∈ (0, 1);
x 7→ cos(x2 ),
x ∈ R+
are not uniformly continuous on their denition domains.
4. Show that the function
(x1 , x2 ) 7→ x1 x2 ,
(x1 , x2 ) ∈ [0, 1] × (−1, 2)
is uniformly continuous on [0, 1] × (−1, 2).
77
7.5 Properties of uniformly continuous mappings on compact sets
78
5. Prove that the function
(x1 , x2 ) ∈ R2
(x1 , x2 ) 7→ x1 x2 ,
is not uniformly continuous on R2 .
6. Show that the mappings
t 7→ (cos(t), sin(t)),
t 7→ (1, t, t2 ),
t ∈ R;
t ∈ [0, 1]
are uniformly continuous on their denition domains.
7. Prove that the mapping
(x1 , x2 ) 7→ (|x1 |, x1 +
√
x2 ),
(x1 , x2 ) ∈ R × R+
0
is uniformly continuous on (x1 , x2 ) ∈ R × R+
0.
8. Prove that the mapping
(x1 , x2 ) 7→ (1, x1 x2 ),
(x1 , x2 ) ∈ R2
is not uniformly continuous on R2 .
9. Show that the mapping
(x1 , x2 , x3 ) 7→ (x1 , x22 , x33 ),
(x1 , x2 , x3 ) ∈ Z × Z × Z
is uniformly continuous on Z × Z × Z.
10. Show that the linear map l : Rm → Rk dened in example 7.3.2 is uniformly continuous
on Rm .
Answers
2 For example the functions f = g : x 7→ x, x ∈ R+
0.
7.5 Properties of uniformly continuous mappings on compact sets
In general it is not necessary that a mapping f : (M ⊂ Rm ) → Rk continuous on M must be
bounded on M , must attain its maximum value or minimum value, and must be uniformly
continuous on M . For example, the function considered in example 7.4.1:
f : x 7→
78
1
,
x
x ∈ (0, 1)
7.5 Properties of uniformly continuous mappings on compact sets
79
is continuous on (0, 1), however, it is unbounded, does not reach maximal neither minimal
value and is not uniformly continuous on (0, 1). We will show that if M ⊂ Rm is a compact
set, then a function f ∈ C(M ; R) is necessarily bounded, uniformly continuous on M and
also reaches its maximal and minimal value on M .
THEOREM 7.5.1. Let M ⊂ Rm be a compact set and let a mapping f : M → Rk be
continuous on M . Then the range of the mapping f , i.e. the set f (M ), is a compact subset
of Rk .
Proof. Let {Sa }a∈I , where I is an index set, be an open covering of f (M ). Then:
∀τ ∈ M ∃ατ ∈ I : f (τ ) ∈ Sατ
and
[
f (M ) ⊂
[
Sατ ⊂
ατ ∈M
Sα .
α∈I
f is continuous on M , so it is continuous at τ , thus for Sατ exists such a neighborhood O(τ )
of the point τ that for all x ∈ M ∩ O(τ ) we have f (x) ∈ Sατ , i.e. f (M ∩ O(τ )) ⊂ Sατ . We
can assume that O(τ ) is an open ball centered at τ . It is self-evident that
[
M⊂
O(τ ).
τ ∈M
Compactness of M allow for choosing a nite sub-covering:
n
[
M⊂
O(τi ).
i=1
Now,
M=
n
[
M ∩ O(τi ).
i=1
However, f (M ∩ O(τi )) ⊂ Sατi , for i = 1, 2, . . . , n. Thus
f (M ) ⊂
n
[
i=1
Sατi ⊂
[
Sα .
α∈I
{Sατi }i∈{1,2,...,n} is a nite sub-covering of the covering {Sα }α∈I . This means that f (M ) is
a compact set.
The following theorem is a consequence of the previous one and theorem 5.3.4.
THEOREM 7.5.2. Let M ⊂ Rm be a compact set and let a mapping f : M → Rk be
continuous on M . Then f (M ) is a closed and bounded set in Rk .
The special case of this theorem is the following one:
79
7.5 Properties of uniformly continuous mappings on compact sets
80
THEOREM 7.5.3 (Boundedness theorem (1. Weierstrass theorem)). Let M ⊂ Rm be
compact and let a function f : M → R be continuous. Then f is bounded on M .
With respect to the theorems 5.3.4 and 7.5.3, we easily obtain the following corollary
Corollary 7.5.1. Let M ⊂ R be closed and bounded set (e.g. an interval [α, β]) and let
f : M → R be a continuous function on M . Then f is bounded on M .
Furthermore, theorem 7.5.1 (for k = 1) together with theorem 5.3.5 imply the following:
THEOREM 7.5.4 (Extreme value theorem (2. Weierstrass theorem)). Let M ⊂ Rm be a
compact set and let a function f : M → R be continuous on M . Then f attains on M its
maximum value and its minimum value, i.e. there are points c1 , c2 ∈ M such that
f (c1 ) = min f (x),
f (c2 ) = max f (x).
x∈M
x∈M
As a consequence of this theorem and theorem 5.3.3 we obtain:
Corollary 7.5.2. Let f : ([α, β] ⊂ R) → R be continuous function on an interval [α, β].
Then there are numbers c1 , c2 ∈ [α, β] such that
f (c1 ) = min f (x),
f (c2 ) = max f (x).
x∈[α,β]
x∈[α,β]
A function continuous on a compact set can attain its maximal value as well as minimal
value more then once. It can attain these values innitely many times as shown on gure
7.2, where a function continuous on the interval [α, β] attains its minimal value in the entire
interval [c01 , c02 ].
Example 7.5.1. Let M ⊂ Rm be a compact set and let b ∈ Rm \ M . Prove that there is a
point c in M minimally distanced from b, (see Fig. 7.3), i.e.
∃c ∈ M : dm (c, b) = min{dm (x, b); x ∈ M }.
(Let us note that the number dm (c, b) is naturally called distance of the set M and the point
b.
Solution: Example 7.1.11 and theorem 7.2.1 imply that the function
x 7→ dm (x, b),
x∈M
is continuous on M . Compactness of M then tells us (with respect to the Extreme value
theorem) that there is c in M in which our function attains its minimum value, i.e.
dm (c, b) = min dm (x, b).
x∈M
80
7.5 Properties of uniformly continuous mappings on compact sets
81
y
f (c2 )
α = c1
O
c01
c2
c001
c02
β
x
f (c1 )
Figure 7.2:
THEOREM 7.5.5 (Heine-Cantor theorem). Let M ⊂ Rm be a compact set and let a
function f : M → R be continuous on M . Then f is uniformly continuous on M .
Proof. Let ² be arbitrary positive number. Due to continuity of f in any point y ∈ M there
is a positive δy such that:
²
dm (x, y) < δy ⇒ |f (x) − f (y)| < .
2
Let
¨
«
δy
O(y; δy ) = x ∈ R ; dm (x, y) <
.
2
m
The system {O(y; δy )}y∈M is an open covering of M . Due to compactness of M we can
choose a nite sub-covering, i.e. there is a nite sequence y1 , y2 , . . . , yn ∈ M such that
M⊂
n
[
O(yi , δyi ).
i=1
Let
1
min{δy1 , δy2 , . . . , δyn }.
2
Let x1 , x2 ∈ M be such that dm (x1 , x2 ) < δ . The nite system of open sets {O(yi ; δyi )}ni=1
covers the set M . Therefore, there is an index j ∈ {1, 2, . . . n} such that x1 ∈ O(yj , δyj ) and
thus dm (x1 , yj ) < δyj /2. Making use of the triangle inequality we obtain
δ=
dm (x2 , yj ) ≤ dm (x2 , x1 ) + dm (x1 , yj ) < δ +
81
δyj
δy
δy
≤ j + j = δyj .
2
2
2
7.5 Properties of uniformly continuous mappings on compact sets
82
c
b
Figure 7.3:
Furthermore,
²
²
and
|f (x2 ) − f (yj )| <
2
2
and nally we can estimate the dierence between the values of f at x1 and x2 , respectively
²
²
|f (x1 ) − f (x2 )| ≤ |f (x1 ) − f (yj )| + |f (yj ) − f (x2 )| < + = ².
2 2
This shows that f is uniformly continuous on M .
|f (x1 ) − f (yj )| <
Corollary 7.5.3. Let M ⊂ R be closed and bounded set (M = [α, β] for example) and let
f : M → R be continuous on M . Then f is uniformly continuous on M .
Corollary 7.5.4. Let M ⊂ Rm be a compact set and let a mapping f = (f1 , f2 , . . . , fk ) :
M → Rk be continuous on M . Then f is uniformly continuous on M .
Example 7.5.2. Let α, β be a pair of real numbers such that α < β . Let f : (α, β) → R be
monotonous, bounded and continuous on (α, β). Show that such a function f is uniformly
continuous on (α, β).
Solution: There exists the limits
lim f (x) = A ,
x→α
lim f (x) = B,
x→β
since f is monotonous and bounded. Let us consider the following function dened on [α, β]:
8
>
<
f (x) , x ∈ (α, β)
, x=α
g : x 7→ > A
: B
, x = β.
82
7.6 Properties of continuous mappings on connected sets
83
Due to f ∈ C(α, β) and
lim g(x) = g(α) ,
x→α
lim g(x) = g(β)
x→β
we see that g is continuous on [α, β] and therefore (corollary 7.5.3) g is uniformly continuous
on [α, β]. Since f = g|(α,β) , f is uniformly continuous on (α, β).
Problems
1. Construct a continuous and bounded function f dened on R such that f does not
attain neither maximum value nor minimum value.
2. Let f : ([a, b] ⊂ R) → R be continuous on [a, b]. Prove that the functions:
x 7→ min f (t),
a≤t≤x
x ∈ [a, b];
x 7→ max f (t),
a≤t≤x
x ∈ [a, b]
are continuous on [a, b].
3. Let α, β be a given pair of real numbers such that α < β . Let f : (α, β) → R be
continuous on (α, β) and f have limits in the end-points α, β , i.e. let exist real numbers
A, B such that
lim f (x) = A,
lim f (x) = B.
x→α
x→β
Then f is uniformly continuous on (α, β). Prove it!
4. Let M ⊂ Rm be a closed set and let b ∈ Rm \ M . Show that there exists such a point
c ∈ M that
dm (c, b) = min dm (x, b).
x∈M
(Hint: Let a ∈ M and A = {x ∈ M ; dm (x, b) ≤ dm (a, b)}. Then A is compact set and
dm (c, b) = minx∈A dm (x, b).)
7.6 Properties of continuous mappings on connected sets
Lemma 7.6.1. Let f : (M ⊂ Rm ) → Rk be a continuous on M . Let A ⊂ Rk be closed in
Rk and let S ⊂ M be closed in Rm . Then the set SA = {t ∈ S; f (t) ∈ A} is closed in Rm .
Proof. Let a ∈ Rm be an arbitrary limit point of the set SA . Then a is limit point of the set
S , too. Since S is closed, a ∈ S and a belongs also to M . The mapping f is continuous at
a. Theorem 7.1.1 implies
lim f (x) = f (a).
x→a
83
7.6 Properties of continuous mappings on connected sets
84
This means that (with respect to theorem 6.4.1)
lim f (x) = f (a).
SA 3x→a
(1)
The following two cases are possible:
a) f (a) is a limit point of the set A
b) f (a) is not a limit point of the set A .
In the case a) the fact A is closed implies f (a) ∈ A and therefore a ∈ SA . In the case b) we
0 (f (a)) ∩ A = ∅. Then (1) implies:
obtain there exists a neighborhood O0 (a) of a such that O
(f (a)) ∩ SA
for O0 (f (a)) exists such a neighborhood of the point a that that for all x ∈ O
(f (a)) ∩ SA only f (a) belongs to
we have f (x) ∈ O0 (f (a)). Among the values f (x), x ∈ O
(f (a)), thus f (x) = f (a) for all x ∈ O
(f (a)) ∩ SA . This implies that a ∈ SA .
O
The intervals [c, +∞) as well as (−∞, c], c ∈ R are closed sets in R. This fact together
with lemma 7.6.1 imply the following:
Corollary 7.6.1. Let f : (M ⊂ Rm ) → R be a continuous function on M and let S ⊂ M
be a closed set in Rm . Then for all c ∈ R the sets:
{x ∈ S : f (x) ≤ c},
and
{x ∈ S : f (x) ≥ c}
closed in Rm .
Denition 7.6.1. A closed set M ⊂ Rm is called connected if there are no non-empty closed
sets A, B in Rm such that
M =A∪B
and
A ∩ B = ∅.
As a simple example of a connected set we have any one-point set in Rm .
Lemma 7.6.2. Any closed interval [α, β] ⊂ R is a connected set.
Proof. Let us suppose the interval in question [α, β] is not a connected set. This mens there
are two closed non-empty number sets A, B such that
[α, β] = A ∪ B
and
A ∩ B = ∅.
Let a ∈ A and b ∈ B and let us choose them in such a way that a < b. Let
σ = sup{A ∩ [a, b]}.
It is self-obvious that a ≤ σ ≤ b. We will show that σ ∈
/ A ∪ B . If σ ∈ A then σ < b and
(σ, b] ⊂ B . This implies that σ is a limit point of the set B , and therefore σ ∈ B - but this is
84
7.6 Properties of continuous mappings on connected sets
85
impossible because the sets A and B are disjoint. If σ ∈ B then σ ∈
/ A and with respect to
example 5.2.9 σ must be a limit point of the set A. This, however, implies a contradiction
because A was supposed to be closed. So, we have shown that σ ∈
/ A ∪ B , i.e. σ ∈
/ [α, β].
Thus, the number σ obeys a < σ < b where a, b is a pair of numbers belonging to the interval
[α, β]. This means that [α, β] cannot be an interval and this is an contradiction. As a result
we have the interval [α, β] is a connected set.
Lemma 7.6.3. Let φ : ([α, β] ⊂ R) → Rm be continuous on [α, β]. Then the values of φ,
i.e. the set φ([α, β]) is a closed and connected set.
Proof. Theorem 7.5.2 implies that φ([α, β]) is a closed set in Rm . We will show it is also
connected. Let us suppose that φ([α, β]) is not a connected set. Then there are two nonempty closed disjoint sets A, B in Rm such that their union is equal exactly to φ([α, β]).
With respect to lemma 7.6.1 the sets SA and SB are closed in R and obviously SA 6= ∅ as
well as SB 6= ∅, and
[α, β] = SA ∪ SB ,
SA ∩ SB = ∅.
This means the interval [α, β] is not a connected set and this is a contradiction with the
statement of lemma 7.6.2. Thus φ([α, β]) is a connected set.
Now we will extend the notion of connectedness to the sets that are not closed.
Denition 7.6.2. We say that a set M ⊂∈ Rm is connected if for any pair of points a, b ∈ M
exists a closed and connected set S ⊂ M such that a, b ∈ S .
It is obvious that M ⊂ Rm - a closed connected set with respect to denition 7.6.1 is also
a connected set with respect to denition 7.6.2 (it suces to consider S = M ).
Any interval I ⊂ R is a connected set. In fact, for any two points a, b from I is the set
{x ∈ I : a ≤ x ≤ b} closed and connected and containing both a and b (lemma 7.6.2) and
this means (with respect to denition 7.6.2) that the interval I is a connected set.
Example 7.6.1. Show that a connected set M ⊂ R is either an one-point set or an interval.
Solution: If M is an one-point set then it is connected. Let M ⊂ R be connected and
let it contain at least two dierent points. We will show that then M must be an interval
by indirect proof. Let us suppose that M is not an interval. Then there are two points
a, b, a < b in M such that there exists a point c: a < c < b such that c ∈
/ M . Since M is
connected there exists a closed and connected set S ⊂ M such that it contains both points
a and b. With respect to corollary 7.6.1 (for m = 1, f = id ) are the sets
Sc− = {x ∈ S; x ≤ c},
85
Sc+ = {x ∈ S; x ≥ c}
7.6 Properties of continuous mappings on connected sets
86
closed in R, furthermore Sc− 6= ∅, Sc+ 6= ∅ (a ∈ Sc− , b ∈ Sc+ ), and
S = Sc− ∪ Sc+ ,
Sc− ∩ Sc+ = ∅.
However, this is a contradiction with the connectedness of S . This shows that a connected
set M ⊂ R (if not an one-point set) is an interval.
Denition 7.6.3. Let a mapping φ : ([α, β] ⊂ R) → Rm be continuous on [α, β] and let
φ(α) = a and φ(β) = b. Then the set φ([α, β]) is called the path from a to b (or equivalently,
the path connecting a and b).
Denition 7.6.4. A non-empty set M ⊂ Rm is called path-connected if any two points of
M can be connected by a path that belongs to M .
THEOREM 7.6.1. If a set M ⊂ Rm is path-connected then it is connected.
Proof. Let a and b are arbitrary points from M . Then there is a path connecting these two
points, i.e. there is a continuous (on [α, β]) mapping φ : ([α, β] ⊂ R) → Rm such that
φ(α) = a, φ(β) = b and φ([α, β]) ⊂ M . We know (lemma 7.6.3) that φ([α, β]) is closed and
connected and this shows that for any two points a, b ∈ M exists a closed and connected set
φ([α, β]) that contains the points a and b. This means that M is connected.
Example 7.6.2. Let c ∈ Rm and r > 0. Show that the closed ball
B(c, r) = {x ∈ Rm , dm (c, x) ≤ r}
and the open ball
K(c, r) = {x ∈ Rm , dm (c, x) < r}
are connected sets.
Solution: We will show that our closed ball B(c, r) is a connected set. Let c = (c1 , . . . , cm )
and let a = (a1 . . . . , am ) and b = (b1 , . . . , bm ) be any two points in B(c, r). Let us dene a
mapping φ as follows:
φ : [0, 1] → Rm ,
φ(t) = bt + (1 − t)a.
This means we have φ = (φ1 , . . . , φm ) with
φj : t 7→ bj t + (1 − t)aj ,
86
t ∈ [0, 1] (j = 1, 2, . . . , m).
7.6 Properties of continuous mappings on connected sets
87
It is obvious that φ([0, 1]) is the path from a to b. We will show that this path is in B(c, r),
in fact, for any t ∈ [0, 1] we have:
Ì
dm (φ(t), c) =
m
X
Ì
(bj t + (1 − t)aj − cj )2 =
j=1
2
4t2
m
X
(bj − cj )2 + 2t
j=1
”
m
X
[(bj − cj )t + (1 − t)(aj − cj )]2 =
j=1
m
X
(bj − cj )(aj − cj ) + (1 − t)2
j=1
m
X
31/2
(aj − cj )2 5
≤ |Cauchy's ineq.|
j=1
—1/2
t2 d2m (b, c) + 2t(1 − t)dm (b, c)dm (a, c) + (1 − t)2 d2m (a, c)
q
=
[tdm (b, c) + (1 − t)dm (a, c)]2 = tdm (b, c) + (1 − t)dm (a, c) ≤ tr + (1 − t)r = r.
Thus, theorem 7.6.1 tells us that B(c, r) is a connected set. By a similar way one can prove
that our open ball K(c, r) is also a connected set. However, the connectedness of the open
ball K(c, r) can be proven also in the following way: let a, b be any two points in K(c, r) and
let
r1 = max{dm (a, c), dm (b, c)}.
Since the closed ball B(c, r1 ) is connected and {a, b} ⊂ B(c, r1 ) ⊂ K(c, r), the ball K(c, r) is
connected, too.
The connected sets in Rm with m ≥ 2 can have very complicated structure. Figure 7.4
shows a connected set in R2 and gure 7.5 shows a disconnected set in R2 .
THEOREM 7.6.2. Let M ⊂ Rm be a connected set and let f : M → R be continuous
on M . Let x1 , x2 be two points in M such that f (x1 ) < f (x2 ). Then for any y0 such that
f (x1 ) < y0 < f (x2 ) there exists a point x0 ∈ M such that f (x0 ) = y0 .
Proof. Connectedness of M involves existence of a closed and connected set S ⊂ M such
that both points x1 , x2 are in S . Let us consider the following two sets
Sy+0 = {x ∈ S; f (x) ≥ y0 }
Sy−0 = {x ∈ S; f (x) ≤ y0 }.
It is obvious that x1 ∈ Sy−0 and x2 ∈ Sy+0 and therefore both sets Sy±0 are non-empty. Furthermore,
Sy+0 ∪ Sy−0 = S.
Continuity of f (on M ) guarantees (see lemma 7.6.1) that the sets Sy±0 are closed in Rm .
Connectedness of S implies Sy+0 ∩Sy−0 6= ∅. Let x0 ∈ Sy+0 ∩Sy−0 . Then f (x0 ) ≥ y0 and f (x0 ) ≤ y0
and therefore f (x0 ) = y0 .
Corollary 7.6.2. Let I ⊂ R be an interval and let f : I → R be continuous on I . If
x1 , x2 are two points in I such that f (x1 ) < f (x2 ) then the function f attains all the values
between f (x1 ) and f (x2 ). Especially, if f (x1 )f (x2 ) < 0 then there exists such x0 between x1
and x2 that f (x0 ) = 0.
87
7.6 Properties of continuous mappings on connected sets
88
Figure 7.4:
Figure 7.5:
Example 7.6.3. Let us consider the function:
f : x 7→ x − cos(x), x ∈ [0, 1].
Show that there exists x0 ∈ [0, 1] such that x0 = cos(x0 ).
Solution: f ∈ C([0, 1]), f (0) = −1 < 0, f (1) = 1 − cos(1) > 0. With help of corollary
7.6.2 we can state there exists x0 in [0, 1] such that f (x0 ) = 0 and this exactly means that
x0 = cos(x0 ).
Example 7.6.4. Let the function f : ([α, β] ⊂ R) → [α, β] be continuous on [α, β]. Show
that there exists t ∈ [α, β] such that f (t) = t.
Solution: If f (α) = α or f (β) = β then t = α or t = β . Let this trivial case be not our
case, i.e. f (α) 6= α and f (β) 6= β . Then α < f (α) as well as f (β) < β . Let us consider the
function g :
g : x 7→ x − f (x), x ∈ [α, β].
Then g ∈ C([α, β]; R) and g(α) = α − f (α) < 0 and g(β) = β − f (β) > 0. Thus corollary
7.6.2 guarantees the existence of mentioned t in [α, β] at which g(t) = 0, i.e. t − f (t) = 0.
THEOREM 7.6.3. Let I ⊂ R be an interval and let a function f : I → R be continuous
on I . Then the f -image of I (values of f ) is either an one-point set or an interval.
88
7.6 Properties of continuous mappings on connected sets
89
Proof. If f is a constant function then f -image of I is obviously one point. Let f be not
a constant function. We will show that f (I) is an interval. Let y1 , y2 be any two points in
f (I) and let y1 < y2 . Corollary 7.6.2 implies that f attains all the values between y1 and y2 ,
i.e. (y1 , y2 ) ⊂ f (I). This shows that f (I) must be an interval.
Corollary 7.6.3. Let I ⊂ R be an interval and let f : I → R be continuous and strictly
increasing or strictly decreasing on I . Then f (I) is an interval.
THEOREM 7.6.4. Let I ⊂ R be an interval and let f : I → R be monotonic on I such
that f (I) is an interval. Then the function f is continuous on I .
Proof. Let us consider that f : I → R is non-decreasing and that I is an open interval.
(In other cases, the proof can be done in an analogical way). Let f (I) = J . Indirectly: let
f be discontinuous at some point c in I . Theorems 6.5.1 and 6.5.2 imply that f has the
(proper) limit from the left as well as the (proper) limit from the right in the point c. Let
us denote these limits by f (c+ ) and f (c− ), respectively. f is nondecreasing and therefore if
x1 ∈ I , x1 < x < c, we have f (x1 ) ≤ f (x) ≤ f (c) and if x2 ∈ I and x2 > x > c, we have
f (x2 ) ≥ f (x) ≥ f (c). Thus
f (x1 ) ≤ f (c− ) ≤ f (c) ≤ f (c+ ) ≤ f (x2 ).
This shows that the numbers f (c+ ) and f (c− ) belong to the interval J . Discontinuity of f
at c implies f (c− ) < f (c+ ). Let us choose a number d such that f (c− ) < d < f (c+ ) (see
gure 7.6). Necessarily, d ∈ J , however, f does not attains the value d in any point of I ,
but this is in contradiction with our assumption that f maps I into an interval (J ).
We already know that the exponential function x 7→ ax , x ∈ R is continuous on its
domain, i.e. on whole real line (see example 7.1.6), for instance. With respect to theorem
7.6.4 and theorem 4.4.4 one can nd out that the logarithmic function is continuous on its
domain.
THEOREM 7.6.5 (On continuity of inverse function). Let I ⊂ R be an interval and let
f : I → R be continuous and increasing (decreasing) on I . Then f maps I into an interval,
say J . There exists an inverse function f −1 : J → I that is continuous and increasing
(decreasing) on J .
Proof. The statement that f maps I into an interval follows from corollary 7.6.3. The
existence of the inverse function f −1 follows from theorem 4.2.2. This theorem implies also
that f −1 is increasing (decreasing) on J and maps J into I . Thus, theorem 7.6.4 implies
that f −1 is continuous on J .
89
7.6 Properties of continuous mappings on connected sets
90
y
f (c+ )
d
f (c)
f (c− )
O
x
c
Figure 7.6:
Example 7.6.5. Show that the function
"Ê
2
f : x 7→ sin(x ),
x∈I=
3π
,
2
Ê
5π
2
#
has an inverse function dened on J = [−1, 1]. Find this inverse function and prove that it
is increasing and continuous on J .
Solution: We will start with the monotonicity of f on I . Let
Ê
3π
≤ x1 < x 2 ≤
2
Ê
5π
.
2
Then
3π
5π
≤ x21 < x22 ≤
.
2
2
The function sin |[3π/2,5π/2] is increasing and therefore sin(x21 ) < sin(x22 ). This means f is
increasing on I . The continuity of the functions sin and x 7→ x2 on R implies continuity
of its composition x 7→ sin(x2 ) on R (theorem 7.1.2). Theorem 7.2.1 then implies that f is
continuous on I . Furthermore, we have
min f (x) = sin
x∈I
3π
2
= −1,
max f (x) = sin
x∈I
5π
2
= +1.
As a consequence of theorem 7.6.3 we see that the values of f coincide with J . Theorem 7.6.5
guarantees existence of the inverse function f −1 : J → I that is continuous and increasing
90
7.6 Properties of continuous mappings on connected sets
91
on J .
We will nd the function f −1 . We know (theorem 2.2.3) that f ◦ f −1 = iJ , i.e.
sin
h€
Š2 i
f 1 (x)
= x,
x ∈ J.
For all x ∈ J : f −1 (x) ∈ I , therefore
€
f
−1
Š2
(x)
•
π π˜
− 2π ∈ − ,
2 2
holds. Evidently,
”
—
”
—
sin (f −1 (x))2 − 2π = sin (f −1 (x))2 = x,
x ∈ J.
With respect to what has just been said about the values of the function f −1 we obtain the
following formula
€
Š2
f −1 (x) − 2π = arcsin(x),
which implies
f −1 (x) = (2π + arcsin(x))1/2 ,
x ∈ J.
Example 7.6.6. Show that the function
f : x 7→ x − sin(x),
x∈R
has an inverse function that is increasing and continuous on R.
Solution: For any two real numbers x1 , x2 such that x1 < x2 we have
| sin(x1 ) − sin(x2 )| = 2 cos

x1 − x2
x1 + x2 ‹
sin
2
2
≤ 2 sin
x1 − x2
2
< x 2 − x1 .
This implies that
f (x2 ) − f (x1 ) = x2 − x2 − (sin(x2 ) − sin(x1 )) > x2 − x1 − (x2 − x1 ) = 0,
and we see that f is increasing on R. It is evident that f is continuous on R. We will show
that f (R) = R. This will follow from the following: let y be arbitrary real number, then:
f (y − 2) = y − 2 − sin(y − 2) ≤ y − 2 + 1 = y − 1 < y < y + 1 = y + 2 − 1
≤ y + 2 − sin(y + 2) = f (y + 2),
and we can use corollary 7.6.2 to see that there exists x ∈ R such that f (x) = y . This
means that f (R) = R. Now, theorem 7.6.5 guarantees existence of the inverse function
f −1 : R → R and it is continuous and increasing on R.
91
7.6 Properties of continuous mappings on connected sets
92
Continuity of elementary functions
The continuity of goniometric functions on their domains (examples 7.1.4, 7.1.7 and 7.1.12)
together with theorems 4.4.10 and 7.6.5 imply continuity of the inverse trigonometric functions (the functions arcsin, arccos, arctan) on their domains. We have shown (see examples
7.1.6, 7.1.7, 7.1.8 and 7.1.12) that the basic elementary functions as well as any polynomial
of many real variables are continuous on their domains. Theorems 7.1.2 and 7.1.4 allow to
state that all elementary functions of one as well as many real variables are continuous on
their domains.
Such the elementary functions are e.g.:
f : x 7→ arcsin(sin(x)),
see gure 7.7
x∈R
1
x
x ∈ R \ {0}
see gure 7.8
q
È
x
9
−2/3
x7
3 + cos(x2 ) +
h : x 7→ x
+ arcsin
+
2
+
x
ln(1 + x4 ), x ∈ R+
2
1+x
È
7
F : (x1 , x2 ) 7→ x1 − x2 + arctan(x1 x2 + 3x42 ) + x31 + arcsin(x2 ) + sin(ln( 1 + x21 + x22 )) + 2,
g : x 7→ −1 + e ,
(x1 , x2 ) ∈ R × [−1, 1]
G : (x1 , x2 , x3 ) 7→ |x1 + x2 + x3 |
√
2
2
2
2
+ earccos(x1 +x2 +x3 ) ,
(x1 , x2 , x3 ) ∈ G3 (0, 1),
where
G3 (0, 1) = {(x1 , x2 , x3 ) ∈ R3 ; x21 + x22 + x23 ≤ 1}
is the closed ball of unit radius centered at (0, 0, 0).
Problems
1. Find whether the sets M1 , M2 , M3 , M4 are connected in R2 :
M1 = {(x, y) ∈ R2 ; xy = 1},
1
M2 = { x, x sin
∈ R2 ; x ∈ (0, 1]} ∪ {0, 0},
x
M3 = {(x, y) ∈ R2 ; xy = 1},
M4 = {(x, y) ∈ R2 ; x2 + y 2 = 1}.
2. Let A, B be a pair of connected sets in Rm such that A ∩ B 6= ∅. Prove that the union
of the sets A and B is then also a connected set in Rm .
3. Let I ⊂ R be an interval and let x1 , x2 , . . . xn are some points of I . Let f ∈ C(I; R).
Prove that there is x ∈ I such that
f (x) =
f (x1 ) + f (x2 ) + · · · + f (xn )
.
n
92
7.6 Properties of continuous mappings on connected sets
93
y
π
2
−π
π
2
− π2
π
3
2π
2π
x
− π2
Figure 7.7:
y
O
x
Figure 7.8:
4. Show that there is at least one number x0 in (0, 1) such that
3x0 = 5x0 .
5. Show that a polynomial of third order has, at least, one (real) root.
6. Let [a, b] ⊂ R and let f and g be continuous functions on [a, b] such that f (a) < g(a) and
f (b) > g(b). Show that there is at least one c ∈ (a, b) with the property: f (c) = g(c).
7. Let f be a monotonic function mapping the interval [a, b] ⊂ R into [a, b]. Show that
there is at least one c in (a, b) such that c = f (c).
8. Show that the function
f : x 7→ tan(x) − x,
93

π π‹
x∈ − ,
2 2
7.7 Continuity of the mappings in metric spaces
94
has an inverse function dened, continuous and increasing on R.
9. Show that the function
¨
f : x 7→
x
, x ∈ [0, 1]
x − 2 , x ∈ (3, 4]
(that is continuous on D(f ) = [0, 1] ∪ (3, 4]) has an inverse function dened and increasing on [0, 2]. Show that this inverse function is not continuous on [0, 2]. (Compare
this result with theorem 7.6.5.)
10. Prove that the set of points of discontinuity of a monotonic function is at most countable (is nite or innite but countable).
11. Prove that the function
(x1 , x2 ) 7→ cosh(x2 ) sin(x1 ) + sinh(x2 ) cos(x1 ),
(x1 , x2 ) ∈ R2
takes on the value 0 at innitely many points of the plane R2 .
Answers
1. The set M1 is not connected and the sets M2 , M3 and M4 are connected in R2 .
7.7 Continuity of the mappings in metric spaces
In this section we will extend in a natural way the notions, statements and theorems concerning continuity of the mappings from Rm to Rk we were dealing with in the previous
sections of this chapter to the case of mappings dened on a given metric space and with
values in another metric space. We will not repeat proofs of theorems if they are like enough
to their Euclidean space - to Euclidean space counterparts.
7.7.1 Continuity at a point
We know that every metric space can be considered as the Hausdor topological space
equipped with the so-called metric topology. Particulary, every point of a metric space has
a system of (open) neighborhoods.
Denition 7.7.1. Let (X, dX ), (Y, dY ) be a pair of metric spaces. We say that a mapping
f : (M ⊂ X) → Y is continuous at a point a ∈ M if
∀O(f (a))∃O(a)∀x ∈ M ∩ O(a) : f (x) ∈ O(f (a))
holds. We use the notation
lim f (x) = f (a).
x→ai
94
7.7 Continuity of the mappings in metric spaces
95
Let us mention that the set M ∩ O(a) is an open neighborhood of the point a in a relative
topology on M .
We often say that the mapping f : (M ⊂ X) → Y , where (X, dX ) and (Y, dY ) are metric
spaces, is the mapping from the metric space (X, dX ) to the metric space (Y, dY ).
Any neighborhood of a point of a metric space contains an open ball centered at given
point and this ball is also a neighborhood of given point (see theorem 5.2.1), therefore we
can rewrite denition 7.7.1 into the form:
Denition 7.7.2 (An equivalent denition of continuity at a point). Let (X, dX ), (Y, dY )
be a pair of metric spaces. We say that a mapping f : (M ⊂ X) → Y is continuous at a
point a ∈ M if
∀² > 0∃δ > 0∀x ∈ M : (dX (x, a) < δ ⇒ dY (f (x), f (a)) < ²).
It is easy to see that this denition, in the special case X = Rm , dX = dm ; Y =
Rk , dY = dk , reduces to denition of continuity of a mapping from Rm to Rk at given point
(see denition 7.1.2).
Example 7.7.1. Let I ⊂ R be an interval and let (B(I), dB ), (R, d1 ) be metric spaces
dened in example 5.2.1. (B(I) is the linear space of all bounded functions dened on the
interval I equipped with the supremum metric.) Let then x0 ∈ I , f0 ∈ B(I) and let a
function F : B(I) → R be dened as follows
F (f ) = f (x0 ),
f ∈ B(I).
We will show that F is continuous at f0 .
Solution: For any f, f0 ∈ B(I) we have
d1 (F (f ), F (f0 )) = |f (x0 ) − f0 (x0 )| ≤ sup |f (x) − f0 (x)| = dB (f, f0 ).
x∈I
Let ² be an arbitrary positive number and let 0 < δ ≤ ². Then for any f ∈ B(I) such that
dB (f, f0 ) < δ we have
d1 (F (f ), F (f0 )) ≤ dB (f, f0 ) ≤< δ ≤ ².
This shows that F is continuous at f0 .
The notion of the continuity of a mapping at given point is closely related to the limit
of a sequence. This is expressed in details in the following theorem.
95
7.7 Continuity of the mappings in metric spaces
96
THEOREM 7.7.1. Let (X, dX ), (Y, dY ) be metric spaces. The mapping f : (M ⊂ X) → Y
is continuous at a ∈ M if and only if for any sequence {xn }n∈Z ⊂ M with the property
limn→∞ xn = a:
lim f (xn ) = f (a)
n→∞
holds.
Proof. (a) Let the mapping f : (M ⊂ X) → Y be continuous at a ∈ M and let the
sequence {xn }n∈Z ⊂ M converges to a. Continuity of f at a means
∀² > 0∃δ > 0∀x ∈ M : (dX (x, a) < δ ⇒ dY (f (x), f (a)) < ²).
Since
lim xn = a,
n→∞
there exists n0 ∈ N for any δ > 0 such that for all n ≥ n0 we have dX (xn , a) < δ . Then
dY (f (xn ), f (a)) < ²
for all n ≥ n0 . Thus
lim f (xn ) = f (a).
n→∞
(b) Let for any sequence {xn }n∈Z ⊂ M that converges to a ∈ M be limn→∞ f (xn ) = f (a).
Let us suppose that f is not continuous at a, i.e. the following statement
∃² > 0∀δ > 0∃xδ ∈ M : dX (xδ , a) < δ ∧ dY (f (xδ ), f (a)) ≥ ²0
is true. We can choose the numbers δ as 1, 1/2, 1/3, . . . , 1/n, . . . . For every n ∈ N
there exists a point xn ∈ M such that
dX (xn , a) <
1
∧ dY (f (xn ), f (a)) ≥ ²0 .
n
This means that limn→∞ xn = a but the sequence {f (xn )}n∈N has not limit equal to
f (a). This is, in fact, in contradiction with our assumption that limn→∞ f (xn ) = f (a)
if limn→∞ xn = a. Thus the mapping f must be continuous at our point a.
This theorem an the theorems on proper limits (see section 6.3) imply the following
Corollary 7.7.1. Let (X, dX ) be a metric space and let the functions f : (M ⊂ X) → R,
g : (M ⊂ X) → R be continuous ar a ∈ M . Then the functions |f |, λf (λ ∈ R), f + g, f g
are continuous at a. If g(a) 6= 0 then also the fraction f /g is continuous at a.
This corollary implies theorem 7.1.4 and theorem 5.7.1 is a consequence of theorem 7.7.1.
Theorem 7.7.1 and corollary 7.7.1 allow us to decide in certain cases whether considered
mapping is continuous or is not at given point.
96
7.7 Continuity of the mappings in metric spaces
97
Example 7.7.2. Let the function F : (C([0, 1]; R) ⊂ B([0, 1]) → R) be dened as follows
F (f ) =
8
>
0
>
<
1
>2
>
:1
if f attains negative values ,
if f (x) = 0, ∀x ∈ [0, 1] ,
for nonnegative and nonzero function f .
Show that the function F is not continuous at f0 = 0.
Solution: Let us consider the following sequence of functions from C([0, 1]; R):
fn : x 7→
xn
, x ∈ [0, 1]
n
It is easy see that
(n = 1, 2, 3, . . . ).
xn
1
= ,
n
x∈[0,1] n
dB (fn , f0 ) = sup
n ∈ N.
This immediately implies that
lim fn = f0 .
n→∞
Since F (fn ) = 1 for all n ∈ N, we have
lim F (fn ) = 1 6= 0 = F (f0 ).
n→∞
And this means (with respect to corollary 7.7.1) that F is not continuous at f0 .
7.7.2 Continuity on a set
Let (X, dX ) and (Y, dY ) be a pair of metric spaces. We say that f : (M ⊂ X) → Y is
continuous on M if f is continuous at any point of the set M . The set of all continuous
mappings from M to (Y, dY ) is denoted by C(M ; Y ) or briey C(M ).
The mapping f : (M ⊂ X) → Y is called uniformly continuous on M if the statement
∀² > 0∃δ > 0∀x1 , x2 ∈ M : (dX (x1 , x2 ) < δ ⇒ dY (f (x1 ), f (x2 )) < ²).
It is evident that an uniformly continuous mapping on M is continuous on M . The
opposite statement is not true (see example 7.4.1).
In an analogical way as we have used to prove the boundedness theorem as well as Cantor
theorem in section 7.5 the following statements can be proved.
THEOREM 7.7.2. Let (X, dX ), (Y, dY ) be metric spaces and let M ⊂ X be a compact set.
If the mapping f : M → Y is continuous on M then the range f (M ) is a compact subset in
the space (Y, dY ).
97
7.7 Continuity of the mappings in metric spaces
98
Corollary 7.7.2. Let (X, d) be a metric space and let the function f : (M ⊂ X) → R be
continuous on M . If M is a compact set then f is bounded on M and f attains its maximum
and its minimum value, i.e. there exist points c1 , c2 in M such that
f (c1 ) = min f (x),
f (c2 ) = max f (x).
x∈M
x∈M
THEOREM 7.7.3. Let (X, dX ), (Y, dY ) be metric spaces and let M ⊂ X be a compact set.
If the mapping f : M → Y is continuous on M then it is uniformly continuous on M .
Problems
1. Let (X, dX ), (Y, dY ) be metric spaces and M ⊂ X .
a) Show that if a ∈ M is an isolated point then any mapping f : M → Y is
continuous at a.
b) Let a ∈ M be a limit point of the set M . Prove that a mapping f : M → Y is
continuous at a if and only if
lim f (x) = f (a).
x→a
2. Let (B(M ), dB ) be the metric space dened in example 5.2.1.
a) Let a function F : B(M ) → R be dened as:
F (f ) =
1
sup f (x),
2 x∈M
f ∈ B(M ).
Show that the function F is continuous on B(M ) (see example 6.8.1).
b) Let M = [0, 1], A(M ) = {g ∈ C(M ; R); |g| ≤ 1} and let
F0 : B(M ) → R
be such a function that
8
<0
F0 (f ) = :
1
for f ∈ A(M ),
for f ∈ B(M ) \ A(M ) .
Show that the function F0 is not continuous on B(M ).
c) Show that the mapping F : B([0, 1]) → B([0, 1]) given by
F(f ) = f 2 ,
is continuous on B([0, 1]).
98
f ∈ B([0, 1])
7.7 Continuity of the mappings in metric spaces
99
3. Let (X, d) be a metric space and let b ∈ X . Prove that the function
x 7→ d(x, b),
x∈X
is uniformly continuous on X . (Hint: for x1 , x2 ∈ X : |d(x1 , b) − d(x2 , b)| ≤ d(x1 , x2 )
holds.)
4. Let (X, d) be a metric space and let the functions f : (M ⊂ X) → R, g : (M ⊂ X) →
R be uniformly continuous on M . Show that then also the following functions: f + g
and λg (λ ∈ R) are uniformly continuous on M .
5. Let F be the mapping dened in problem 2c) and let
B1 ([0, 1]) = {g ∈ B([0, 1]); |g| ≤ 1}.
Show that the mapping (restriction of the mapping F to the set B1 ([0, 1])): F1 =
F|B1 ([0,1]) is uniformly continuous on B1 ([0, 1]).
6. Let (X, dX ), (Y, dY ) be metric spaces and let a mapping f : X → Y be continuous on
X . Show that if a set E ⊂ Y is open (closed) in (Y, dY ) then the set
f −1 (E) = {x ∈ X; f (x) ∈ E}
is open (closed) in (X, dX ).
99