Lecture 4
Econ 2001
2015 August 13
Lecture 4 Outline
1
Open and Closed Set
2
Continuity
Announcements:
- Tomorrow: …rst test at 3pm, in WWPH 4716. The exam will last an hour.
- Tomorrow: recitation at 1pm, in WWPH 4716.
Open and Closed Sets
A set is open if at any point we can …nd a neighborhood of that point
contained in the set.
De…nition
Let (X ; d) be a metric space. A set A
8x 2 A 9" > 0
X is open if
such that
B" (x)
A
Remember that B" (x) = fy 2 X : d(y ; x) < "g... so openness depends on X .
De…nition
A set C
X is closed if X n C is open.
Draw Pictures
Open and Closed Sets: Examples
Open Interval in R
(a; b) is open in R (with the usual Euclidean metric).
Given x 2 (a; b), a < x < b. Let
" = minfx
a; b
xg > 0
Then
y 2 B" (x) ) y 2 (x
Hence B" (x)
"; x + ")
(x
(x
a); x + (b
x)) = (a; b)
(a; b), so (a; b) is open.
" must depend on x; in particular, " gets smaller as x nears the boundaries of
the interval.
Closed Interval in R
[a; b] is closed in R (with the usual Euclidean metric).
R n [a; b] = ( 1; a) [ (b; 1) is the union of two open sets, which must be
open (prove this later).
Open and Closed Sets: Examples
An open ball is always an open set
Suppose y 2 B" (x). Then d(x; y ) < ".
Let
="
d(x; y ) > 0
If d(z; y ) < , then
d(z; x)
d(z; y ) + d(y ; x)
<
= "
+ d(x; y )
d(x; y ) + d(x; y )
= "
Hence B (y )
B (x), so B" (x) is open.
This is very usefuly since it holds for any X ; d.
Open and Closed Sets: Examples
Openness and closedness depend on the underlying metric space
In the metric space X = [0; 1] (with standard metric), [0; 1] is open.
Since [0; 1] is the underlying metric space,
B" (0) = fx 2 X : d(x; 0) < "g = fx 2 [0; 1] : jx
0j < "g = [0; ")
Most sets are neither open nor closed
[0; 1] [ (2; 3) is neither open nor closed.
An open set may consist of a single point
If X = N and d(m; n) = jm
nj, then
B1=2 (1) = fm 2 N : jm
1j < 1=2g = f1g
Since 1 is the only element of the set f1g and B1=2 (1) = f1g
f1g is open.
f1g, the set
Open and Closed Sets: Examples
In any metric space (X ; d), ; and X are both open... and closed
To see that ; is open, note that the statement
8x 2 ; 9" > 0
such that
B" (x)
;
is vacuously true since there aren’t any x 2 ; (any statement about points in
the empty set is true).
To see that X is open, note that B" (x) = fz 2 X : d(z; x) < "g is trivially
contained in X .
Since ; is open, X is closed; since X is open, ; is closed.
Open and Closed Sets: Results
Theorem
Let (X ; d) be a metric space. Then
1
2
3
; and X are both open and closed.
The union of an arbitrary (…nite, countable, or uncountable) collection of
open sets is open.
The intersection of a …nite collection of open sets is open.
Proof.
1
Already done.
2
Suppose fA g
S
x2
so [
2
2
2
is a collection of open sets.
A
) 9 0 2 such that x 2 A
A is open.
)
9" > 0 such that B" (x)
0
A
0
S
2
A
Prove that the intersection of a …nite collection of open sets is open.
Proof.
Suppose A1 ; : : : ; An
If x 2
\ni=1 Ai ,
X are open sets.
then
so
x 2 A1 ; x 2 A2 ; : : : ; x 2 An
9"1 > 0; : : : ; "n > 0 such that B"1 (x)
Let
An
" = minf"1 ; : : : ; "n g > 0
Then
B" (x)
A1 ; : : : ; B"n (x)
B"1 (x)
A1 ; : : : ; B" (x)
B"n (x)
An hence B" (x)
n
\
Ai
i =1
which proves that \ni=1 Ai is open.
This needs …nite intersection as the in…mum of an in…nite set of positive
numbers could be zero and the proof would fail.
The intersection of an in…nite collection of open sets need not be open.
Interior, Closure, Exterior and Boundary
Let (X ; d) be a metric space and A
X.
De…nition
The interior of A, denoted intA, is the largest open set contained in A
(alternatively, the union of all open sets contained in A).
De…nition
The closure of A, denoted A, is the smallest closed set containing A
(alternatively, the intersection of all closed sets containing A).
De…nition
The exterior of A, denoted ext A, is the largest open set contained in X n A.
Note that ext A = intX n A.
De…nition
The boundary of A, denoted @A is equal to (X n A) \ A.
Interior, Closure, Exterior and Boundary
Let (X ; d) be a metric space and A
X.
FACTS
A point is interior if and only if it has an open ball that is a subset of the set
x 2 intA
,
9" > 0; B" (x)
A
A point is in the closure if and only if any open ball around it intersects the set
x 2A
,
8" > 0; B" (x) \ A 6= ?
A point is exterior if and only if an open ball around it is entirely outside the
set
x 2 extA
,
9" > 0; B" (x) X n A
A point is on the boundary if any open ball around it intersects the set and
intersects the outside of the set
B" (x) \ A 6= ?
and
x 2 @A , 8" > 0;
B" (x) \ X n A 6= ?
Interior, Closure, Exterior and Boundary
Interior, Closure, Exterior and Boundary Example
Let A = [0; 1] [ (2; 3). Then
int A = (0; 1) [ (2; 3)
A = [0; 1] [ [2; 3]
ext A
= int (X n A)
= int (( 1; 0) [ (1; 2] [ [3; +1))
=
@A =
( 1; 0) [ (1; 2) [ (3; +1)
(X n A) \ A
= (( 1; 0] [ [1; 2] [ [3; +1)) \ ([0; 1] [ [2; 3])
=
f0; 1; 2; 3g
Sequences and Closed Sets
We can characterize closedness also using sequences: a set is closed if it
contains the limit of any convergent sequence within it, and a set that
contains the limit of any sequence within it must be closed.
Theorem
A set A in a metric space (X ; d) is closed if and only if
fxn g A
and
)x 2A
xn ! x 2 X
We will prove the two directions in turn.
A set is closed if it contains the limit of any convergent sequence within it.
Proof.
Let A be closed. Then X n A is open.
Consider a convergent sequence xn ! x 2 X , with xn 2 A for all n.
We need to show that x 2 A. Suppose not.
If x 62 A, then x 2 X n A, so there is some " > 0 such that B" (x)
the de…nition of open set).
Since xn ! x, there exists N(") such that
n > N(") ) xn 2 B" (x)
) xn 2 X n A
This is a contradiction. Therefore,
as desired.
fxn g
) xn 62 A
A; xn ! x 2 X ) x 2 A
X n A (by
A set that contains the limit of any sequence within it must be closed
Proof.
fxn g A
and
) x 2 A.
xn ! x 2 X
We need to show that A is closed (or equivalently, that X n A is open).
Suppose
Suppose not: X n A is not open. Hence, there exists x 2 X n A such that for
every " > 0,
B" (x) 6 X n A
So there exists y 2 B" (x) such that y 2
6 X n A. Then y 2 A, hence
\
B" (x) A 6= ;
Construct a sequence fxn g as follows: for each n, choose
xn 2 B n1 (x) \ A
Given " > 0, by the Archimedean Property we can …nd N(") such that
1
N(") > "1 , so n > N(") ) n1 < N (")
< ", therefore xn ! x.
Then fxn g A, xn ! x, so x 2 A, a contradiction. Therefore, X n A is open,
and A is closed.
Familiar (maybe) Terminology
De…nitions
Let (X ; d) be a metric space and E
X.
A point x is a limit point of E if every B" (x) contains a point y 6= x such that
y 2 E.
If x 2 E and x is not a limit point of E , then x is called an isolated point of E .
E is dense in X if every point of X is a limit point of E , or a point of E (or
both).
Results
E is closed if every limit point of E is a point of E .
E is open if every point of E is an interior point of E .
Limits of Functions in Metric Spaces
Yesterday we de…ned the limit of a sequence, and now we extend those ideas
to functions from one metric space to another.
For functions from reals to reals: f : (c; d) ! R, y is the limit of f at x0 if
for each " > 0 there is a (") > 0 such that 0 < jx x0 j < (") ) jf (x) y j < "
We extend this to metric spaces by replacing each absolute value with a
metric.
De…nition
Let (X ; d) and (Y ; ) be metric spaces with A
say that f has limit y0 as x ! x0 if
X , f : A ! Y , and x0 2 A. We
8" > 0 9 (") > 0 such that 0 < d(x; x0 ) < (") ) (f (x); y0 ) < "
Notation
When f has limit y0 as x goes to x0 we write
f (x) ! y0 as x ! x0
Notice the di¤erent metrics.
or
lim f (x) = y0
x!x0
Limits of Functions and Sequences
Standard results like the uniqueness of limits theorem hold. Also, the results
we saw on sequences about sums, multiplication by a scalar, products, and
division extend to functions (when appropriate).
The limit of a function can also be characterized using sequences.
Theorem
Let (X ; d) and (Y ; ) be metric spaces, f : X ! Y , and x0 2 X . Then
lim f (x) = y0
x!x0
if and only if
for each fxn g ! x0 in (X ; d)
the sequence ff (xn )g
with xn 6= x0
converges to y0 in (Y ; )
For every sequence that converges in the domain, the corresponding sequence
given by the function converges to in the range.
Prove these results as part of Problem Set 4.
Continuity in Metric Spaces
For functions from reals to reals: f : (a; b) ! R is continuous at x0 means
for each " > 0 there is a (") > 0 such that jx x0 j < (") ) jf (x) f (x0 )j < "
which is easy to generalize.
De…nition
Let (X ; d) and (Y ; ) be metric spaces. A function f : X ! Y is continuous at a
point x0 2 X if
8" > 0 9 (x0 ; ") > 0 such that d(x; x0 ) < (x0 ; ") ) (f (x); f (x0 )) < "
Notice that di¤erent metrics are used when appropriate.
Continuity at x0 requires:
f (x0 ) is de…ned;
and either
x0 is an isolated point of X (i.e. 9" > 0 such that B " (x0 ) = fx0 g); or
lim f (x) exists and equals f (x0 )
x!x0
De…nition
f is continuous if it is continuous at every element of its domain.
(x0 ; ") means that
can depend on x0 . What if it does not? Later.
Continuity in Metric Spaces
Remember, given f : X ! Y and A Y the inverse image is a subset of X
de…ned as:
f 1 (A) = fx 2 X : f (x) 2 Ag X
The inverse image is used to provide a characterization of continuous
functions.
Theorem
Let (X ; d) and (Y ; ) be metric spaces, and f : X ! Y . Then
f is continuous
if and only if
f
1
(A) is open in X for every A
Y such that A is open in Y
Continuity is equivalent to the fact that the inverse image of every open set in
the range is an open set in the domain.
NOTE: an equivalent statement of the theorem
f is continuous if and only if f
1
(C ) is closed in X for every closed C
Y.
1
f is continuous ) f
(A) is open in X for every A
Y such that A is open in Y .
Proof.
Suppose f is continuous.
Given A Y , with A open, we must show that f
1
Suppose x0 2 f
1
(A) is open in X .
(A). Let y0 = f (x0 ) 2 A.
Since A is open, we can …nd " > 0 such that B" (y0 )
Since f is continuous, there exists
d(x; x0 ) <
> 0 such that
(f (x); f (x0 )) < "
)
) f (x) 2 B" (y0 )
) f (x) 2 A
So B (x0 )
f
1
A.
) x2f
(A), and therefore f
1
1
(A)
(A) is open.
The inverse image of any open set in the range is an open set in the domain ) f
is continuous.
Proof.
Suppose
f
1
(A) is open in X for each A
Y such that A is open in Y
We need to show that f is continuous.
Let x0 2 X , with " > 0, and let A = B" (f (x0 )).
A is an open ball, hence an open set, so f
x0 2 f
1
(A), so there exists
d(x; x0 ) <
)
)
)
)
1
(A) is open in X .
> 0 such that B (x0 )
f
1
(A).
x 2 B (x0 )
x2f
1
(A)
f (x) 2 A which is nothing but B" (f (x0 ))
(f (x); f (x0 )) < "
Thus, we have shown that f is continuous at x0 .
Since x0 is an arbitrary point in X , f is continuous.
Algebra of Continuity
The composition of continuous functions is continuous:
Theorem
Let (X ; dX ), (Y ; dY ) and (Z ; dZ ) be metric spaces. If f : X ! Y and g : Y ! Z
are continuous, then g f : X ! Z is continuous.
Proof.
Suppose A
Z is open.
g is continuous, thus g
is open in X .
1
I claim that
f
Observe
(A) is open in Y ; f is continuous, thus f
1
x2f
(g
1
(g
1
1
(A)) = (g
f)
1
f (x) 2 g
(A))
,
,
,
,
(A)
1
(A)
g (f (x)) 2 A
(g
f )(x) 2 A
x 2 (g
f)
1
(A)
which establishes the claim.
This shows that (g
f)
1
(A) is open in X , so g
f is continuous.
1
(g
1
(A))
Uniform Continuity
The de…nition of continuity at a point allows for
to depend on that point.
When it does not, the function is, for lack of a better expression, more
continuous.
De…nition (Uniform Continuity)
Let (X ; d) and (Y ; ) be metric spaces and f : X ! Y . We say f is uniformly
continuous if
8" > 0 9 (") > 0 such that 8x0 2 X ; d(x; x0 ) < (") ) (f (x); f (x0 )) < "
Contrast with the earlier de…nition of continuity at a point x0 :
8" > 0 9 (x0 ; ") > 0 such that d(x; x0 ) < (x0 ; ") ) (f (x); f (x0 )) < ". This
allows to depend on x0 and ".
Uniform continuity requires
continunity.
to depend only on ". It is more restrictive than
Uniform Continuity: Example
A continuous function that is not uniformly continuous
Consider f : (0; 1] ! R de…ned as f (x) = x1 .
f is continuous (check this) but not uniformly continuous.
Fix " > 0 and x0 2 (0; 1]. Let x =
1
x
1
x0
>
x0
1+"x0 .
Clearly, x < x0 . Then
0
1
1
1
1
1 + "x0
1
"x0
=
=
="
=
x
x0
x
x0
x0
x0
x0
Thus, in the de…nition of continuity (x0 ; ") must be chosen small enough so
that
"(x0 )2
x0
(x0 ; ") x0
=
< "(x0 )2
1 + "x0
1 + "x0
which converges to zero as x0 ! 0.
jf (x)
f (x0 )j =
Hence there is no (") > 0 that will work for all x0 2 (0; 1].
Uniform Continuity: Example
An f : [a; b] ! R that has a bounded derivative is uniformly continuous on [a; b].
However, a function with an unbounded derivative can also be uniformly
continuous.
A function with an unbounded derivative that is uniformly continuous
p
Consider f : [0; 1] ! R de…ned as f (x) =
x.
f is continuous (verify this) and also uniformly continuous.
Given " > 0, let
= "2 (notice this does not depend on x).
Then given any x0 2 [0; 1], jx
of Calculus)
jf (x)
x0 j <
f (x0 )j =
Z
Z
implies (by the Fundamental Theorem
x
x0
jx
1
p dt
2 t
x0 j
0
<
p
=
"
=
p
p
1
p dt = jx
2 t
x0 j
"2
Thus, f is uniformly continuous on [0; 1], even though f 0 (x) ! 1 as x ! 0.
Lipschitz Continuity
De…nition
Let X ; Y be normed vector spaces. A function f : X ! Y is Lipschitz on E
9K > 0 such that kf (x)
f is locally Lipschitz on E
f (z)jjY
X if
K kx
X if
zkX 8x; z 2 E
8x0 2 E 9" > 0 such that f is Lipschitz on B" (x0 ) \ E
The function is locally Lipschitz if every point in its domain has a
neighborhood on which the function is Lipschitz.
Remark
Lipschitz continuity is stronger than either continuity or uniform continuity:
locally Lipschitz
Lipschitz
)
)
continuous
uniformly continuous
Every C 1 function is locally Lipschitz (a function f : Rm ! Rn is said to be C 1 if
all its …rst partial derivatives exist and are continuous).
Homeomorphisms
De…nition
Let (X ; d) and (Y ; ) be metric spaces. A function f : X ! Y is called a
homeomorphism if it is one-to-one, onto, continuous, and its inverse function is
continuous.
Note
Suppose that f is a homeomorphism and U
X , and let g = f
1
: Y ! X.
Then
and
U open in X ) g
y2g
1
1
(U) , g (y) 2 U , y 2 f (U)
(U) is open in (f (X ); ) ) f (U) is open in (f (X ); )
Therefore (X ; d) and f (X ); jf (X ) are identical in terms of properties that
can be characterized solely in terms of open sets.
Such properties are called “topological properties.”
Remark
Topological properties are invariant under homeomorphisms.
Tomorrow
We study more properties of functions from R to R.
1
Boudedness and Extreme Value Theorem
2
Intermediate Value Theorem and Fixed Points
3
Monotonicity
4
Complete Spaces and Cauchy Sequences
5
Contraction Mappings