ADDENDUM B: CONSTRUCTION OF R AND THE COMPLETION
OF A METRIC SPACE
ANDREAS LEOPOLD KNUTSEN
Abstract. These notes are written as supplementary notes for the course MAT211Real Analysis, taught at the University of Bergen, Fall 2015.
They are meant to
supplement [Ru, 2.1-2.3 and 3.8-3.12].
1. Relations and equivalence relations
We start with a couple of denitions (almost) left out in [Ru].
following terminology not used in [Ru]: We denote a function
f
We also recall the
from a set
A
to a set
B
f : A → B and say that
• f is injective or an injection if it is one-to-one;
• f is surjective or a surjection if it is onto;
• f is bijective or a bijection if it is both one-to-one and onto.
We also recall that if f is bijective, there is a well-dened inverse mapping
by
g : B → A dened
Applying
f
rst and then
g
as
g(y) = the
unique
simply brings
x
x∈A
back to
such that
x.
f (x) = y.
The inverse map
g
is often
−1 , which should not be confused with the notation of inverse images as in
denoted by f
[Ru, Def. 2.2].
Denition 1.1.
Let X be a set. A relation R on X is a subset of the Cartesian product
R ⊂ X × X.
One often denotes xRy , x ∼ y , or x ≡ y to mean (x, y) ∈ R. We will in these notes
stick to the notation x ∼ y .
X × X,
that is,
Denition 1.2.
equivalence relation is a relation satisfying the following properties:
An
x ∼ x for all x ∈ X .
x ∼ y , then y ∼ x, for all x, y ∈ X .
Transitivity: if x ∼ y and y ∼ z , then x ∼ z , for all x, y, z ∈ X .
(i) Reexivity:
(ii) Symmetry: if
(iii)
Example 1.3.
The
order relation
x ∼ y ⇐⇒ x < y
on
R
dened in [Ru, Def. 1.5] is not an equivalence relation, as (i) fails (x
hold), and also (ii) fails (if
Example 1.4.
x<y
it is not true that also
y < x).
<x
fails to
However, (iii) holds.
The relation on the set of all sets in [Ru, Def. 2.3] dened by
A ∼ B ⇐⇒ ∃ bijection f : A → B
(meaning in the language of [Ru, Def. 2.3] that the sets
to-one correspondence) is an equivalence relation:
A
and
B can be put in one-
Reexivity holds because the identity
id : A → A mapping x to x is a bijection. Symmetry holds because a bijection
f : A → B has an inverse map f −1 : B → A that is still a bijection. Transitivity holds
because if f : A → B and g : B → C are bijections, then the composed map
map
g◦f :A→C
dened by
1
(g ◦ f )(x) = g(f (x))
2
ANDREAS LEOPOLD KNUTSEN
is a bijection.
Example 1.5.
Fix any
m ∈ N.
Let
X = Z.
The relation on
Z
dened by
x ∼ y ⇐⇒ m divides x − y
is an equivalence relation.
Example 1.6.
The relation on
C
given by
z1 ∼ z2 ⇐⇒ |z1 | = |z2 |
is an equivalence relation.
Exercise 1.7.
Prove that the relations in the last two examples are indeed equivalence
relations.
Denition 1.8.
Given an equivalence relation
equivalence class [x] of x to be the set:
∼
on a set
X
and
x ∈ X,
we dene the
[x] := {y ∈ X | y ∼ x}.
representative of the equivalence class.
set of equivalence classes by X/ ∼.
Any element in the equivalence class is called a
We often denote the
x∈X
x ∼ x, so that x ∈ [x],
whence x lies in at least one equivalence class. If x ∈ [y1 ] and x ∈ [y2 ], then x ∼ y1 and
x ∼ y2 . By symmetry, we have y1 ∼ x. But this, together with x ∼ y2 and transitivity,
yields that y1 ∼ y2 , that is, [y1 ] = [y2 ]. Hence, X/ ∼ is a collection of disjoint subsets of
X whose union is the whole of X . (Such a collection is called a partition.)
Using the properties of equivalence relations, it is not dicult to prove that any
lies in precisely one equivalence class.
Example 1.9.
Indeed, by reexivity,
Consider Example 1.5 and let
n ∈ Z.
Then the equivalence class of
n
is
[n] = {k ∈ Z m divides n − k} = {n, n ± m, n ± 2m, n ± 3m, . . .}.
Clearly
[n]
contains a unique number
of equivalence classes
Z/ ∼,
n0
satisfying
which is often denoted
0 ≤ n0 ≤ m − 1.
by Z/mZ or Zm , is
Thus the set
in one-to-one
correspondence with the set
{0, 1, 2, . . . , m − 1}.
Example 1.10.
Consider Example 1.6 and let
z ∈ C.
Then the equivalence class of
z
is
[z] = {c ∈ C | |c| = |z|},
which consists of all points lying on the circle of radius
|z| about the origin in the complex
plane. Thus the set of equivalence classes is in one-to-one correspondence with the set
of circles about the origin in the complex plane (including the circle with radius zero,
that is, the origin itself ), which is again in one-to-one correspondence with the set of
nonnegative real numbers.
2. Cantor's construction of
R
from
Q
by means of Cauchy sequences
In this section, we will go through Cantor's construction of the real numbers by using
Cauchy sequences, and this idea will be generalized when we construct completions of
arbitrary metric spaces in the next section. The aim is to give a dierent, independent
proof of [Ru, Thm. 1.19]:
Theorem 2.1. There exists an ordered eld R with the least-upper-bound property and
containg Q as an ordered subeld.
CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE
3
Before starting the proof, we remark that we do not need to know beforehand what
is to talk about distances in
Q
and Cauchy sequences: Indeed, the number
R
|p − q| ∈ Q
for any p, q ∈ Q and satises the denition of a metric. Moreover, we say that a sequence
{pn } in Q is a Cauchy sequence if for any rational > 0, there is an integer N such that
|pm − pn | < whenever m, n ≥ N and that it converges to x ∈ Q if for any rational
> 0, there is an integer N such that |pn − x| < whenever n ≥ N .
We recall the following:
Proposition 2.2. There are Cauchy sequences in Q that do not converge to any element
in Q.
Proof.
This can be proved with the same technique used in [Ru, Ex.
to prove that
Q
1.1 and 1.9(a)]
neither satises the least-upper-bound nor the greatest-lower bound-
property. The association in [Ru, (3)] of a rational number
denes a sequence
{pn }
q
to any rational number
p
recursively by
p2n − 2
2
=2−
pn + 2
pn + 2
pn+1 = pn −
(1)
p21 < 2 and decreasing if p21 > 2.
The sequence is a Cauchy sequence. If the limit exists, say L = lim pn , then taking limits
2
in (1) shows that L = 2, whence the limit cannot exist in Q, by [Ru, Ex. 1.1].
which is easily checked (by induction) to be increasing if
Exercise 2.3.
Prove that
{pn }
in the proof of Proposition 2.2 is a Cauchy sequence
(without using the knowledge that it converges in
Remark 2.4.
L,
R).
There are other recursive relations yielding Cauchy sequences whose limits
if they exist, must satisfy
L2 = 2,
for instance
xn+1 =
xn
2
+
1
xn .
The moral of Cantor's construction is the following: since
Q lacks limits of Cauchy
Q, we add those limits representing them simply by the Cauchy sequences
themselves. The elements of Q can be viewed as constant Cauchy sequences in Q. In
√
2 would be represented by the Cauchy sequence {pn }
this language, the real number
sequences in
in the proof of Proposition 2.2.
However, we have to take into account that dierent
Cauchy sequences may yield the same limits, for instance dierent starting points
above yield dierent Cauchy sequences
for the sequences
xn
{pn }
√
still representing
2,
p1
and the same applies
in Remark 2.4. It is to avoid this ambiguity that equivalences enter
the picture:
Denition 2.5.
Two Cauchy sequences {pn }
{|pn − qn |} in Q converges to 0.
{pn } ∼ {qn } if {pn } and {qn } are
and
{qn }
in
Q
are said to be
equivalent if
the sequence
We write
Exercise 2.6.
in
Show that
equivalent.
∼ is an equivalence relation on the set of all Cauchy sequences
Q.
We denote by
[pn ]
the equivalence class of
equivalence classes of Cauchy sequences on Q,
{pn }
and dene
R
to be the
set of all
that is, with the notation of Denition
1.8, we have
R := {Cauchy
Q}/ ∼ .
We note that Q can be identied with a subset of R by identifying q ∈ Q with the
constant Cauchy sequence Cq := {q, q, q, q, . . .}. It is clear that Cp ∼ Cq if and only if
p = q , so Q is in a natural way a subset of the newly dened set R.
We dene two operations + and · on R that are compatible with the ones on the
subset Q:
[pn ] + [qn ] := [pn + qn ] and [pn ] · [qn ] := [pn · qn ].
sequences in
4
ANDREAS LEOPOLD KNUTSEN
We have to prove that these operations are well-dened, that is, that they are independent of the choice of representative for the equivalence classes of Cauchy sequences. This
is taken care of by the following:
Lemma 2.7. If
{p0n
·
qn0 }
Proof.
{p0n } ∼ {pn } and {qn0 } ∼ {qn }, then {p0n + qn0 } ∼ {pn + qn } and
∼ {pn · qn }.
{p0n } ∼ {pn }
By denition,
lim |qn0
− qn | = 0.
and
{qn0 } ∼ {qn }
means that
lim |p0n − pn | = 0
and
By the triangle inequality,
|(p0n + qn0 ) − (pn + qn )| ≤ |p0n − pn | + |qn0 − qn |
whence
lim |(p0n + qn0 ) − (pn + qn )| = 0.
It follows that
{p0n + qn0 } ∼ {pn + qn },
as desired.
To prove the remaining part, we write
|p0n qn0 − pn qn | = |qn0 (p0n − pn ) + pn (qn0 − qn )| ≤ |qn0 ||p0n − pn | + |pn ||qn0 − qn |,
and us the fact that any Cauchy sequence is bounded (see Exercise 2.8 below) to conclude
again that
lim |p0n qn0 − pn qn | = 0.
Exercise 2.8.
Prove that a Cauchy sequence (in any metric space) is bounded.
One can verify that the set
element for addition
0R
R
with the operations
1R
and identity element (for multiplication)
Cauchy sequence
C1 .
+
and
·
is a eld, with neutral
being the equivalence class of the constant Cauchy sequence
C0
being the equivalence class of the constant
Similarly, the additive inverse element of
[pn ]
is
[−pn ]
and the
[pn ] 6= C0 is [1/pn ]. (Here we need to be a bit careful due to
pn = 0 for some n. On the other hand, as [pn ] 6= C0 , we have that
for any > 0, the inequality |pn | ≥ holds for large enough n. Thus we may pick a
representative qn of the equivalence class such that qn 6= 0 for all n.) The operations are
compatible with the ones on Q, so R contains Q as a subeld.
multiplicative inverse of
the possibility that
Exercise 2.9.
Work out the details above to prove that
R
with the operations
+
and
·
is indeed a eld.
We now want to dene an order on
dening, for any two Cauchy sequences
{pn } < {qn } ⇐⇒ there
R compatible with the
{pn } and {qn } in Q:
N >0
is a integer
one on
Q.
We start by
r>0
n ≥ N.
and a rational number
such that
pn + r < q n
for all
Lemma 2.10. Assume that {pn } < {qn }.
If {p0n } ∼ {pn } and {qn0 } ∼ {qn }, then {p0n } < {qn0 }.
Proof.
{pn } < {qn }.
r > 0 such that
Assume that
rational number
qn > pn + r
(2)
0
Since {pn }
Then by denition, there is an integer
∼ {pn }
for all
N1 > 0
and a
n ≥ N1 .
0
and {qn }
∼ {qn } there are integers N2 and N3 such that
r
r
for all n ≥ N2
(3)
− ≤ pn − p0n ≤
3
3
r
r
(4)
− ≤ qn − qn0 ≤
for all n ≥ N3 .
3
3
Hence, for n ≥ N := max{N1 , N2 , N3 }, we have by (2)-(4) that
r
r
2r
r
2r
r
qn0 ≥ qn − > (pn + r) − = pn +
≥ (p0n − ) +
= p0n + ,
3
3
3
3
3
3
0
0
whence {qn } > {pn } by denition.
CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE
By the lemma, we may therefore dene, for any
[pn ] < [qn ] ⇐⇒ {pn } < {qn }
5
[pn ], [qn ] ∈ R,
for any representatives
Lemma 2.11. The relation < is an order relation [Ru, Def.
{pn }
1.5]
{qn }.
and
on R satisfying
• [pn ] < [qn ] ⇒ [pn ] + [rn ] < [qn ] + [rn ],
• [pn ], [qn ] > 0R ⇒ [pn ] · [qn ] > 0R ,
for all [pn ], [qn ], [rn ] ∈ R.
Exercise 2.12.
Prove Lemma 2.11.
Thus, we have made
R into an ordered eld [Ru, Def.
1.17] containing
Q as an ordered
subeld.
We nally would like to prove that
R
satises the least upper bound property [Ru,
Def. 1.10]. This would conclude the proof of Theorem 2.1.
Assume therefore that
Let
[un ] ∈ R
E
is a nonempty subset of
R
that is bounded above.
be an upper bound. Since the representative
there is an integer
N
such that, for all
{un }
is a Cauchy sequence,
n ≥ N:
un ≤ |un | ≤ |un − uN | + |uN | < 1 + |uN | ∈ Q.
Hence, setting
U := |un | + 2 ∈ Q,
we have
un + 1 < U ∈ Q,
so that
CU ≥ [un ] in R and CU
is an upper bound of
E
as well. (The point is to substitute
the original upper bound with an upper bound represented by a
E is nonempty, there is at least one
n ≥ N for some integer N , whence
Since
for all
element
[pn ] ∈ E .
constant sequence.)
As above,
|pn | < 1 + |pN |
pn − 1 > −2 − |pN | =: L ∈ Q,
so that
CL < [pn ]
CL
and
is
not an upper bound of E .
{xn } and {yn } in Q recursively as follows.
x0 = U and y0 = L. Then, having dened x0 , . . . , xn and y0 , . . . yn , we proceed
xn +yn
and yn+1 := yn .
follows: If C xn +yn is an upper bound of E , let xn+1 :=
2
We now dene two sequences
We set
as
2
xn +yn
and xn+1 := xn .
2
Hence, at each step, one of xn and yn remains the same, whereas the other one takes
1
the value of the intermediate point between the two. Therefore |xn+1 −yn+1 | = |xn −yn |,
2
Otherwise, let
yn+1 :=
so that
|xn − yn | =
(5)
|U − L|
2n
It is also easy to see that
yn ≤ yn+k ≤ xn+k ≤ xk
(6)
Moreover, both
{xn }
and
{yn }
for all
n, k ≥ 0.
are Cauchy, since (using (6)) one has, for
m ≥ n:
|xm −xn | ≤ |xm −yn |+|yn −xn | = (xm −yn )+(xn −yn ) ≤ (xn −yn )+(xn −yn ) = 2(xn −yn ),
n→∞
{yn }.
which tends to
0
{xn }
are equivalent, thus dene the same element
and
{yn }
as
We will prove that
[xn ]
by (5). Similarly for
is the least upper bound of
By (5) again, and Denition 2.5,
[xn ] = [yn ] ∈ R.
E.
To do so, we will need:
Lemma 2.13. For any k ≥ 0, one has that Cxk is an upper bound for E and Cyk is not.
6
ANDREAS LEOPOLD KNUTSEN
Proof.
We prove this by induction on
Cx0 = CU is
that Cxk is an
k.
E.
E , for k ≥ 0.
< xk , then, by the
By construction,
an upper bound of
Assume now
upper bound of
Cxk+1 = Cxk
is an upper bound of
E.
If
xk+1
If
xk+1 = xk ,
then
recursive denition of
the sequence, we must have
xk+1 =
so that
Cxk+1 = C xk +yk
2
The proof concerning
xk + yk
2
and
C xk +yk
is an upper bound of
C yk
We will now prove that
is an upper bound of
E,
2
E.
is similar.
[xn ]
is an upper bound of
[qn ] ∈ E
E.
[qn ] > [xn ]. This means
qn > xn + r for all
n ≥ N1 . Since {xn } is Cauchy and decreasing by (6), there is an integer N2 so that
0 ≤ xm − xn < 12 r for all n ≥ m ≥ N2 . Hence, for n ≥ m ≥ N := max{N1 , N2 }, we have
If it were not, there would exist some
that there is an integer
[qn ] > CxN ,
whence
N1
and a rational number
such that
r>0
1
qn > xn + r > xm + r,
2
contradicting the fact that CxN is an
such that
upper bound of
E
by Lemma
2.13.
Finally, we will prove that
[xn ]
E.
[qn ] of E
is a least upper bound of
If it were not, there would exist some upper bound
such that [qn ] < [xn ] =
[yn ]. This means that there is an integer N1 and a rational number r > 0 such that yn >
qn + r for all n ≥ N1 . Since {yn } is Cauchy and increasing by (6), there is an integer N2
1
so that 0 ≤ yn − ym < r for all n ≥ m ≥ N2 . Hence, for n ≥ m ≥ N := max{N1 , N2 },
2
we have
whence
1
qn < yn − r < ym − r,
2
contradicting the fact that CyN is
[qn ] < CyN ,
not an upper bound of
E
by
Lemma 2.13.
We have therefore concluded the proof that
that
R
[xn ]
is a least upper bound of
E,
whence
has the least-upper-bound property, thus nishing the proof of Theorem 2.1.
3. Completion of a metric space
The procedure of the preceding section of constructing
R from Q by adding equivalence
classes of Cauchy sequences can be generalized to arbitrary metric spaces. We will follow
closely the treatment in [Sh] and [Li], adjusted to t to the notation and contents of [Ru].
We start with the following denition, which is very important in its own right
Denition 3.1.
An
Let
isometry from
X and Y be metric spaces, with metrics dX
X to Y is a bijection f : X → Y such that
and
dY ,
respectively.
dY (f (x), f (y)) = dX (x, y).
−1
(The inverse map f
We say that
X
and
: Y → X is automatically also an isometry.)
Y are isometric if there exists an isometry between
them.
The point is that an isometry, besides inducing a one-to-one-correspondence between
X
and
Y
as sets, also
preserves all distances.
Thus, we may regard
X
and
Y
to be the
same space for all practical purposes.
Recall the denition of a
subset from [Ru, Def.
complete metric space
2.18(j)].
from [Ru, Def. 3.12] and of a
dense
CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE
Denition 3.2.
If
metric space
with metric
(i)
X
X ∗,
X
is a metric space, with metric
is a subspace of
d∗ ,
X ∗;
d,
a
completion
of
is a complete
such that
X ⊂ X ∗ and the metric on X
d(x, y) = d∗ (x, y) for all x, y ∈ X );
that is,
∗
one on X (meaning that
∗
(ii) X is dense in X .
X
7
is induced by the
We will prove that any metric space admits a completion and that all such completions
are isometric:
Theorem 3.3. Any metric space admits a completion that is unique up to isometry (that
is, any two completions of the same metric space are isometric).
By the uniqueness property we will talk about
By [Ru, Thm.
3.11(c)] we know that
R
the completion of a metric space.
(constructed by Dedekind cuts as in [Ru,
App. to Chp. 1] or by Cauchy sequences as in the previous section) is complete, and
by [Ru, Thm. 1.20(b)] it contains
completion of
Q
Q
as a dense subset. Hence, by uniqueness,
R
is the
(in its usual metric).
The proof of Theorem 3.3 is interesting in itself because the construction behind is
typical in mathematics.
We start by proving the uniqueness statement of Theorem 3.3.
Proposition 3.4. If Y and Z are completions of a metric space X , then Y and Z are
isometric.
Proof.
and
We have both
dZ ,
X⊂Y
and
X ⊂ Z.
Denote the metrics in
X, Y
and
Z
by
dX , dY
respectively.
y ∈ Y is a limit point of X or a point of X . By [Ru,
Thm. 3.2(d)], there is a sequence {xn } in X converging to y . This is a Cauchy sequence
in X (by [Ru, Thm. 3.11]), whence also in Z ⊃ X . As Z is complete, the sequence {xn }
converges to some z ∈ Z .
Since
X
Y,
is dense in
any point
We dene a mapping
ι:Y →Z
by
ι(y) = z.
{x0n }
This is well-dened: if we choose another sequence
in
dZ (xn , x0n ) = dX (xn , x0n ) = dY (xn , x0n ) → 0
0
and xn
xn → y
xn → z in Z , we
since
→ y
in
have, for any
Y . Hence,
> 0, that
X
as
converging to
y,
then
n → ∞,
by the triangle inequality and the fact that
dZ (x0n , z) ≤ dZ (x0n , xn ) + dZ (xn , z) → 0,
x0n → z
{xn } in X .
so that
in
Z.
Thus, our mapping does not depend on the choice of the sequence
ι preserves distances. Let y, y 0 ∈ Y with {xn } and {x0n } sequences
0
0
0
0
in X such that xn → y and xn → y . By denition of ι, we have xn → ι(y) and xn → ι(y )
in Z . Hence
We next prove that
dZ (ι(y), ι(y 0 ) = lim dZ (xn , x0n ) = lim dX (xn , x0n ) = lim dY (xn , x0n ) = dY (y, y 0 ),
so that
ι
preserves distances.
Finally, we prove that
ι
is a bijection.
Since distance is preserved, it is immediate that
To prove that
ι is
=
dY (y, y 0 ), whence
ι
is injective: indeed, if
ι(y) = ι(y 0 ),
y 0 by the property of metrics.
y=
z ∈ Z . As X is dense in Z , there exists by [Ru, Thm.
3.2(d)] again a sequence {xn } in X converging to z . Since {xn } is a Cauchy sequence in
X by [Ru, Thm. 3.11], it is a Cauchy sequence also in Y ⊃ X . As Y is complete, the
sequence {xn } converges to some y ∈ Y . By construction of ι, we have that ι(y) = z . then
0 = dZ
(ι(y), ι(y 0 ))
surjective, let
8
ANDREAS LEOPOLD KNUTSEN
The following generalizes Denition 2.5
Denition 3.5.
be
equivalent
{pn }
and
{qn }
Exercise 3.6.
in
Two Cauchy sequences
if the sequence
{d(pn , qn )}
{pn } and {qn } in a metric space X are said to
in R converges to 0. We write {pn } ∼ {qn } if
are equivalent.
Show that
∼ is an equivalence relation on the set of all Cauchy sequences
X.
As in the previous section, we denote the equivalence class of
X ∗ := {Cauchy
sequences in
X}/ ∼,
{pn }
by
[pn ].
We dene
the set of all equivalence classes. This set will turn
out to be the one fullling the conditions in Denition 3.2, once we dene a metric on
it. To do so, we need:
Lemma 3.7. If {pn } and {qn } are Cauchy sequences in X , then the sequence {d(pn , qn )}
converges in R.
Moreover, if {p0n } ∼ {pn } and {qn0 } ∼ {qn }, then lim d(pn , qn ) = lim d(p0n , qn0 ).
Proof.
m, n:
We have, by the triangle inequality, for any
d(pn , qn ) ≤ d(pn , pm ) + d(pm , qm ) + d(qm , qn ),
whence
|d(pn , qn ) − d(pm , qm )| ≤ d(pn , pm ) + d(qn , qm ).
Since
{pn }
and
{qn }
are Cauchy, the sum on the right is arbitrarily small for
enough, whence also the sequence
{d(pn , qn )}
in
is Cauchy. As
R
R
m, n
large
is complete by [Ru,
Thm. 3.11(c)], it converges.
Assume now that
{p0n } ∼ {pn }
and
{qn0 } ∼ {qn }.
Then
0 ≤ d(pn , qn ) ≤ d(pn , p0n ) + d(p0n , qn0 ) + d(qn0 , qn ),
and the rst and third term on the right tend to
sequences. Hence
lim d(pn , qn ) ≤ lim d(p0n , qn0 )
0
by denition of equivalent Cauchy
and the reverse inequality is obtained by
symmetry.
By this lemma, the function
∆([pn ], [qn ]) = lim d(pn , qn )
is well-dened on
X∗ × X∗
(that is, independent of the choice of representative in the
equivalence class) and takes values in
R.
Lemma 3.8. ∆ is a metric on X ∗ .
Exercise 3.9. Prove Lemma 3.8.
X can be identied with a subspace of X ∗ . To do so, we
mimic the way we identied Q with a subset of R in the previous section, and dene,
for each point p ∈ X , the constant Cauchy sequence {p, p, p, p, . . . , } and denote its
∗
equivalence class in X by Cp .
Next we want to prove that
Lemma 3.10. The map p 7→ Cp from X to X ∗ is injective and distance-preserving.
Proof.
The map is injective, since if
sequences,
lim(p − q) = 0,
Cp = Cq , then by denition of equivalence of Cauchy
p = q (as the sequences are constant).
which happens only if
The map preserves distances because
∆(Cp , Cq ) = lim d(p, q) = d(p, q),
again because the sequences are constant.
CONSTRUCTION OF R AND THE COMPLETION OF A METRIC SPACE
Hence we can identify
X
with a subspace of
(Formally speaking, we should say that
Lemma 3.11.
Proof.
X
is
X∗
and treat
as a subspace.
isometric to a subspace of X ∗ .)
X is dense in X ∗ .
We must show that any
[pn ] ∈ X ∗
if
n, m ≥ N .
X or a point of X , cf.
> 0, there is an N such
is a limit point of
Def. 2.18(j)]. By denition of Cauchy sequence, for any
d(pm , pn ) < X ⊂ X∗
9
[Ru,
that
Then,
∆([pn ], CpN ) = lim d(pn , pN ) ≤ .
n→∞
This means that any
Cp N .
If
CpN = [pn ],
denition.
-neighborhood of [pn ] intersects X in at least one point, namely
[pn ] ∈ X and if CpN 6= [pn ], then [pn ] is a limit point of E , by
then
To nish the proof of Theorem 3.3 we only have left to prove that
do so, we exploit the density of
X
X∗
is complete. To
∗
in X and need the following lemma.
Lemma 3.12. Any Cauchy sequence in X ⊂ X ∗ converges to an element in X ∗ .
Proof.
{Ck } be a Cauchy sequence in X (k = 1, 2, 3, . . .), that is, we have, Ck = Cpk ,
pk ∈ X for any k , recalling that Cpk is the equivalence class of the constant
sequence with terms pk . Since, for any m, n:
Let
for a point
Cauchy
d(pm , pn ) = ∆(Cpm , Cpn ) = ∆(Cm , Cn ),
{pn } is a Cauchy sequence in X , thus dening
{Ck } converges to [pn ], as k → ∞. Indeed,
we have that
claim that
an element
[pn ] ∈ X ∗ .
We
∆(Ck , [pn ]) = ∆(Cpk , [pn ]) = lim d(pk , pn ),
n→∞
k → ∞,
which tends to
0
Lemma 3.13.
X ∗ is complete.
Proof.
as
since
{pk }
is Cauchy.
{xn } be a Cauchy sequence in X ∗ . Since X is dense in X ∗ (through the
identication p 7→ Cp , the equivalence class of the constant Cauchy sequence at p), we
∗
have that any neighborhood of any xn ∈ X contains a point of X . Hence, for any n,
1
∗
there is a point yn ∈ X such that ∆(Cyn , xn ) <
n . The sequence {Cyn } in X is a
Cauchy sequence, since for any m, n:
Let
∆(Cyn , Cym ) ≤ ∆(Cyn , xn ) + ∆(xn , xm ) + ∆(xm , Cym )
1
1
<
+ ∆(xn , xm ) + ,
n
m
and we can make this sum arbitrarily small by choosing
m
and
n
large enough, as
{xn }
is Cauchy.
By Lemma 3.13, the sequence
{Cyn }
converges to an element in
X ∗,
say
P ∈ X ∗.
Then
∆(xn , P ) ≤ ∆(xn , Cyn ) + ∆(Cyn , P ) <
and this can be made arbitrarily small for large
converges to
P ∈
X ∗ , as desired.
This concludes the proof of Theorem 3.3.
n
as
1
+ ∆(Cyn , P ),
n
{Cyn } tends to P .
Thus,
{xn }
10
ANDREAS LEOPOLD KNUTSEN
References
[Li]
T. Lindstrøm,
Mathematical Analysis,
notes used in the course MAT2400 at the University of
Oslo.
[Ru]
W. Rudin,
Principles of mathematical analysis,
International Series in Pure and Applied Mathe-
matics, Third Edition, McGraw-Hill, 1976.
[Sh]
G. E.Shilov,
Elementary real and complex analysis.
Revised English edition translated from the
Russian and edited by Richard A. Silverman. Corrected reprint of the 1973 English edition. Dover
Publications, Inc., Mineola, NY, 1996.
Andreas Leopold Knutsen, Department of Mathematics, University of Bergen, Postboks 7800, 5020 BERGEN, Norway
E-mail address :
[email protected]