Lecture
§ 34. Complete metric spaces. Baire category theorem.
34.1. Definition. A sequence (xn) of points of a metric space (X, ρ) is called a Cauchy sequence or a fundamental sequence, if for
any > 0 there exists n0 ∈ N such that ρ(xn, xm) < for n, m > n0.
A metric space (X, ρ) is called complete, if every Cauchy sequence
converges. (Note that every convergent sequence is a Cauchy
sequence.)
34.2. Examples.
1. The spaces R
n in any of the following metrics
(ρ1) ρ1(x, y) = Pn
i=1 |xi − yi
|;
(ρ2) ρ2(x, y) = pPn
i=1(xi − yi)
2;
(ρ∞) ρ∞(x, y) = max{|xi − yi
| : i = 1, . . . , n}
are complete, where x = (x1, . . . , xn), y = (y1, . . . , yn).
In the one-dimensional case all of the metrics above on R coincide, and the completeness of R is proved in the course of mathematical analysis.
2. A closed subset Y of a complete metric space (X, ρ) is complete in the metric ρ|Y
(see Remark 12.8 of Lecture 2).
3. A metric space (X, ρ) is complete if and only if X is complete
in the metric ρ
0
(x, y) = min{ρ(x, y), 1}.
4. Let (Xn, ρn) — be complete metric spaces, n ∈ N. The space X =
Q
{Xn :
n ∈ N} is complete in the metric ρ(x, y) = sup{
ρ
0
n(xn,yn)
n
: n ∈ N}, where x = (xn), y = (yn), ρ
0
n
(xn, yn) =
min{ρn(xn, yn), 1}.
Let x
k
, k ∈ N, — be a Cauchy sequence in (X, ρ). Since ρ
0
(prn(x), prn(y)) = ρ
0
(xn, yn) ≤
nρ(x, y) for any points x = (xn), y = (yn), then prn(x
k
) = x
k
n
, k ∈ N, — is a Cauchy
sequence in (Xn, ρ0
n
) for any fixed n ∈ N. The space (Xn, ρ0
n
) is complete, so the
sequence x
k
n
, k ∈ N, has a limit x
∗
n
, n ∈ N. Then the sequence x
k
, k ∈ N,
converges to the point x
∗ = (x
∗
n
). Indeed, any neighborhood of the point x
∗ has, by Theorem 26.5
of Lecture 7, the form Ox∗ =
TN
n=1 pr−1
n
(Oε(x
∗
n
)); for any n = 1, . . . , N there exists mn ∈ N such that
ρ
0
n
(x
k
n
, x∗
n
) < for k > mn; for m = max{mn : n = 1, . . . , N} we have x
k ∈ Ox∗ for k > m.
5. A set that is a countable intersection of open sets of a topological space is called a Gδ-set. If (X, ρ) — is a complete metric space, Y —
a Gδ-set of X, then on Y there exists a complete metric equivalent to the metric ρ|Y .
X \ Y =
S
{Fk : k ∈ N}, where the sets Fk — are closed in X. Define the mapping
f = ∆{fj : j ∈ {0} ∪ N} : Y →
Y{Zj : j ∈ {0} ∪ N},
where X0 = X, f0 = i : Y → X — the embedding, Zj = R, fj : Y → R, fj (x) = 1
ρ(x,Fj )
, j ∈ N.
Obviously, f — is an embedding into the product of spaces metrizable by complete metrics. Hence on the product Q
{Zi
: i ∈ {0} ∪ N} there exists a complete metric ρ
0
(Example 34.2.4). If
we prove that the image f(Y ) is closed in Q
{Zj : j ∈ {0} ∪ N}, then ρ
0
|f(Y ) — is a complete metric on the space f(Y ), homeomorphic to Y .
Let x = (xj ) ∈
Q
{Zj : j ∈ {0} ∪ N} \ f(Y ). Consider the case x0 ∈ Y . There exists
j ∈ N such that fj (x0) 6= xj . Let O1 and O2 — be disjoint neighborhoods of the points xj and fj (x0)
respectively. Since the function fj is continuous, there exists a neighborhood V x0, of the point x0 such
that fj (V x0) ⊂ O2. Then
x = (xj ) ∈ pr−1
0
(V x0) ∩ pr−1
j
(O1) ⊂
Y{Zj : j ∈ {0} ∪ N} \ f(Y ).
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The case x0 6∈ Y . Then x0 ∈ Fj for some j ∈ N. Let r > 0 be such that xj + 1 <
1
r
,
O1 = [0, xj + 1). Then
x = (xj ) ∈ pr−1
0
(Orx0) ∩ pr−1
j
(O1) ⊂
Y{Zj : j ∈ {0} ∪ N} \ f(Y ).
6. A compact metrizable space X is complete in any metric generating its topology.
34.3. Definition. A subset Y is nowhere dense in X, if the set X \Cl(Y ) is everywhere
dense in X.
A subset Y is nowhere dense in X if and only if for every nonempty
open subset O there exists a nonempty open subset V ⊂ O such that
V ∩ Y = ∅.
34.4. Theorem (Baire category theorem). In a complete metric space the complement of a
countable union of nowhere dense subsets is everywhere dense.
Proof. Let Yn, n ∈ N, — be nowhere dense subsets, O — a nonempty open
subset. There exists a closed ball Br1
(x1), of radius r1 ≤ 1 such that Br1
(x1) ⊂ O and
Br1
(x1)∩Y1 = ∅, there exists a closed ball Br2
(x2), of radius r2 ≤
1
2
such that Br2
(x2) ⊂ Br1
(x1),
Br2
(x2) ∩ Y2 = ∅, and so on.
Brn
(xn), n ∈ N, — is a sequence of nested closed balls, whose radii tend to zero. Since for any n < m ρ(xn, xm) ≤ rn, and limn→∞ rn = 0, then (xn) — is a Cauchy sequence. There exists x = limn→∞ xn. Since every ball Brk
(xk) is closed
in X, and the point x is the limit of the subsequence (xn : n ≥ k) of points from Brk
(xk), then
x ∈ Brk
(xk). Hence x ∈ O, x 6∈ Yn, n ∈ N, and x 6∈
S∞
k=1 Yn.
34.5. Remark. Spaces that are a countable union of nowhere dense subsets are called sets of the first category.
§ 35. Totally bounded metric spaces.
35.1. Definition. A metric space (X, ρ) is called totally bounded, if
for any ε > 0 a finite subcover {Oxj : j =
1, . . . , k} can be chosen from the cover {Ox : x ∈ X}.
The set E = {xj : j = 1, . . . , k} is called a finite ε-net.
35.2. Properties. 1. A totally bounded metric space is bounded. For a finite 1-net x1, . . . , xn let M = max{ρ(x1, xk) : k = 1, . . . , n}. Then X = OM+1(x1).
2. A totally bounded space is separable. As the everywhere dense set
one can take the union A =
S∞
n=1 A 1
n
of finite 1
n
-nets A 1
n
, n ∈ N.
3. A compact metrizable space is totally bounded in any metric generating the topology.
35.3. Examples. 1. A subspace of a totally bounded space is totally bounded.
Indeed, let yx ∈ Oε
2
(x)∩Y , x ∈ E, (we consider only those balls of the finite ε
2
-net
{Oε
2
(x) : x ∈ E}, which intersect Y ). Since Oε
2
(x) ⊂ Oε(yx), then {Oε(yx) : x ∈ E} —
is a (finite) ε-net for Y .
2. In n-dimensional Euclidean space R
n boundedness (the possibility of enclosing a set in a cube) coincides with total boundedness.
Indeed, if a cube with edge a is divided into small cubes with edge a
k
(k
n small cubes), then the diameter
of each small cube of the partition equals a
√
n
k
, i.e. the vertices of the small cubes form a finite a
√
n
k
-net.
3. Let (Xn, ρn) — be totally bounded metric spaces, n ∈ N. The space
X =
Q
{Xn : n ∈ N} is totally bounded in the metric ρ(x, y) = sup{
ρ
0
n(xn,yn)
n
: n ∈ N}, where x =
(xn), y = (yn), ρ
0
n
(xn, yn) = min{ρn(xn, yn), 1}.
Let ε > 0, N be such that 1
N <
ε
2
. Let the subset XN ⊂ X consist of the points whose
only the first N coordinates are nonzero. Then for any point x = (xn) ∈ X there exists
a point x
0 = (x
0
n
) from XN such that ρ(x, x0
) <
ε
2
(one must zero out all coordinates of x starting
from N + 1). We may assume that XN =
Q
{Xn : n = 1, . . . , N}. Take in each space
(Xn, ρn) a finite ε
2
-net An, n = 1, . . . , N. It is easy to verify that Q
{An : n = 1, . . . , N} —
is a finite ε
2
-net on XN , which will be an ε-net for X.
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4. The sphere {(xn) : P∞
n=1 x
2
n = 1} in Hilbert space `
2
(`
2 = {(xn) : P∞
n=1 x
2
n < ∞} with
norm ||x|| =
pP∞
n=1 x
2
n
, where x = (xn)) is not totally bounded.
It contains points of which only one coordinate is 1, and the rest are 0. The distance between
them is √
2. Hence a finite number of balls of radius less than 1
2
cannot cover it.
35.4. Corollary. A metrizable space is metrizable by a totally bounded metric
if and only if it is separable.
Proof. Necessity is explained in Property 35.2.2.
Sufficiency follows from Corollary 33.2, Property 35.2.3 and Example 35.3.1.
§ 36. Criterion for compactness of a metrizable space in terms of metrics.
36.1. Theorem (Criterion for compactness of a metrizable space in terms of
metrics). A metrizable space (X, ρ) is compact if and only if X
is totally bounded and complete in some (and hence any) metric.
Proof. Necessity follows from Property 35.2.3 and Example 34.2.6.
Sufficiency. By Theorem 33.3 of Lecture 10 it suffices to show that any sequence contains a convergent subsequence. Consider an arbitrary sequence (xn). Since X is totally bounded, from the sequence (xn) one can choose a fundamental
subsequence (xnk
) as follows. Let An — be a finite 1
n
-net,
n ∈ N, for X. Then {O 1
n
(y) : y ∈ An} — is a finite cover of X, n ∈ N. There exists O1(y1), containing an infinite number of terms of the sequence (xn). In the finite cover {O1
2
(y)∩ O1(y1) :
y ∈ A2} of the set O1(y1) there exists a subset O1
2
(y2) ∩ O1(y1), containing an infinite
number of terms of the sequence (xn), and so on. The desired Cauchy subsequence
is defined as follows xn1 ∈ O1(y1), xn2 ∈ O1
2
(y2) ∩ O1(y1), n2 > n1, and so on.
Since X is complete, there exists a limit x ∈ X of the Cauchy sequence (xnk
).
Hence X is compact.
36.3. Corollary. A subset Y of a metric space (X, ρ) is compact if and
only if Y is totally bounded and complete in the metric ρ|Y .
In particular, if X is a complete metric space, then Y is compact if and only
if Y is totally bounded and closed.
§ 37. Uniform continuity of mappings of metrizable compact spaces.
37.1. Lemma (Lebesgue covering theorem). Let X — be a metrizable compact space, Ω — an
open cover of X, ρ — a metric on X. Then there exists ε > 0 such that the cover
{Oεx : x ∈ X} (of open balls of radius ε) refines Ω.
Proof. For every point x ∈ X there exists εx > 0 such that O2εx x ⊂ U ∈ Ω.
Choose from the cover {Oεx x : x ∈ X} a finite subcover {Oεxk
xk : k = 1, . . . , n}. Then
ε = min{εxk
: k = 1, . . . , n} is the desired number, called the Lebesgue number of the cover.
A mapping f : X → M of metric spaces is uniformly continuous, if for
any ε > 0 there exists δ > 0 such that f(Oδx) ⊂ Oεf(x) for any point x ∈ X (the image
of a set of diameter 2δ has diameter 2ε).
37.2. Theorem (Uniform continuity of mappings of metrizable compacta). Let X — be a metrizable compact space, f : X → M — a continuous mapping into a metric space M, ρ — a metric on X. Then f is a uniformly continuous mapping
of metric spaces.
Proof. Let Ω = {Oε
2
t : t ∈ M}, u = f
−1
(Ω) = {f
−1
(Oε
2
t) : t ∈ M}. By Lemma
37.1. there exists δ > 0 such that the cover {Oδx : x ∈ X} refines u. Then for any point
x ∈ X
f(Oδx) ⊂ Oε
2
t, and t ∈ Oε
2
f(x).
Hence f(Oδx) ⊂ Oεf(x).
§ 38. Uniform metric on a product. Spaces of mappings in the topology of uniform convergence. Topology of pointwise convergence.
62
Recall that on the set Y
X (of mappings of the set X into a topological space
Y or the Cartesian product of |X| copies of the space Y , or the |X|-th power of Y ) is defined the
Tychonoff topology T (§ 17 of Lecture 4).
38.1. Definition. Let (Y, ρ) — be a metric space, ρ
0
(x, y) = min{ρ(x, y), 1} —
a bounded metric on Y . On the product Y
X define the uniform metric
d
0
(f, h) = sup{ρ
0
(f(x), h(x)) : x ∈ X}.
The metric topology Td0 on Y
X is called the topology of uniform convergence.
Checking the correctness of this definition is not difficult.
38.2. Proposition. If (Y, d) is a metric space, then Td0 ≥ T (i.e. Td0 ⊃
T ).
Proof. It suffices to show that the identity mapping id : (Y
X, Td0 ) →
(Y
X, T ) is continuous. Continuity of the mapping id is equivalent to continuity of the mappings
prx ◦ id, x ∈ X (see Corollary 17.3 of Lecture 4). Let f ∈ Y
X, Oεf(x) — be a neighborhood of the point f(x)
in Y . Then for the neighborhood
Oεf = {g ∈ Y
X : ρ
0
(f(x), g(x)) < ε for any point x ∈ X}
f(Oεf) ⊂ Oεf(x).
Assignment N 11
1. Prove that for any metrics ρ1 and ρ2 on a metrizable compact space X the following condition holds: for any > 0 there exist δ1 > 0 and δ2 > 0 such that for any points x, y ∈ X
ρ2(x, y) < as soon as ρ1(x, y) < δ1, and ρ1(x, y) < as soon as ρ2(x, y) < δ2.
2. Prove that the spaces R
n in any of the metrics of Example 34.2.1 are complete.
3. Give an example of a set and topologically equivalent metrics on it such that
one metric is complete and the other is not complete.
Prove that on a locally compact, non-compact metrizable space there exist a complete metric and an incomplete metric.
4. Let two metrics ρ1 and ρ2 be given on a set X. Prove that if there exist real numbers k1 > 0 and k2 > 0 such that ρ1(x, y) ≤ k2ρ2(x, y) and ρ2(x, y) ≤ k1ρ1(x, y) for
any x, y ∈ X, then the metrics ρ1 and ρ2 are topologically equivalent. Moreover, if one of the
metrics is complete (totally bounded), then the other metric is also complete (totally bounded).
5. Prove that Hilbert space `
2 is complete.
6. Prove that a metric space (X, ρ) is complete if and only if
every decreasing sequence of nonempty closed subsets A1 ⊃ A2 ⊃ . . . of the space X such that limn→∞ diamAn = 0, has nonempty intersection.
Prove that an at most countable intersection of open everywhere dense sets of a complete
metric space is everywhere dense.
7. Let (X, ρ) — be a complete metric space, Y ⊂ X. Prove that (Y, ρ|Y ) — is a complete
metric space if and only if Y is closed in X.
8. A metric space (X, ρ) is complete (totally bounded) if and only
if (X, ρ0
) is complete (totally bounded), where ρ
0
(x, y) = min{ρ(x, y), 1}.
9. Prove that on a countable product of complete (totally bounded) metric spaces there exists a complete (totally bounded) metric.
10. Let (X, ρ) — be a complete metric space, Y ⊂ X. Prove that Y is metrizable
by a complete metric if and only if Y is a Gδ-subset of X (i.e. Y
is the intersection of countably many sets open in X).
Prove that on the space of irrational numbers there exists a complete metric.
11. Prove that on the space of rational numbers no metric is complete.
12. (Banach fixed-point theorem.) A mapping f : X → X of a metric
space is called contracting, if there exists 0 ≤ α < 1 such that ρ(f(x), f(y))) <
αρ(x, y) for any x, y ∈ X.
63
Prove that any contracting mapping of a complete metric space into itself
has a fixed point (i.e. a point that maps to itself).
13. Prove that a subset Y of a metric space (X, ρ) is totally bounded if and
only if from every infinite sequence one can choose a Cauchy subsequence.
14 (Criterion of compactness in `
2
). A subset X in `
2 is compact if and only
if it is closed, bounded, and the condition
lim
N→∞
( sup
x∈X
X∞
n=N
|xn|
2
) = 0
holds.
Prove that the subset {x = (x1, x2, . . .) ∈ `
2
:
P∞
n=1 |anxn|
2 ≤ 1}, where an > 0, an → +∞
is compact.
Additional problems for Assignment N 11
15. Check the validity of the following statement. For any metrics ρ1 and ρ2 on a metrizable compact space X the following condition holds: there exist numbers k1 and k2 such that for any
points x, y ∈ X
k1ρ1(x, y) ≤ ρ2(x, y) ≤ k2ρ1(x, y).
16. Prove that any metric on a metrizable space is totally bounded if and
only if the space is compact.
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