Download METRIC SPACES AND COMPACTNESS 1. Definition and examples

METRIC SPACES AND COMPACTNESS
1. Definition and examples
Definition 1.1. Let X be a set and d : X × X → [0, ∞i a function subject to the conditions:
(1) d(x, x) = 0 for any x ∈ X and d(x, y) = 0 only if x = y.
(2) d(x, y) = d(y, x) for all x, y ∈ X.
(3) d(x, z) ≤ d(x, y) + d(y, z) for any x, y, z ∈ X.
Then the function d is called a metric on X and the pair (X, d) is called a metric space. Each
metric space comes equipped with the metric topology Td generated by the basis Bd defined as
Bd = {Bx (r) | x ∈ X, r > 0}
As in the case of the Euclidean spaces, Bx (r) are the open balls of radius r centered at x:
Bx (r) = {y ∈ X | d(x, y) < r}
Examples: 1. For any p ≥ 1, the function dp : Rn × Rn → [0, ∞i
p
dp (x, y) = p |x1 − y1 |p + .. + |xn − yn |p
with x = (x1 , ..., xn ) and y = (y1 , ..., yn ), is a metric on Rn . For p = 2 this becomes the familiar Euclidean distance function. All of these metrics induce the same topology on Rn , namely the Euclidean
topology.
2. Let (X, d) be a metric space and let A be a subset of X. Then (A, d|A ) is also a metric space. For
example, any surface that embeds into R3 is a metric space - it inherits any of the metrics dp from
example 1.
Metric spaces share many nice properties with Euclidean spaces. For example, a subset U of a
metric space X is open if and only if for every x ∈ U there is some ε > 0 such that Bx (ε) ⊆ U .
Likewise, a subset A is closed if and only if it contains the limits of all of its convergent sequences
(prove these two facts!).
2. Metrizability
Definition 2.1. We say that a topological space (X, T ) is metrizable if there is a metric on X such
that the metric topology Td agrees with T .
It is easy to see that metrizability is preserved by homeomorphisms. That is to say, if (X, TX ) is a
metrizable topological space and f : X → Y is a homeomorphism to a topological space (Y, TY ), then
the latter is also metrizable. If dX is the metric on X inducing the topology TX , then the function
dY (a, b) = dX (f −1 (a), f −1 (b))
a, b ∈ Y
is a metric on Y and it induces the topology TY (check that this is true!).
Since every metric space carries a topology (the metric topology), it is natural to ask if in fact the
converse is true. That is, is every topological space (X, T ) metrizable? As we shall see, this is not so.
Theorem 2.2. A metrizable topological space (X, T ) is Hausdorff.
Proof. Let d be a metric on X such that T = Td . Let a, b ∈ X be two arbitrary points and let
D = d(x, y). Since x 6= y it must be that D 6= 0. But then the sets Ua = Ba (D/2) and Ub = Bb (D/2)
are two disjoint open sets containing a and b respectively.
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Example 3: We already saw that the topological space (R, Tf c ) (where Tf c is the finite complement
topology) is not Hausdorff. By theorem 2.2 it cannot be metrizable.
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METRIC SPACES AND COMPACTNESS
3. Completions of metric spaces
Definition 3.1. Let (X, d) be a metric space. We say that a sequence xi ∈ X is a Cauchy sequence
if for every ε > 0 there is some n0 ∈ N such that for every m, n ≥ n0 we get d(xm , xn ) < ε. We say
that (X, d) is a complete metric space if every Cauchy sequence in X is a convergent sequence.
For example, (Rn , d2 ) is a complete metric space but (h0, 1i, d2 ) is not: the sequence xi = 1/(i + 1)
is Cauchy but has not limit in h0, 1i. This last example shows that subsets of complete metric spaces
are not necessarily themselves complete.
Theorem 3.2. For every metric space (X, d) there exists a complete metric space (X 0 , d0 ) with the
properties:
(1) X is contained in X 0 as a dense subset: X̄ = X 0 .
(2) d0 extends d: if x, y ∈ X then d0 (x, y) = d(x, y).
Furthermore, if (X 00 , d00 ) is another such metric space, then there is a homeomorphism F : X 0 → X 00
such that d00 (F (a), F (b)) = d0 (a, b) for every a, b ∈ X 0 .
A metric space (X 0 , d0 ) as in the theorem above is called a complete extension of (X, d). The
theorem tells us that complete extensions always exist and are unique (up to metric preserving homeomorphisms) if we require that X be dense in the extension.
4. The Sobolev spaces
For n ∈ {0, 1, 2, ...}, let C n [a, b] be the set of all n times differentiable functions from [a, b] to R.
When n = 0, C 0 [a, b] is the set of all continuous function from [a, b] to R. Given any real number
p ≥ 1 and an integer q with 0 ≤ q ≤ n, we define metrics dp,q on C n [a, b] as:
ÃZ
!1/p
q
bX
(i)
(i)
p
dp,q (f, g) =
|f (x) − g (x)| dx
a i=0
(i)
where f (x) is the i-th derivative of f (x) and f (0) (x) = f (x).
Example 4: Consider the simplest case of dp,q where q = 0 = n and p = 2: Then
s
Z b
d2,0 (f, g) =
(f (x) − g(x))2 dx
a
0
Show that d2,0 is a metric on C [a, b].
These metric spaces of functions (C n [a, b], dp,q ) are not complete but by theorem 3.2 we know that
they have a unique completion with C n [a, b] being dense.
Definition 4.1. The Sobolev space Lp,q (C n [a, b]) is the unique complete metric space (X, D) which
contains C n [a, b] as a dense subset and where D extends dp,q .
The various Sobolev spaces play an important role in theory of partial differential equations. Observe that in the definition of the metrics dp,q it was essential that we could subtract two functions
f, g (since terms of the form f (i) (x) − g (i) (x) entered the definition of dp,q ). These definitions don’t
generalize for functions into arbitrary topological spaces Y . However, we will see soon how one can in
general put a topology on the set C 0 (X, Y ) of continuous functions from X to Y using compact sets.
5. Compactness for metric spaces
Definition 5.1. We say that a metric space (X, d) is totally bounded if for every ε > 0 there exist
finitely many points xi ∈ X, i = 1, ..., N such that
X=
N
[
i=1
Bxi (ε)
METRIC SPACES AND COMPACTNESS
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Typically the number N of points xi needed to cover X by radius ε balls will increase as ε becomes
small. But now matter how small ε gets, N stays finite.
Proposition 5.2. If a metric space (X, d) is totally bounded then X is separable.
Proof. Let εn = 1/n for n ∈ N. For each such εn , there is a finite set of points An = {xn1 , ..., xn`n }
such that
`n
[
Bxi (1/n)
X=
i=1
Let A = ∪n∈N An . Clearly A is countable and we claim that A = X. Let x ∈ X be an arbitrary
point. For any n ∈ N, there is some yn ∈ An so that x ∈ Byn (1/n). This gives a sequence yn ∈ A
with d(x, yn ) < 1/n. Thus lim yn = x. This shows that every x ∈ X lies in the closure of A and so
A = X.
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Proposition 5.3. A metric space (X, d) is second countable if and only if it separable.
In exercise 3 you are being asked to prove this proposition. Recall that we already know that
a second countable space is separable. The content of the proposition is that for metric spaces the
converse is also true.
Lemma 5.4. Let (X, d) be a metric space with the property that every sequence xi ∈ X has a convergent subsequence. Then X is totally bounded.
Proof. If X is not totally bounded, then there is some ε > 0 so that X cannot be covered by a finite
number of balls of radius ε. We use this to construct a sequence xi ∈ X: let x1 be arbitrary and choose
consecutive elements of the sequence so that xi+1 ∈
/ ∪ij=1 Bxj (ε). Let yn = xin be some convergent
subsequence of xi with lim yn = y0 . Let U = By0 (ε/2) and let yk , y` ∈ U be any two elements (which
must exist by the convergence property) with k < `. Then ik < i` and
d(xik , xi` ) ≤ d(xik , y0 ) + d(y0 , xi` ) < ε/2 + ε/2 = ε
This implies that xi` ∈ Bxik (ε) which is a contradiction (to the way we constructed the sequence xi ).
Thus X must be totally bounded.
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Theorem 5.5. A metric space (X, d) is compact if and only if every sequence xi ∈ X has a convergent
subsequence.
Proof. =⇒ Suppose X is compact and let xi ∈ X be an arbitrary sequence. If there was no
convergent subsequence of xi we could for each y ∈ X find an open set Uy which contains only finitely
many elements of the sequence xi . But then F = {Uy | y ∈ X} is a cover of X implying that there
are finitely many points y1 , ..., ym ∈ X so that X = Uy1 ∪ ... ∪ Uy1 . This would tell us that the set
{xi | i ∈ N}. In particular, some value must occur infinitely many times giving rise to a convergent
subsequence.
⇐= Suppose that every sequence xi ∈ X has a convergent subsequence and let F be a cover of
X. Find a countable open cover F 0 = {Ui ⊆ X | Ui is open , i ∈ N} with the property that for every
Ui ∈ F 0 there is some V ∈ F with Ui ⊆ V (see exercise 2).
Claim: F 0 has a finite subcover.
Ãn−1 !
[
To prove the claim, suppose the opposite. Then by picking an arbitrary xn ∈ X −
Ui we
i−1
obtain a sequence. By the assumption, there must be a subsequence yi = xni with converges to y0 .
Since F 0 is a cover, there is some m ∈ N with y0 ∈ Um . But then yj ∈
/ Um for all j ≥ m which is a
contradiction. This concludes the proof of the claim.
Let thus {U1 , .., UN } be a finite cover of X and let Vi ∈ F be such that Ui ⊆ Vi for all i = 1, ...., N .
The {V1 , ..., Vn } is also a finite cover of X.
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METRIC SPACES AND COMPACTNESS
Corollary 5.6. Every compact metric space X is totally bounded, separable and second countable.
Theorem 5.7. Let (X, d) be a metric space. Then X is compact if and only if X is complete and
totally bounded.
Proof. =⇒ Suppose X is compact, we need to show that X is complete and completely bounded.
To see completeness, let xi ∈ X be any Cauchy sequence in X. By theorem 5.5 there is a convergent
subsequence of xi with limit x0 . It is then easy to see that xi converges with limit x0 (see exercise 3).
Thus X is complete.
On the other hand, consider the open cover Fε of X defined for any ε > 0 as
Fε = {Bx (ε) | x ∈ X}
By compactness of X there must be a finite subcover of Fε . This shows that X is totally bounded.
⇐= Let X be complete and totally bounded. We will show that X is compact by showing that
any sequence xi ∈ X has a convergent subsequence (see theorem 5.5). Let εk = 1/k for k ∈ N. For
ε1 , there is a finite cover of X with balls of radius ε1 . Since xi has infinitely many terms, one of the
these radius εk balls contains infinitely many elements of the said sequence. Let B1 be that ball. For
ε2 there is again a finite cover of X by ε2 balls. Since B1 contains infinitely many xi , there is some
radius ε2 ball B2 such that B2 ∩ B1 contains infinitely many of the xi . One then continues this process
indefinitely producing a sequence of radius εk balls Bk so that Bk ∩ bk−1 ∩ ... ∩ B1 contains infinitely
many elements of the sequence xi .
Pick now indices n1 < n2 < n3 < ... such that xn` ∈ B` and set yi = xni . It is easy to see that
yi is Cauchy and so by the completeness assumption on X, it must have a convergent subsequence.
This completes the proof.
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Corollary 5.8. Consider Rn equipped with the Euclidean topology. Then a subset A of Rn is compact
if and only if it is closed and bounded.
Proof. By exercise 4, A is totally bounded and by exercise 5, A is complete. Now use theorem 5.7.
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6. The one point compactification
Definition 6.1. The compactification of a topological space X is a compact topological space Y
containing X as a dense subset.
Given any non-compact topological space X, compactifications always exist, in fact a compactification of any non-compact topological space can be achieved by adding only one additional point to
X and extending the topology in a suitable way. This particular way of compactifying X is called the
one-point compactification of X. Here are the details:
Given any non-compact topological space X, define Y = X ∪ {p} where p is some abstract point
(and p ∈
/ X). Let TX be the existent topology on X. We define a new topology TY on Y as follows:
A subset U of Y is an open set of Y if either
(1) p ∈
/ U and U ∈ TX or
(2) p ∈ U and X − U is a compact closed subset of X.
Let’s show first that X is dense in Y . By definition of the topology on Y , the set {p} is not open
(since X is non-compact) and so X ⊂ Y is not closed. Therefor X̄ 6= X which forces X̄ = Y .
Now let’s show that Y is compact. Let F be an open cover of Y . There must be some set U0 ∈ F
which contains p. But then X − U0 is a compact set and so there is a finite number of elements
U1 , ..., Un ∈ F whose union is all of X − U0 . But then F 0 = {U0 , U1 , ..., Un } is a finite subcover of F
showing that Y is compact.
7. Exercises
(1) Prove proposition 5.3.
METRIC SPACES AND COMPACTNESS
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(2) Let (X, d) be a second-countable metric space. Then for every open cover F of X there exists
a countable open cover F 0 = {Ui ⊆ X | Ui is open , i ∈ N} with the property that for every
Ui ∈ F 0 there is some V ∈ F with Ui ⊆ V .
(3) Let (X, d) be a metric space and xi ∈ X a Cauchy sequence which has a convergent subsequence with limit x0 . Show that then xi is also convergent with limit x0 .
(4) Consider Rn equipped with the Euclidean topology. Show that A ⊆ Rn is bounded if and
only if it is totally bounded.
(5) Let X be a complete topological space and let A be a subset of X endowed with the relative
topology. Then A is also complete.