Royden once commented that the structure
of a metric space by itself is too barren to be fruitful in the study of
interesting problems in analysis and topology.[1]
However, we can add one more property, namely completeness, to establish many
theorems that are topological in character. This chapter is devoted to some of
the most used examples of complete compact spaces and their properties.
First we recall the notion of Cauchy
sequence, which is directly related to the definition of completeness. A
sequence {xn} in a metric space (X, d) is Cauchy if for all ε>0,
there is an index N such that if n, m≥0, then d(xn, xm)<ε.
A metric space is said to be complete
if every Cauchy sequence in X converges to a point in X. Equivalently, we can
say that metric space X is complete if every Cauchy sequence in X has a
convergent subsequence. We can prove the equivalence easily by showing that the
subsequence has the same limit point with the original sequence.
Some metric spaces we are familiar with
are actually complete. For example, Euclidean space Rk is complete
either in Euclidean metric or in square metric. Let d_bar denote the standard
bounded metric on R. If J is countable then the metric D(x,y)=sup{d_bar (xi, yi)/i}
induces the product topology on Rω, and it is complete under this
metric. When J is not countable, RJ is complete given the uniform
metric.
Given topological spaces X and Y, we use
C(X, Y) to denote the set of all continuous maps from X to Y. When X is a
topological space and Y is a complete metric space, C(X, Y) is complete in the
uniform metric. C(X, Y) is rather important in this chapter’s study because it
turns out that we can impose many fruitful topologies on this space and derive
a deep theorem, namely the Ascoli’s theorem. Before we get that far, we should
empower ourselves with some new notions and relate them to what we have
learned, such as compactness.
One thing noticeable is that completeness
is not a topological property because it is not preserved by homeomorphisms.[2]
For example, on (-1, 1) with Euclidean metric, the sequence {xn} defined
by xn=1-1/n is Cauchy but doesn’t converge (because limit point 1 is
not in the space). However, R, which is homeomorphic to (-1, 1), is complete
with Euclidean metric. This is also an example showing that in general, a
subspace of a complete metric space is not complete. However, if the subset is
closed, then it is complete and vice versa.
Proof:
suppose Y is a closed subset of X, which is complete and metrizable, and let
the sequence {yn} to be a Cauchy sequence in Y. Because {yn}
is also Cauchy in X, by definition of X being complete, {yn}
converges to a point in X, say y. Since Y is closed, it contains the limit
point of {yn}, which is y. By definition Y is complete.
Conversely, assume Y is complete and we
prove that it is closed by showing the limit of a convergent sequence in Y
belongs to Y. Let {yn} be a sequence in Y and converges to y in X.
Because X is a metric space, {yn} is also Cauchy, and thus by
completeness of Y it contains limit point of {yn}, which is exactly
y. So Y is closed. (Q.E.D)
An interesting application of completeness
is the Peano space-filling curve, which passes through every point of the unit
square. Roughly speaking, this curve is a continuous mapping from unit interval
onto the unit square. The discovery was motivated by Cantor’s theory that R and
R2 has the same cardinality, and it also had historical importance.
For example, it motivated the study of algebraic topology by showing that
result like Jordan Curve Theorem and invariance of domain might need more
careful proofs[3].
After having learned something about
completeness, a natural question might emerge: is there any way that we can
relate this definition to materials we have known before? The answer is “yes”
because compactness and completeness can be combined together to form a new
kind of metric space.
Let us first recall a very important equivalence
theorem with regard to metric space. Namely, if X is a compact metric space,
then X is also limit point compact and sequentially compact. [4]
Previously we have shown that a metric space where every Cauchy sequence has a
convergent subsequence is complete, and thus it is clear that every compact metric space is complete
(sequential compactness).
Note that the converse is not true in
general. One straightforward counterexample is the real line R under Euclidean
metric (it is complete, but cannot be covered by finitely many open sets). Here
comes a new question: what extra requirements should we put on a complete
metric space so that it can also be compact? It turns out that we need to introduce
a new notion: total boundedness.
Definition: a metric space X is totally bounded if for each ε>0, X
can be covered by a finite number of open balls of radius ε. A subset E of X is
totally bounded provided that E,
considered as a subspace of the metric space X, is totally bounded.
A metric
space (X, d) is compact if and only if it is complete and totally bounded.
Munkres[5]
laid out a detailed proof in section 45, so interested readers can refer to the
book. The main reason we need total boundedness is because we can inductively
construct a Cauchy sequence in X by choosing a decreasing radius epsilons, and
with property of completeness, we can show that X is sequentially compact.
Now that we know when a complete metric
space is compact, our next step is to find the compact subspaces of C(X, Rn)
in the uniform topology. By Heine-Borel theorem, a subspace of Euclidean space
Rn is compact if and only if it is closed and bounded. However, in a
general metric space, the condition of being closed and bounded is necessary
but not sufficient. For example, the closed unit ball of C[0,1] is closed and
bounded in C[0,1], but it fails to be sequentially compact and thus compact. As
a remedy, we need to impose another property, i.e. equicontinuity, upon the
subspace of C(X, Rn).
Definition: Let (Y, d) be a metric space.
Let F be a collection of real-valued functions of C(X, Y). If x0
belongs to X, then F is equicontinuous
at x provided that for all ε>0, there is a neighbourhood U of x such that
for all x’ in U, d(f(x), f(x’))<ε for all f in F.
Note that equicontinuity is a stronger condition
than continuity because it requires all functions in F to be continuous given
the same choice of epsilon. In general, an infinite collection of continuous
functions is not equicontinuous. For instance, define fn(x)=xn
(0≤x≤1) for all n belongs to N. Then {fn} is a countable collection
of continuous functions on [0, 1]. However, it is not equicontinuous at x=0.
It turns out that if X is a space and (Y, d)
is a metric space, and the subset F of C(X, Y) is totally bounded under the
uniform metric corresponding to d, then F is equicontinuous under d[6]. If
we further assume that X and Y are compact, the converse is also true. One
thing worth mentioning about these two proofs is that I think when it comes to
something related to equicontinuity, we can often use the ε/3 trick. Namely, we
first use continuity of one function, say f, to create an ε/3 distance between
f(x) and f(x’), in which x’ is in U(x). Then we can use corresponding uniform
metric of continuous functions to define the other two inequality, and finally
we use triangle inequality to wrap up the whole proof.
Before we prove the classical version of
Arzela-Ascoli theorem, we introduce two more definitions on a family of
functions. A sequence of functions {fn} of C(X, Y) is pointwise bounded under d if for each x
in X, the subset Fa={f(a)|f in F} of Y is bounded. {fn} is
uniformly bounded if there is M≥0
such that |fn|≤M for all n.
Now we can combine things together and
prove the classical version of Ascoli’s theorem: Let X be a compact space; let
(Rn, d) denote Euclidean space in Euclidean metric; give C(X, Rn)
the corresponding uniform topology. A subspace F of C(X, Rn) has
compact closure if and only if F is equicontinuous and pointwise bounded under
d.
At
first glance this is a dense theorem; however, we can piece things together to
handle it. First assume G to be the compact closure of F. Since X is compact,
by extreme value theorem all functions in C(X, Rn) have extreme
values, and thus sup metric is well-defined. Since G is compact and metrizable,
it is totally bounded[7],
and total boundedness further implies it is equicontinuous under d. Compactness
of G also implies it is bounded under sup metric, and thus G is pointwise
bounded under d by definition. Since F is contained in G, F is also
equicontinuous and pointwise bounded under d.
Conversely, assume that F is equicontinuous
and pointwise bounded under d. To show that G is compact, we can equivalently
show that it is complete and totally bounded. Completeness is easy to prove
since G is closed in the complete metric space (C(X, Rn), sup
metric). To prove total boundedness, the book used two steps. Firstly, we prove
that G is also equicontinuous and pointwise bounded under d by using the ε/3
trick. Then we show that there is a compact subspace Y of Rn such that
G in contained in C(X, Y). Since F is equicontinuous, we can apply the previous
lemma and claim that G is totally bounded.
Now I tend to prove Arzela’s theorem given
as an exercise in page 280.
Theorem
(Arzela’s theorem). Let X be compact;
let fn C(X, Rk). If the
collection {fn} is pointwise bounded and equicontinuous, then the
sequence fn has a uniformly convergent subsequence. [8]
Proof:
The topology of pointwise convergence is
simply the subspace topology that C(X, Y) receives from the product topology on
YX. Since {fn} is pointwise bounded and equicontinuous,
by Ascoli’s theorem, {fn} has compact closure. As a result C|({fn})
is sequentially compact, which implies that { fn } has a subsequence
{fnk} which converges under the uniform metric. By the very
definition of uniform metric, {fnk } is uniformly convergent.[9] (Q.E.D)
Arzela-Ascoli’s theorem can be applied to determine
whether certain integrals have minimizers or to solve some non-linear
differential equations like Peano’s theorem. However, if X is not compact in
the first place, then this theorem loses its power because we cannot even define
sup metric on C(X, Rn)! So in terms of Arzela-Ascoli theorem, it’s
all about compactness.
Now we should explore some interesting
topologies on C(X, Y) in addition to the uniform topology. Specifically, we
look at three topologies: the topology of pointwise convergence, the topology
of compact convergence, and the compact-open topology.
Given spaces X and Y, a sequence of
functions {fn} is said to be pointwise convergent if for every point
x in X, the sequence fn(x) in Y is convergent. Given a point x of X
and an open set U of Y, let S(x, U) = {f | f belongs to Y and f(x) is contained
in U}. Then the sets S(x, U) are subbasis for topology of pointwise convergence on YX.
A typical basis element about the function
f is all functions that are epsilon close to it at finitely many points. Simply
put, the topology of pointwise convergence is just the subspace topology that
C(X, Y) inherits from the product topology on YX. The reason why we
call it pointwise convergent topology is because if for each x in X, fn(x)
converges to f(x), then fn converges to f in this topology and
vice versa.
A convergent sequence of continuous functions
in the topology of pointwise convergence doesn’t necessary have a continuous
limit. A classic example is fn(x)=xn , 0≤x≤1, whose
convergence function is piecewise. We know that in uniform topology, if {fn}
is continuous and convergent to f, then f is also continuous. So naturally a
new question pops up: is there a topology coarser than uniform topology but
satisfies that the limit point of convergent sequence of continuous functions
is continuous? It turns out we can find one, namely the topology of compact
convergence.
Definition: Let (Y, d) be a metric space;
let X be a topological space. Given an element f of YX, a compact
space C of X, and a number >0, let BC
(f, ε) denote the set of all those elements g of YX for which:
sup{d(f(x), g(x)) | x belongs to C} < ε. The sets BC (f, ε) form
a basis for topology on YX, and it’s called the topology of compact convergence.
Note that this topology is stronger than
the one on pointwise convergence because it demands basis element containing f
to consists of function that are ε close to f at all points at some specified
compact set. The reason why we call it topology of compact convergence is due
to the theorem that fn converges to f in the topology of compact
convergence if and only if for each compact subspace C of X, the sequence fn
| C converges uniformly to f | C.
A space X is compactly generated provided that set A C being open in C for each compact subspace C
of X implies A to be open in X.[10]
It turns out that if X is locally compact or has a countable basis at each of
its point, then X is compactly generated.
Compactly generated spaces have many
useful properties. For example (assume X is compacted generated), if we can
show that for each compact subspace of X, the restricted function f | C is
continuous, then we can actually conclude that f is continuous. Using this
property, we can prove that in terms of X and (Y, d), C(X, Y) is closed in YX
in the topology of compact convergence.
If
X is compact, then uniform convergence on compact set implies the uniform
convergence upon X. If X is discrete, the only compact sets are the finite
sets, so compact convergence coincides with pointwise convergence. If X is
arbitrary, then uniform convergence implies compact convergence, which further
indicates pointwise convergence.
Since both uniform and compact convergence
topologies depend on metrics, we are curious whether these topologies can work
when Y is arbitrary. We focus our exploration on C(X, Y) and it turns out that compact-open topology on C(X, Y),
generated by S(C, U) = {f | f belongs to C(X, Y) and f(C) is contained in U}, where
C is a compact subspace of X and U is an open subset of Y, is a generalization
of the compact convergence topology to arbitrary X and Y. Furthermore, when Y
is metrizable, compact convergence topology is the same as compact-open
topology. The proof is just to show that
one topology is finer than the other and vice versa. As a corollary, the
compact convergent topology on C(X, Y) doesn’t depend on metric of Y. (The
reason is that we didn’t say anything about metric when defining compact-open
topology)
Another desirable property of
compact-open topology is that it is jointly continuous because the expression
f(x) is continuous both in x and f. In fact, if X is locally compact Hausdoff,
and let C(X, Y) have the compact-open topology, then the map
e:
X × C(X, Y) → Y
defined by e(x, f)=f(x)
is continuous. (e is called the evaluation
map)
Local compactness is the crucial property
in order for C(X, Y) to be jointly continuous. A counter-example is to let X= Q,
set of rational numbers.[11]
Compact-open topology is widely applied
in algebraic topology. Its connection with homotopy is briefly discussed in
page 288.
Using the three topologies on C(X, Y),
along with Tychonoff theorem, we can pprove a more general Ascoli’s theorem, in
which C(X, Y) is given the topology of compact convergence. Interested readers
can refer to section 47 of the book. Munkres certainly does a better job laying
out details of proof than I do!
[1] Real Analysis, 4th edition.
H.L. Royden and P.M. Fitzpatrick
[2] Topology, 2nd edition. J.R.
Munkres
[3] A.P.M
Kupers, “On Space-Filling Curves and the Hahn-Mazurkiewicz Theorem.” Notes for
a Stanford SUMO talk on space-filling curves.
[4] A
space X is limit point compact if every infinite subset of X has a limit point,
and X is said to be sequentially compact if every sequence of points X has a
convergent subsequence.
[5] Topology 2nd edition, J.R.
Munkres, pp.276
[6] Again,
the detailed proofs are in page 277 and 278.
[7]
Refer to previous discussion.
[8]
For
whatever reason, this is what most people think about when it comes to
discussion of Arzela-Ascoli theorem. I read Introduction
to Real Analysis by Stephen Abbott and Real
Analysis by Royden for references, and their discussion of Arzela-Ascoli
theorem is this Arzela’s theorem. I guess the reason is that in terms of
analysis, finding a criterion of a uniform subsequence is more important than
determining a compact closure.
[9] This
is probably the easiest proof since we can cite Ascoli’s theorem right away.
Another proof with more analysis flavor is to use Cantor diagnonalization argument.
I read this proof in Royden’s Real
Analysis.
[10]
This notion can be equivalently defined by using B as a closed set.
[11]
We have proved that Q is not locally compact.
