ADDENDUM C: COMPACT SETS AND THE ARZELÀ-ASCOLI
THEOREM
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, 2.31-2.41, 3.5-3.6 and 7.19-7.25].
1. An alternative description of compact sets
The aim of this section is to prove the following important result, which gives three
equivalent notions of compactness:
Theorem 1.1. Let X be a metric space. The following conditions are equivalent:
(i) X is compact;
(ii) any infinite subset of X has a limit point in X;
(iii) any sequence in X contains a convergent subsequence.
Proof. The fact that (i) implies (ii) is [Ru, Thm. 2.37], and the fact that (ii) implies (iii)
is proved in the proof of [Ru, Thm. 3.6(a)].
We will now prove that (iii) implies (i).
Assume therefore that X satisfies condition (iii) and that {Uα } is an open cover of X.
We want to find a finite subcover. We will assume that X is nonempty, otherwise there
is nothing to prove.
We give two different proofs of this.
PROOF ALTERNATIVE 1: This follows the lines in [Ru, Exc. 23, 24, 26, p. 45].
Step I: X has an at most countable dense subset.
Indeed, fix δ > 0. Pick x1 ∈ X. If Nδ (x1 ) $ X, choose x2 ∈ X \ Nδ (x1 ). Having
chosen x1 , . . . , xk−1 ∈ X, pick
xk ∈ X \ (Nδ (x1 ) ∪ · · · ∪ Nδ (xk )) ,
if possible, that is, if the set on the right is nonempty. If the process does not terminate,
we would have a sequence {xk } in X such that d(xi , xj ) ≥ δ for all i 6= j. No subsequence
of this sequence can therefore be Cauchy, whence (by [Ru, Thm. 3.11]) no subsequence
can converge, a contradiction to assumption (iii). Hence, the process must terminate,
which means that there are finitely many points x1 , . . . , xM ∈ X so that
X = Nδ (x1 ) ∪ · · · ∪ Nδ (xM ).
Both M and the points x1 , . . . , xM depend on δ. Hence, using δ = 1/n, we obtain for
any n ∈ Z+ that there is an integer Mn and finitely many points x1,n , . . . , xMn ,n ∈ X so
that
(1)
X = N1/n (x1,n ) ∪ · · · ∪ N1/n (xMn ,n ).
The set E := {xi,n }n∈Z+ ,1≤i≤Mn is at most countable (by [Ru, Thm. 2.12]; our set may
be finite due to repetitions among the xi,n ) and we claim that it is dense: Indeed, if
x ∈ X and δ > 0 are given, then pick any n so that 1/n < δ. By (1), x must lie in one
of the neighborhoods N1/n (xi,n ), whence d(x, xi,n ) < 1/n < δ for some i, which means
1
2
ANDREAS LEOPOLD KNUTSEN
that Nδ (x) ∩ E 6= ∅. Hence, by definition, x is a limit point or a point of E, so that E
is dense in X by definition (see [Ru, Def. 2.18(j)]).
Step II: X can be covered by a countable subcollection of the Uα s.
To prove this, let x ∈ X. Then x ∈ Uαx for some index αx , and since Uαx is open,
there is a positive rational number rx such that
(2)
N2rx (x) ⊂ Uαx .
Let E ⊂ X be the at most countable dense subset of X that exists by Step I. Then,
as E is dense, x is either a point of E or a limit point of E (see [Ru, 2.18(j)]. Hence,
Nrx (x) ∩ E 6= ∅, which means that there is a p ∈ E such that d(p, x) < rx . We have
(3)
x ∈ Nrx (p) ⊂ Uαx ,
since for any z ∈ Nrx (p), we have d(z, x) ≤ d(z, p) + d(p, x) < rx + rx = 2rx , so that
z ∈ N2rx (x) ⊂ Uαx , using (2).
To summarize, we have for any x ∈ X found a Uαx in {Uα }, a p ∈ E and a rational
number rx such that (3) holds. Since
{Nr (p)}p∈E,r∈Q
is a countable collection of subsets of X by [Ru, Thm. 2.12], we see from (3) that a
countable collection of the subsets in {Uα } are sufficient to cover X.
Step III: Conclusion of the proof.
By Step II, we can find a countable subcollection {U1 , U2 , U3 , . . .} of sets in {Uα } so
that X = ∪∞
i=1 Un . We want to prove that X can be covered by finitely many of the open
sets Un . If this were not the case, we would have that
Vn := U1 ∪ · · · ∪ Un $ X for all n ∈ Z+ ,
which means that the complements Fn := (Vnc ) = X \ Vn satisfy
(4)
Fn+1 ⊂ Fn and Fn 6= ∅ for all n.
Now we can construct a sequence {xn } in X by picking one xn ∈ Fn for each n ∈ Z+ .
By assumption (iii), this sequence contains a convergent subsequence {xnk }. Let p ∈ X
be its limit. Then, as X is covered by the sets Un , there is an integer M such that
p ∈ UM ⊂ VM . As VM is open, there is a δ such that Nδ (p) ⊂ VM . As VM = (FM )c , we
have that Nδ (p) ∩ FM = ∅, whence by (4), also
Nδ (p) ∩ Fn = ∅ for all n ≥ M,
which means that Nδ (p) can contain only finitely many xn , so that p cannot be a limit
of any subsequence of {xn }, a contradiction.
PROOF ALTERNATIVE 2: This follows closely the approach in [Li].
We start by defining a function
f : X → [0, 1]
by
f (x) = sup{r ∈ R | 0 < r < 1 and Nr (x) ⊂ Uα for some α}.
Step I: f is positive, that is, f (x) > 0 for all x ∈ X.
Indeed, for all x ∈ X, there exists an index α0 such that x ∈ Uα0 , as X = ∪α Uα . Since
Uα0 is open, there exists an r such that 0 < r < 1 and Nr (x) ⊂ Uα0 . Thus, f (x) ≥ r > 0.
Step II: |f (x) − f (y)| ≤ d(x, y) for all x, y ∈ X.
Indeed, as f (x) ≥ 0, the inequality is immediate if f (x), f (y) ≤ d(x, y). Hence, we
may without loss of generality assume that f (x) ≥ f (y) and f (x) > d(x, y). Then, by
definition of f , there exists an r > d(x, y) and an index α0 such that
(5)
Nr (x) ⊂ Uα0 .
COMPACT SETS AND THE ARZELÀ-ASCOLI THEOREM
3
For any z ∈ Nr−d(x,y) (y) we have
d(z, x) ≤ d(z, y) + d(y, x) < (r − d(x, y)) + d(x, y) = r;
it follows that
Nr−d(x,y) (y) ⊂ Nr (x) ⊂ Uα0 ,
for any r satisfying (5), so that
f (y) ≥ r − d(x, y) for any r satisfying (5).
Hence, by definition of f , one has f (y) ≥ f (x)−d(x, y). As we assumed that f (x) ≥ f (y),
we obtain that
|f (x) − f (y)| = f (x) − f (y) ≤ d(x, y),
as desired.
Step III: f is continuous on X.
Indeed, f is even uniformly continuous: given > 0, choose δ = ; by Step II, if
d(x, y) < δ, then |f (x) − f (y)| ≤ d(x, y) < δ = .
Step IV: f attains a minimum r0 on X, that is, there is a p ∈ X such that
0 < r0 = f (p) ≤ f (x) for all x ∈ X.
Let E := {f (x) | x ∈ X} ⊂ (0, 1] be the range of E. Then by [Ru, Thm. 2.28 and
3.2(c)], there is a sequence {yn = f (xn )} in E converging to inf E. By assumption (iii),
there is a subsequence {xnk } of xn converging to some p ∈ X, and by the continuity
of f (Step III), the sequence {f (xnk )} in E converges to f (p) by [Ru, Thm. 4.2]. But
any subsequence of a convergent sequence has the same limit as the sequence, whence
f (p) = lim f (xn ) = inf E. Thus, f (x) ≥ f (p) for all x ∈ X, as desired.
Step V: There are finitely many points x1 , . . . , xn ∈ X such that X = Nr0 (x1 )∪
· · · ∪ Nr0 (xn ), where r0 is as in Step IV.
Indeed, if this were not true, we could pick an x1 ∈ X, an x2 ∈ X \ Nr0 (x1 ), and
continuing this way, having picked x1 , . . . , xn−1 we could pick
xn ∈ X \ Nr0 (x1 ) ∪ · · · ∪ Nr0 (xn−1 ) ,
and the process would never stop. Hence, for all m, n, we would have d(xm , xn ) ≥ r0 , so
that no subsequence of {xn } is Cauchy, whence by [Ru, Thm. 3.11], no subsequence of
{xn } converges, a contradiction to assumption (iii).
Step VI: Conclusion of the proof.
By definition of f and of r0 as its minimal value, for each of the points x1 , . . . , xn ∈ X
of the previous step, there is an index αi such that Nr0 (xi ) ⊂ Uαi . Hence,
X = Nr0 (x1 ) ∪ · · · ∪ Nr0 (xn ) ⊂ Uα1 ∪ · · · ∪ Uαn ,
and we have found a finite subcover, proving that X is compact.
We note that we also proved the following, which is used in [Ru, pf. of Thm. 7.25]:
Proposition 1.2. Any compact metric space X contains an at most countable dense
subset.
Proof. In Step I of alternative 1 in the last proof, we proved that condition (iii) in
Theorem 1.1 implies the existence of an at most countable dense subset.
We conclude this section with the following easy result, left out in [Ru] (but proved
in the lectures), which will be used in the proof of Theorem 2.7 below.
Proposition 1.3. A compact metric space is bounded.
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ANDREAS LEOPOLD KNUTSEN
Proof. Pick any x ∈ X (if X is empty, there is nothing to prove). Since any point in X
lies within finite distance from x, the collection of neighborhoods
{Nn (x)}n∈Z+
of X is an open cover of X. Since X is compact, there must exist a finite subcover,
which means that there is an M ∈ Z+ so that
X = ∪M
i=1 Ni (x) = NM (x).
This implies that d(x, y) < M for all y ∈ X, whence X is by definition bounded (see
[Ru, Def. 2.18(i)]).
2. Spaces of functions and the Arzelà-Ascoli theorem
There are several variants of this theorem, introduced at around the same time by the
two Italian mathematicians Cesare Arzelà (1847-1912) and Giulio Ascoli (1843-1896)
around 1883.
One weaker version of this theorem is [Ru, Thm. 7.25]. Another version is the one
giving necessary and sufficient conditions for a subspace of C(X) to be compact when X
is a compact metric space, given in [Ru, Exc. 19, p. 168]. The aim of this section is to
give a proof of this result, in Theorem 2.7 below, as well as summarizing some important
results from [Ru].
We recall that, for any metric space X, one defines the set
C(X) := {f : X → C | f is continuous and bounded on X}
(and the condition of being bounded is redundant when X is compact, by [Ru, Thm.
4.15]) and makes this set a metric space by the metric
d(f, g) := ||f − g|| := sup |f (x) − g(x)|,
x∈X
called the supremum norm metric.
Following [Ru], we want to describe:
• convergent sequences in C(X);
• closed subsets of C(X);
• bounded subsets of C(X);
• compact subsets of C(X).
The first three are taken care of in [Ru], although not as explicitly as one might wish,
so we will summarize the results here. The last item, concerning compact subsets, is
taken care of by the Arzelà-Ascoli theorem 2.7 below (but only for compact X).
We recall the following important result, mentioned briefly on [Ru, p. 151]:
Proposition 2.1. A sequence in C(X) converges (with respect to the supremum norm
metric) if and only if it converges uniformly as a sequence of functions on X.
Proof. By [Ru, Thm. 7.9], a sequence of functions {fn : X → C} converges uniformly to
a function f : X → C if and only if
(6)
sup |fn (x) − f (x)| −→ 0 as n → ∞,
x∈X
which by definition of the supremum norm metric is equivalent to saying that
||fn − f || −→ 0 as n → ∞.
Therefore, the “only if” part follows, while the “if” part follows once we prove that
f ∈ C(X) if all fn ∈ C(X), that is, if all fn are continuous and bounded, then so is the
limit function f .
COMPACT SETS AND THE ARZELÀ-ASCOLI THEOREM
5
The fact that f is continuous follows from [Ru, Thm. 7.12]. The fact that it is bounded
follows as there by (6) is an N such that
sup |fN (x) − f (x)| ≤ 1
x∈X
and, as fN is bounded on X, an M ∈ R so that
|fN (x)| ≤ M for all x ∈ X.
Hence
|f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ 1 + M for all x ∈ X,
and f is bounded, as desired.
This result makes it possible to describe closed subsets of C(X) explicitly. Indeed, we
have the following general result valid on any metric space X:
Lemma 2.2. Let E be a subset of a metric space X. Then its closure E (recall [Ru,
Def. 2.26]) can be described as
(7)
E = {p ∈ X | p is the limit of a sequence in E}.
In particular, E is closed if and only if it contains all limits of sequences in E.
Proof. We first prove the containment “⊂” in (7).
If p ∈ E, then p ∈ E or p is a limit point of E. If p ∈ E, it is the limit of the constant
sequence {p, p, p, . . .} in E. If p is a limit point of E, then by [Ru, Thm. 3.2(d)], there
exists a sequence in E converging to p.
We then prove the containment “⊃” in (7).
If p ∈ X is the limit of a sequence {pn } in E, then by definition of convergence, for
any > 0, the neighborhood N (p) contains all but finitely many of the terms of {pn }.
In particular, N (p) ∩ E 6= ∅, so that either p ∈ E or p is a limit point of E by definition.
Hence p ∈ E.
Since E = E if and only if E is closed, by [Ru, Thm. 2.27(b)], the last assertion
follows.
We recall the following definitions (cf. [Ru, Def. 7.28], where the definition is made
only for algebras of functions):
Definition 2.3. A family of complex functions F defined on a set X is uniformly closed
if it contains all complex functions f satisfying the property that there is a sequence
{fn } of functions in F converging uniformly to f .
The uniform closure of F is the set of complex functions f satisfying the property
that there is a sequence {fn } of functions in F converging uniformly to f .
Therefore, an immediate consequence of Proposition 2.1 and Lemma 2.2 is the following.
Proposition 2.4. A subset S ⊂ C(X) is closed (with respect to the supremum norm
metric in C(K)) if and only if it is uniformly closed as a family of functions on X.
The closure of S equals its uniform closure as a family of functions.
To describe bounded subsets of C(X), we recall the following definitions (cf. [Ru, Def.
7.19], where the definition is made only for countable sets, that is, for sequences):
Definition 2.5. A family of complex functions F defined on a set X is pointwise bounded
if the set {f (x)}f ∈F is bounded in C for every x ∈ X, that is, for every x ∈ X, there is
an M (x) ∈ R (depending on x) such that
|f (x)| ≤ M (x) for all f ∈ F.
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ANDREAS LEOPOLD KNUTSEN
A family of complex functions F defined on a set X is uniformly bounded if the set
{f (x)}x∈X,f ∈F is bounded in C, that is, there is an M ∈ R such that
(8)
|f (x)| ≤ M for all x ∈ X and f ∈ F.
In particular, a uniformly bounded set of functions is automatically pointwise bounded.
We have:
Proposition 2.6. A subset S ⊂ C(X) is bounded (with respect to the supremum norm
metric in C(K)) if and only if it is uniformly bounded as a family of functions on X.
Proof. We first prove the “only if” part.
Assume that S is bounded, with respect to the supremum norm metric in C(K). By
definition (see [Ru, Def. 2.18(i)]), there is a g ∈ C(X), and an M1 ∈ R, such that
||f − g|| ≤ M1 , for all f ∈ S,
whence by definition of the metric,
(9)
|f (x) − g(x)| ≤ M1 , for all f ∈ S and all x ∈ X.
Since g ∈ C(X), it is bounded, so there is an M2 ∈ R such that |g(x)| ≤ M2 for all
x ∈ X. It follows that |f (x)| ≤ |f (x) − g(x)| + |g(x)| ≤ M1 + M2 for all x ∈ X, so that
(8) is satisfied for S = F, with M = M1 + M2 .
We next prove the “if” part.
Assume that there is an M ∈ R such that (8) is satisfied. Then one has
|f (x) − g(x)| ≤ |f (x)| + |g(x)| ≤ 2M, for all f, g ∈ S and all x ∈ X,
whence
||f − g|| := sup |f (x) − g(x)| ≤ 2M,
x∈X
so that S is bounded (see again [Ru, Def. 2.18(i)]).
Finally, we are ready to prove the main result (cf. [Ru, Exc. 19, p. 168]). We refer to
[Ru, Def. 7.22] for the definition of equicontinuity.
Theorem 2.7 (Arzelà-Ascoli). Let K be a compact metric space and S ⊂ C(K).
Then S is compact if and only if it is uniformly closed, pointwise bounded and equicontinuous.
Proof. We first prove the “only if” part.
Any compact subset of a metric space is closed ([Ru, Thm. 2.34]) and bounded
(Proposition 1.3). Hence, if S is compact, it is closed and bounded with respect to the
metric in C(K), that is, uniformly closed and uniformly bounded by Propositions 2.4 and
2.6, in particular pointwise bounded. We have left to prove that S is equicontinuous if it
is compact.
If S were not equicontinuous, there would exist an > 0 such that for all δ > 0, there
exist x, y ∈ K and f ∈ S satisfying
d(x, y) < δ and |f (x) − f (y)| ≥ .
In particular, setting δ = 1/n, we find for each n ∈ Z+ , points xn , yn ∈ K and a function
fn ∈ S such that
d(xn , yn ) < 1/n and |fn (xn ) − fn (yn )| ≥ .
This shows that no subsequence of {fn } is equicontinuous. By [Ru, Thm. 7.24], no
subsequence of {fn } converges uniformly on K, meaning that no subsequence converges
in C(K) with respect to the metric on C(K). But then S cannot be compact, by Theorem
1.1, a contradiction.
COMPACT SETS AND THE ARZELÀ-ASCOLI THEOREM
7
We next prove the “if” part and will again use the characterization of compact spaces
given in Theorem 1.1(iii).
Assume therefore that S is uniformly closed, pointwise bounded and equicontinuous and {fn } is a sequence (of functions) in S. Since S is pointwise bounded and
equicontinuous, so is the sequence {fn }, whence by [Ru, Thm. 7.25], {fn } contains a
subsequence {fnk } that converges uniformly on K. By Proposition 2.1, this means that
{fnk } converges as a sequence in C(K) (with respect to the metric in C(K)). Since S is
by assumption uniformly closed, the limit function f lies in S, whence {fn } contains a
convergent subsequence in S. Therefore, S is compact by Theorem 1.1.
References
[Li]
[Ru]
T. Lindstrøm, Mathematical Analysis, notes used in the course MAT2400 at the University of
Oslo.
W. Rudin, Principles of mathematical analysis, International Series in Pure and Applied Mathematics, Third Edition, McGraw-Hill, 1976.
Andreas Leopold Knutsen, Department of Mathematics, University of Bergen, Postboks 7800, 5020 Bergen, Norway
E-mail address: [email protected]