## Topology: Sequentially Compact Spaces and Compact Spaces

We’ve arrived at possibly the most confusing notion in topology/analysis. First, we wish to fulfil an earlier promise: to prove that if C is a closed and bounded subset of R and fR → R is continuous, then f(C) is closed and bounded. [ As a corollary, if f : R → R is continuous, then the image f([a, b]) is closed and bounded and thus attains a minimum and maximum. We saw that this has huge implications in optimisation problems. ]

To do that, we define the following.

Definition. A topological space X is said to be sequentially compact if every sequence has a convergent subsequence.

Thus, a sequentially compact space is one which is so constrained that an infinite sequence must have infinitely many terms clustering around a point.

For example, R is not sequentially compact since xn clearly has no convergent subsequence. Also, X = (0, 1) is not sequentially compact since x= 1/n has no subsequence which converges in X. In the former case, problem occurs because the points keep getting further. In the latter case, the sequence $x_n = 1/n$ really ought to converge to 0, but that limit is not found in the space.

This observation inspires the following result.

Proposition 1. Let (X, d) be a metric space.

• If X is sequentially compact, then it’s bounded.
• If Y is a sequentially compact subspace of X, then Y is closed in X.

Proof.

Suppose X is not bounded; fix $x\in X.$

• We claim that the collection of d(xy), for y in X, is not bounded; indeed, if it were bounded by B, any y and z in X would give d(yz) ≤ d(yx) + d(xz) ≤ 2B, which is a contradiction.
• Pick $x_n \in X, d(x, x_n) > n.$
• The resulting sequence is not even Cauchy since for any n, the set of $d(x_m, x_n)$ for various m is not bounded, so there exists m>n for which $d(x_m, x_n) >1.$

For the second statement, by theorem 1 here, we only need to show any sequence $(x_n)$ in Y converging to $a\in X$ must satisfy $a\in Y.$ But by sequential compactness of Y we must have a subsequence $(x_{n_k})$ which converges to an element of Y. Thus $a\in Y$ indeed. ♦

Exercise

Attempt to prove the second statement of proposition 1 for a topological space X. What additional assumption do you need? [ Answer: X has to be Hausdorff. ]

The next result tells us how to construct new sequentially compact spaces out of old ones.

Proposition 2. Let X and Y be topological spaces.

• If f:X→Y is continuous and X is sequentially compact, then so is f(X).
• X and Y are sequentially compact if and only if X × Y is.
• If X is sequentially compact and $Y\subseteq X$ is closed, then Y is sequentially compact.

Proof.

• Let (yn) be any sequence in f(X). Write yn = f(xn) for xn in X. By sequential compactness of X, there’s a converging subsequence $(x_{n_k})\to a.$ Then $(f(x_{n_k}))\to f(a)$ is also a convergent subsequence of (yn).
• Suppose X and Y are sequentially compact. If (xn, yn) is a sequence in X × Y, then there is a subsequence $(x_{n_k})$ of (xn) which converges to a in X. In the sequence $(y_{n_k}),$ there’s also a subsequence $(y_{n_j})$ which converges to b in Y. Since $(x_{n_j})$ is a subsequence of $(x_{n_k}),$ it also converges to a. Thus $(x_{n_j}, y_{n_j}) \to (a,b).$ To prove the converse apply the property above to projection maps X × Y → X and X × Y → Y.
• Let (yn) be a sequence in Y. By sequential compactness of X, it has subsequence which converges to some a in X. Since Y is closed in X, theorem 2 here tells us a lies in Y. ♦

It turns out for R, the converse to proposition 1 is true.

Theorem 3. If X is a closed and bounded subspace of Rn, then X is sequentially compact.

Proof (Sketchy)

Via scaling, one may assume $X\subseteq [0, \frac 2 3]^n$, in which case proposition 2 tells us we only need to show $[0, \frac 2 3]^n$ is sequentially compact and for that, it suffices to show X = [0, 2/3] is sequentially compact. For a sequence (xn) in X, write out the decimal expansion of each xn; if more than one expression exists, pick any one. Each xn may be written in the form (0.abcd…).

For each xn, take the first digit after the decimal point; since it has only 10 possibilities, one of them (0.a…) must occur infinitely many times. Pick an infinite subsequence with first digit = a. Repeat the argument with this new sequence and find another subsequence with the same first two decimal digits (0.ab…). Iteratively, at the k-th step, we get an infinite subsequence with the same first k decimal digits $(0.d_1 d_2 \ldots d_k).$ This gives $r = (0.d_1 d_2 d_3\ldots) \in [0, \frac 2 3].$ Clearly, there’s a subsequence of (xn) converging to r. Since xn ≤ 2/3, we also have r ≤ 2/3. ♦

Together with propositions 1 and 2, we get:

Corollary 4. A subset X of R is sequentially compact iff it’s bounded, and closed in R.

If X is a closed and bounded subset of R, and f : R → R is continuous, then f(X) is also closed and bounded.

## Compact Spaces

We shall prove that for metric spaces, sequential compactness is equivalent to another topological notion.

Definition. A topological space X is said to be compact if

• whenever {Ui} is a collection of open subsets of X satisfying $X = \cup_i U_i,$ we can find a finite set of indices i1, i2, … in, such that $X = \cup_{k=1}^n U_{i_k}.$

Thus, every open cover of X has a finite subcover, where by “open cover of X”, one means a collection of open subsets of X whose union equals X.

Even though “sequential compactness” and “compactness” are both topological concepts which happen to be equivalent for metric spaces, they are not equivalent for general topological spaces. Indeed, the problem lies with our usage of sequences in sequential compactness. If we replace sequences with nets, the result will be true.

Big Theorem. A metric space (X, d) is sequentially compact if and only if it’s compact.

Step 1 : Prove that compact implies sequentially compact.

Suppose $(x_n)$ is a sequence of elements in a compact metric space (Xd) which has no convergent subsequence. Fix x in X; there’s an open ball N(x, ε(x)) which contains only finitely many terms of the sequence. [ Otherwise, each N(x, ε) contains infinitely many terms of the sequence. Thus we can pick n1 < n< n3 < … such that $d(x_{n_k}, x) < \frac 1 k$ and $(x_{n_k}) \to x.$ ]

Now the collection of all these open balls is an open cover for X so it has a finite subcover $N(x_i, \epsilon(x_i)),$ which contradicts the fact that each $N(x_i, \epsilon(x_i))$ contains only finitely many terms of the sequence.

Step 2 : If X is sequentially compact and {Ui} is an open cover of X, then there is an ε>0 such that any open ball N(x, ε) is contained in some Ui.

If not, then for ε = 1/n (n = 1, 2, …), some open ball N(xn, 1/n) is not contained in any Ui. By sequential compactness, we can find a convergent subsequence $(x_{n_k}) \to a.$ Replacing (xn) with $(x_{n_k}),$ we may assume that (xn) → a and N(xn, 1/n) is not contained in any Ui (since 1/nk < 1/n).

Now this a lies in some Ui, so N(a, 1/m) lies in Ui for some m. Since (xn) → a, there’s an index N such that if n > N then d(a, xn) < 1/(2m). If we pick n > max(N, 2m), then d(axn) < 1/(2m) gives:

$N(x_n, \frac 1 n) \subseteq N(x_n, \frac 1{2m}) \subseteq N(a, \frac 1 m) \subseteq U_i.$

This contradicts our choice of (xn).

Definition. This ε>0 is called a Lebesgue number of the open cover {Ui}.

Step 3 : If X is sequentially compact, then for any ε>0, we can cover X with finitely many open balls N(xε).

Suppose this is not true for some ε>0. Let’s construct a sequence (xn) such that any two distinct terms satisfy d(xn, xm) ≥ ε. Suppose we already have {x1x2, …, xn}. To find the next term, since the union of N(xi, ε) is not the whole of X, we can still find xn+1 outside this union. Thus, d(xn+1xi) ≥ ε for i = 1, 2, …, n. Such a sequence has no convergent subsequence.

Definition. A metric space X in which for any ε>0, X can be covered by finitely many N(x, ε), is said to be totally bounded.

Step 4 : Prove that sequentially compact implies compact.

If {Ui} is an open cover of X, by step 2, pick ε>0 such that any open ball N(x, ε) is contained in some Ui. By step 3, the whole of X can be covered by finitely many N(xi, ε), i = 1, …, n. Each N(xi, ε) is contained in some Ui; and so X can be covered by finitely many Ui. ♦

In the process, we’ve also proved the following:

Corollary. If (X, d) is a compact metric space, then X is totally bounded and has a Lebesgue number ε>0.

## Replacing Sequences with Nets

If we replace the condition “every sequence has a convergent subsequence” with “every net has a subnet”, then this is indeed equivalent to the notion of compactness for arbitrary topological spaces.

Before we state the result, though, we should be clear what we mean by “subnets”. In the case of sequences, we don’t want people to cheat by just taking the first 10 terms. Likewise, in the case of subnets, we force them to contain terms which are indexed by arbitrarily large indices.

Definition. If $(x_i)_{i\in I}$ is a net indexed by directed set I, then a subnet is given by $(x_j)_{j\in J}$ for some subset J of I such that:

• for any $i\in I$, there exists an index j ≥ i contained in J (note that this condition implies J is a directed set).

Now we’re ready to state our theorem.

Theorem. A topological space X is compact if and only if every net has a convergent subnet.

Forward Implication.

Suppose X is compact and $(x_i)_{i\in I}$ is a net indexed by directed set I. Assume it has no convergent subnet, so every $a\in X$ is not a limit of some subnet. Thus there’s an open subset U(a) containing a, and an index i, such that U(a) does not contain xj for all j ≥ i. [ If not, for each open subset U containing a and index i, there’s a j ≥ i such that $x_j\in U.$ This gives a subnet (xj) which converges to a. Contradiction. ]

Since $X = \cup_{a\in X} U(a)$ we can find a finite subset {a1a2, …, an} of X such that $X = \cup_{k=1}^n U(a_k).$ Also for each k=1, …, n, there’s an index ik such that U(ak) does not contain xfor all j ≥ ik. Since I is a directed set, we’ll pick j ≥ all ik‘s, and so $X = \cup_{k=1}^n U(a_k)$ does not contain xj which is absurd.

Backward Implication

Suppose every net in X has a convergent subnet and $\{U_j\}_{j\in J}$ is an open cover of X indexed by J. We construct a net as follows: the index set is I, the collection of all finite subsets of J, ordered by inclusion. Thus, if $S, T\in I,$ then $S\le T \iff S\subseteq T.$ For each $S\in I,$ define the open subset $U_S := \cup_{j\in S} U_j.$ If {Uj} has no finite subcover, then each US ≠ X so we can pick $x_S\in X - U_S.$ This gives a net (xS).

By assumption, (xS) has a convergent subnet (xT) → a. Now a lies in some Uj; there exists an index T’ such that for all T ≥ T’, $x_T\in U_j.$ If we pick T containing j, then $x_T \not\in U_T \supseteq U_j$ so $x_T\not\in U_j$ which is a contradiction. ♦

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### 4 Responses to Topology: Sequentially Compact Spaces and Compact Spaces

1. Shahid Naqvi says:

thanks for an excellent article. I have some things not clear to me. The union of open covers of X seems to cover more than X in the figure below the definion of compact spaces. So how is the union equal to X? If X is replaced by X minus its boundary, how is it not compact? The open circles still seem to cover X-Bd.X? Thanks.

• limsup says:

Hi. The diagrams are really meant for illustrating the proof – actual topological spaces can be much weirder than subsets of the Euclidean plane.

You can imagine the open subsets as the intersection of the circles and X, instead of the circles themselves.

2. Priscilla Pacifica says:

Excellent article. The diagrammatic explanations are very much understandable. Thank you