Compactness and Convergence of Filters

We first define the following.

Definition 1 A topological space \(X\) is called limit point compact if every infinite subset of \(X\) has a limit point. (§Interior, Closure, and Boundary, ⁋Definition 8)

In general, the following holds.

Proposition 2 Every compact space is limit point compact.

Proof

Suppose, for the sake of contradiction, that \(X\) is a compact space that is not limit point compact. Then there exists an infinite subset \(A\) having no limit point, and therefore by the argument after §Interior, Closure, and Boundary, ⁋Definition 8 we must have \(\cl(A)\setminus A=\emptyset\). That is, \(A\) must be closed, and hence compact. On the other hand, since each \(a\in A\) is also not a limit point of \(A\), there exists a suitable open neighborhood \(U_a\) of \(a\) such that \(A\cap U_a=\{a\}\). Then \((U_a)_{a\in A}\) is an open cover of \(A\) with no finite subcover, which is a contradiction.

However, the converse does not hold in general.

Example 3 Consider a two-point space \(X\) with the trivial topology, and an infinite set \(Y\) with the discrete topology. Then \(X\times Y\) is limit point compact but not compact.

We also define the following.

Definition 4 A topological space \(X\) is called sequentially compact if every sequence of points in \(X\) has a convergent subsequence.

Proposition 5 Every sequentially compact space is limit point compact.

Proof

Let \(X\) be a sequentially compact space, and suppose there exists an infinite subset \(A\) having no limit point. Then we can choose a suitable countable subset \(A'\) of \(A\) and regard it as a sequence \((x_n)_{n\geq k}\). Since \(X\) is sequentially compact, this sequence has a convergent subsequence; if this subsequence converges to \(x\), then one can verify that \(x\) is a limit point of \(A'\), and hence of \(A\).

Convergence of Sequences

Meanwhile, the converse of Proposition 5 also fails. This may seem somewhat surprising, because if an arbitrary sequence \((x_n)\) in a limit point compact space is given, the set \(A=\{x_n\mid n\geq 1\}\) is either finite, in which case it trivially has a convergent subsequence, or infinite, in which case it has a limit point. The problem is that there may not exist a subsequence converging to a limit point of the set \(A\).

Example 6 Consider the collection of subsets of \(\mathbb{R}\)

\[\mathcal{B}=\{(a,\infty)\mid a\in \mathbb{R}\}\]

This collection satisfies the conditions of §Bases of a Topological Space, ⁋Corollary 6, and thus defines a topology on \(\mathbb{R}\).

For the topological space \(\mathbb{R}\) defined in this way, define a sequence \((x_n)_{n\geq 1}\) by the formula

\[x_n=-n\]

Then \((x_n)\) has no convergent subsequence. On the other hand, the set \(A=\{x_n\mid n\geq 1\}\) does have a limit point; for instance, \(-2\) is a limit point of \(A\), because any open set containing \(-2\) must also contain \(-1\in A\).

The above example shows that convergence of sequences is not a very good notion for detecting limit points. On the other hand, by §Interior, Closure, and Boundary, ⁋Proposition 6, any limit point of a set \(A\) belongs to the closure of \(A\).

Lemma 7 Let \(X\) be a topological space and \(A\subseteq X\) an arbitrary subset. If there exists a sequence in \(A\) converging to \(x\in X\), then \(x\in \cl(A)\).

Proof

Choose an arbitrary open neighborhood \(U\) of \(x\). Since there exists a sequence \((x_n)\) converging to \(x\), there exists a suitable \(N\in \mathbb{N}\) such that \(x_n\in U\) for all \(n\geq N\). Then \(x_N\in U\cap A\), so \(U\cap A\neq \emptyset\), and therefore \(x\in \cl(A)\).

Therefore, for a topological space \(X\) and a subset \(A\), if we define the sequential closure \(\scl(A)\) of \(A\) by

\[\scl(A)=\{x\in X\mid \text{there exists a sequence in $A$ that converges to $x$}\}\]

then it is clear that \(\scl(A)\subseteq \cl(A)\).

Example 8 Consider the uncountable product \(\mathbb{R}^J\) of copies of \(\mathbb{R}\), equipped with the product topology. Define the set \(A\) by

\[A=\{(x_j)\in \mathbb{R}^J\mid x_j=1\text{ for all but finitely many $j$}\}\]

Then the origin of \(\mathbb{R}^J\) belongs to the closure of \(A\). This is because any basic open set containing the origin has all but finitely many coordinates equal to \(\mathbb{R}\), and by setting those finitely many coordinates to \(0\) and the remaining coordinates to \(1\), we obtain a point lying in the intersection of this basic open set with \(A\). However, no sequence in \(A\) converges to the origin. Indeed, given any sequence in \(A\), since \(J\) is uncountable one can show that there exists \(j\in J\) such that the \(j\)th coordinate of every term of the sequence is \(1\); then the open neighborhood of the origin whose \(j\)th coordinate is \((-1,1)\) and whose remaining coordinates are \(\mathbb{R}\) contains no term of the sequence.

That is, the converse of Lemma 7 also fails in general. Or, using the language above, for a topological space \(X\) and a subset \(A\) we have \(\scl(A)\subsetneq \cl(A)\) in general. If \(\scl(A)=\cl(A)\) holds for every subset \(A\), then \(X\) is called a sequential space.

Meanwhile, the following proposition, although slightly generalized, is still familiar.

Proposition 9 Let \(f:X \rightarrow Y\) be a continuous function and \((x_n)\) an arbitrary sequence in \(X\). If \((x_n)\) converges to \(x\in X\), then \((f(x_n))\) converges to \(f(x)\).

Proof

Choose an arbitrary open neighborhood \(V\) of \(f(x)\). Since \(f\) is continuous, \(f^{-1}(V)\) is an open neighborhood of \(x\). Hence there exists a suitable \(N\in \mathbb{N}\) such that \(x_n\in f^{-1}(V)\) for all \(n\geq N\). Then \(f(x_n)\in V\), so \((f(x_n))\) converges to \(f(x)\).

On the other hand, if the converse of Lemma 7 holds in the space \(X\), then using that result together with the second condition of §Continuous Functions, ⁋Theorem 4, we can also prove the converse of Proposition 9. That is, if for every sequence \((x_n)\) converging to any \(x\in A\), the sequence \(f(x_n)\) converges to \(f(x)\), then \(f\) is continuous at the point \(x\).

Suppose \(X\) is a space in which the converse of Lemma 7 holds. Then for any \(x\in \cl(A)\) we can choose a sequence \((x_n)\) in \(X\) converging to \(x\). Since the sequence \(f(x_n)\) in \(Y\) converges to \(f(x)\), by Lemma 7 we have \(f(x)\in \cl(f(A))\), and the desired result follows from §Continuous Functions, ⁋Theorem 4.

Countability Axioms

The above contents show that convergence of sequences is not an appropriate notion for expressing the concepts we have dealt with so far. Looking at the proof of Proposition 11, one can see in what direction it is good to generalize this.

Definition 10 For a topological space \(X\), we define the following.

1. \(X\) is called first countable if for every point \(x\in X\) there exists a countable local base at \(x\).
2. \(X\) is called second countable if there exists a countable base for \(X\).

Proposition 11 If \(X\) is first countable and \(T_1\), and limit point compact, then \(X\) is sequentially compact.

Proof

As mentioned before, given an arbitrary sequence \((x_n)\), the set \(A=\{x_n\mid n\geq 1\}\) is either finite, in which case it trivially has a convergent subsequence, or infinite, in which case it has a limit point \(x\). If \(x=x_n\) holds for infinitely many \(n\), then again for trivial reasons we can extract a subsequence converging to \(x\); thus we may assume that there are only finitely many \(n\) satisfying \(x_n=x\), and since this does not affect convergence of the sequence we may assume without loss of generality that \(x_n\neq x\) for all \(n\).

Since \(X\) is first countable, we can consider a countable local base \(\mathcal{B}(x)\) at \(x\). Writing the elements of \(\mathcal{B}(x)\) as \(B_1,B_2,\ldots\), we may replace \(B_n\) by \(B_1\cap\cdots\cap B_n\) so that \(B_{n+1}\subseteq B_n\) holds for all \(n\).

Now \(B_1\) is an open set containing \(x\), and since \(x\) is a limit point of \(A\), there exists a suitable \(n_1\) such that \(x_{n_1}\in B_1\). Since \(X\) is \(T_1\), there exists an open set \(U_2\) containing \(x\) but not containing \(x_1,\ldots,x_{n_1}\). Then \(U_2\cap B_2\) is again an open set containing \(x\), and since \(x\) is a limit point of \(A\), there exists a suitable \(n_2\) such that \(x_{n_2}\in U_2\cap B_2\). Repeating this process, we can extract a subsequence of \(A\) converging to \(x\).

Moreover, one can easily show that every first countable space is a sequential space.

Proposition 12 Every second countable space \(X\) is Lindelöf. That is, every open cover of \(X\) has a countable subcover.

Proof

Fix a countable base \(\mathcal{B}=\{B_1, B_2, \ldots\}\) of \(X\), and let \((U_\alpha)_{\alpha\in A}\) be an arbitrary open cover of \(X\). For each point \(x\in X\) there exists \(\alpha\) with \(x\in U_\alpha\), and by the definition of a base there exists \(B_n\in\mathcal{B}\) such that \(x\in B_n\subseteq U_\alpha\). Collecting all such \(B_n\)’s, we obtain a countable subcollection \(\{B_{n_1}, B_{n_2}, \ldots\}\) of \(\mathcal{B}\) covering \(X\). Now for each \(k\) choose one \(\alpha_k\) with \(B_{n_k}\subseteq U_{\alpha_k}\); then \((U_{\alpha_k})_{k\geq 1}\) is a countable subcover of \((U_\alpha)_{\alpha\in A}\).

Convergence of Filters

What played the essential role in the proof of Proposition 11 was that \(x\) possesses a decreasing sequence of open sets

\[B_1\supseteq B_2\supseteq\cdots\]

and by first countability, for any given open set \(U\), one can arrange that \(B_n\subseteq U\) for all sufficiently large \(n\). That is, in a certain sense, the above open sets themselves can be regarded as converging to \(x\). Based on this observation, we define the following.

Definition 13 Consider a topological space \(X\) and a filter \(\mathcal{F}\) defined on it. (§Other Definitions of Topological Spaces, ⁋Definition 3) Then \(\mathcal{F}\) is said to converge to \(x\in X\) if \(\mathcal{N}(x)\subseteq \mathcal{F}\) holds. (§Open Sets, §§Neighborhood filter) In this case, \(x\) is called a limit point of \(\mathcal{F}\).

Definition 13 generalizes the convergence of sequences. To verify this, we first need to define the following.

Definition 14 Let a filter \(\mathcal{F}\) be defined on a set \(X\), and let \(Y\) be a topological space. For a function \(f:X \rightarrow Y\), we say that \(y\in Y\) is a limit point of \(f\) with respect to \(\mathcal{F}\) if \(y\) is a limit point of the filter \({\uparrow}f(\mathcal{F})\). (§Other Definitions of Topological Spaces, ⁋Proposition 7)

Then by definition, \(y\in Y\) being a limit point of \(f\) with respect to \(\mathcal{F}\) means that for every neighborhood \(V\) of \(y\), there exists \(F\in \mathcal{F}\) such that \(f(F)\subseteq V\). In particular, the following holds.

Proposition 15 For a sequence \((x_n)_{n\geq 1}\) in a topological space \(X\), the sequence \((x_n)_{n\geq 1}\) converging to \(x\in X\) is equivalent to the filter generated by the image of the Fréchet filter \(\mathcal{F}\) on \(\mathbb{N}\) under the map \(n\mapsto x_n\) converging to \(x\). (§Other Definitions of Topological Spaces, ⁋Example 4)

Thus we see that convergence of filters generalizes convergence of sequences. Moreover, through this notion we can prove the following proposition.

Proposition 16 For a topological space \(X\) and an arbitrary subset \(A\subseteq X\), the following are equivalent: \(x\in\cl(A)\), and there exists a filter \(\mathcal{F}\) on \(A\) converging to \(x\).

Proof

First, suppose there exists a filter \(\mathcal{F}\) on \(A\) converging to \(x\). That is, \(\mathcal{F}\) contains \(A\) as an element and contains \(\mathcal{N}(x)\) as a subset. Hence by the definition of a filter, for every neighborhood \(U\in \mathcal{N}(x)\) we have \(U\cap A\neq\emptyset\).

Conversely, suppose \(x\in \cl(A)\). Then for every neighborhood \(U\) of \(x\) we have \(U\cap A\neq\emptyset\), so the following expression

\[\mathcal{N}(x)\vert_A=\{U\cap A\mid U\in \mathcal{N}(x)\}\]

defines a filter base. Now letting \(\mathcal{F}\) be the filter defined by \(\mathcal{N}(x)\vert_A\), we obtain the desired result.

Therefore, by the argument after Lemma 7, the following proposition is also obvious.

Proposition 17 For a function \(f:X \rightarrow Y\) between topological spaces, \(f\) is continuous at \(x\in X\) if and only if for every filter \(\mathcal{F}\) converging to \(x\), the filter defined by the filter base \(f(\mathcal{F})\) converges to \(f(x)\).

Tychonoff theorem

The arbitrary product of compact spaces is again compact. If the product is finite, this can be shown in a more intuitive manner, but for an infinite product we need the following lemma. It generalizes Proposition 5 to the language of filters.

Lemma 18 A topological space \(X\) is compact if and only if every ultrafilter converges.

Proof

First, assume \(X\) is compact and let an arbitrary ultrafilter \(\mathcal{F}\) be given. Suppose for contradiction that \(\mathcal{F}\) has no limit point. Then for every \(x\in X\), there exists an open neighborhood \(U_x\) such that \(U_x\not\in \mathcal{F}\). By compactness of \(X\), there exists a finite subcover \(U_{x_1},\ldots, U_{x_n}\) of \(X\).

On the other hand, by [Set Theory] §Filters, Ideals, and Galois Connections, ⁋Proposition 5, \(\mathcal{F}\) is prime. That is, for any subset \(A\subseteq X\), exactly one of \(A\in \mathcal{F}\) or \(X\setminus A\in \mathcal{F}\) holds. Then for any \(A\in \mathcal{F}\),

\[A=A\cap X=(A\cap U_{x_1})\cup \cdots\cup (A\cap U_{x_n})\in \mathcal{F}\]

and since \(U_{x_i}\not\in \mathcal{F}\) by assumption, each \(A\cap U_{x_i}\) is also not in \(\mathcal{F}\); since \(\mathcal{F}\) is maximal, we must have \(X\setminus (A\cap U_{x_i})\in \mathcal{F}\). Then their finite intersection

\[X\setminus A=(X\setminus (A\cap U_{x_1}))\cap\cdots\cap (X\setminus (A\cap U_{x_n}))\]

must also belong to \(\mathcal{F}\), which contradicts the maximality of \(\mathcal{F}\).

Conversely, suppose that every ultrafilter \(\mathcal{F}\) has a limit point, and let \(\mathcal{A}\) be a family of closed subsets of \(X\) satisfying the finite intersection property. Then we can consider an ultrafilter \(\mathcal{F}\) containing the filter generated by \(\mathcal{A}\), and by assumption \(\mathcal{F}\) has a limit point \(x\). That is, \(\mathcal{N}(x)\subseteq \mathcal{F}\), and therefore for every \(F\in \mathcal{F}\) there exists a neighborhood \(U\) of \(x\) such that \(U\cap F\neq\emptyset\). In particular, for any \(A\in \mathcal{A}\) there exists a neighborhood \(U\) of \(x\) such that \(A\cap U\neq\emptyset\), and thus \(x\in \cl(A)=A\) always holds. From this we know that \(x\in\bigcap_{A\in \mathcal{A}}A\), and hence we obtain the desired result by §Compact Spaces, ⁋Proposition 11.

Then the following holds.

Theorem 19 (Tychonoff) The product \(X=\prod_{i\in I} X_i\) of compact spaces \((X_i)_{i\in I}\) is compact. Conversely, if the product space \(X\) is compact, then each \(X_i\) is compact.

Proof

If \(X\) is compact, then each \(X_i\) is compact by the continuity of \(\pr_i\) and §Compact Spaces, ⁋Proposition 8.

For the converse, given any ultrafilter \(\mathcal{F}\) on \(X\), we first verify that \(\pr_i(\mathcal{F})\) defines an ultrafilter base on \(X_i\); then from the assumption that \(X_i\) is compact and Lemma 18 we obtain a limit point \(x_i\) of this ultrafilter, and we can show that \(x=(x_i)_{i\in I}\) is a limit point of \(\mathcal{F}\). Thus the proof is complete by Lemma 18.

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References

[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.