This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
We first define the following.
Definition 1 A topological space \(X\) is limit point compact if every infinite subset of \(X\) has a limit point. (§Interior, Closure, and Boundary of Sets, ⁋Definition 8)
In general, the following holds.
Proposition 2 Every compact space is limit point compact.
Proof
Assume for contradiction that \(X\) is a compact space that is not limit point compact. Then there exists an infinite subset \(A\) with no limit point, so by the argument after §Interior, Closure, and Boundary of Sets, ⁋Definition 8 we must have \(\cl(A)\setminus A=\emptyset\). That is, \(A\) must be closed, and hence compact. Meanwhile, 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 sequentially compact if every sequence 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 assume there exists an infinite subset \(A\) with no limit point. Then we can choose a suitable countable subset \(A'\) of \(A\) and make it into a sequence \((x_n)_{n\geq k}\). Since \(X\) is sequentially compact, this sequence has a convergent subsequence; if this subsequence converges to \(x\), one can verify that \(x\) becomes a limit point of \(A'\) and therefore a limit point of \(A\).
Convergence of Sequences
Meanwhile, the converse of Proposition 5 also does not hold. This may seem somewhat surprising, because given an arbitrary sequence \((x_n)\) in a limit point compact space, 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 \(A\).
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 Topological Spaces, ⁋Corollary 6 and therefore defines a topology on \(\mathbb{R}\).
Define a sequence \((x_n)_{n\geq 1}\) in this topological space \(\mathbb{R}\) by the formula
\[x_n=-n\]Then \((x_n)\) has no convergent subsequence. On the other hand, \(A=\{x_n\mid n\geq 1\}\) has a limit point; for example, \(-2\) is a limit point of \(A\). This is because any open set containing \(-2\) must also contain \(-1\in A\).
The above example shows that convergence of sequences is not such a good concept for checking limit points. Meanwhile, by §Interior, Closure, and Boundary of Sets, ⁋Proposition 6, any limit point of \(A\) belongs to the closure of \(A\).
Lemma 7 For a topological space \(X\) and any subset \(A\subseteq X\), if there exists a sequence in \(A\) converging to \(x\in X\), then \(x\in \cl(A)\).
Proof
Pick 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 \(x_n\in U\) whenever \(n\geq N\). 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 \(A\), 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 obvious that \(\scl(A)\subseteq \cl(A)\).
Example 8 Consider the uncountable product \(\mathbb{R}^J\) with the product topology. Define the set \(A\) by
\[A=\{(x_j)\in \mathbb{R}^J: x_j=1\text{ for all but finitely many $j$}\}\]Then the origin of \(\mathbb{R}^J\) belongs to the closure of \(A\). This is because a basic open set in \(\mathbb{R}^J\) containing the origin is \(\mathbb{R}\) in all but finitely many indices, and the point whose coordinates at these finitely many indices are \(0\) and whose coordinates at the remaining indices are \(1\) lies in the intersection of this base element and \(A\). However, no sequence in \(A\) converges to the origin. This is because, given any sequence in \(A\), using the fact that \(J\) is uncountable we can show that there exists \(j\in J\) such that the \(j\)-th coordinate of every term in the sequence is \(1\), and then the open neighborhood of the origin whose \(j\)-th coordinate is \((-1,1)\) and whose other coordinates are \(\mathbb{R}\) contains no element of this sequence.
That is, the converse of Lemma 7 also does not hold 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 \(A\), we call \(X\) a sequential space.
Meanwhile, the following proposition, though slightly generalized, is still familiar.
Proposition 9 For a continuous function \(f:X \rightarrow Y\) and any sequence \((x_n)\) in \(X\), if \((x_n)\) converges to \(x\in X\), then \((f(x_n))\) converges to \(f(x)\).
Proof
Pick 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 \(x_n\in f^{-1}(V)\) whenever \(n\geq N\). Then \(f(x_n)\in V\), so \((f(x_n))\) converges to \(f(x)\).
Meanwhile, if the converse of Lemma 7 holds in the space \(X\), we can also show the converse of Proposition 9 using this result together with the second condition of §Continuous Functions, ⁋Theorem 4. That is, if for any sequence \((x_n)\) converging to any point \(x\), \(f(x_n)\) converges to \(f(x)\), then \(f\) is continuous at \(x\).
Let \(X\) be a space in which the converse of Lemma 7 holds. Then for any \(x\in \cl(A)\), we can pick a sequence \((x_n)\) in \(X\) converging to \(x\). Then the sequence \(f(x_n)\) in \(Y\) converges to \(f(x)\), so by Lemma 7 we have \(f(x)\in \cl(f(A))\), and we obtain the desired result from §Continuous Functions, ⁋Theorem 4.
Countability Axioms
The above shows that convergence of sequences is not an adequate concept for expressing the notions we have dealt with so far. Looking at the proof of Proposition 11, we can see in what way it is best to generalize this.
Definition 10 We define the following for a topological space \(X\).
- \(X\) is first countable if for every point \(x\in X\), there exists a countable local base at \(x\).
- \(X\) is second countable if there exists a countable base for \(X\).
Proposition 11 If \(X\) is first countable, \(T_1\), 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 the convergence of the sequence, we may assume without loss of generality that \(x_n\neq x\) for all $$
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