Audit the Hypotheses Before Using the Argument
Many topological proof gaps arise when a valid method is applied outside the setting that justifies it. A sequence may not detect closure in a space without a countable neighborhood base. A compact set need not be closed unless the ambient space has a suitable separation property. And a finite subcover can be extracted only from an open cover of a compact space. The aim here is to check these hypotheses at the exact point where an argument needs them.
For a point \(x\) in a topological space \((X,\mathcal{T})\), an open neighborhood is an open set containing \(x\). A countable neighborhood base at \(x\) is a countable collection of open neighborhoods such that every open neighborhood of \(x\) contains at least one member of the collection. The space is first countable if every point has such a base. A space is compact if every open cover of the space has a finite subcover. A space is Hausdorff if every pair of distinct points has disjoint open neighborhoods.
When Can a Sequence Detect Closure?
In every topological space, if a sequence \((a_n)\) in \(A\) converges to \(x\), then \(x\in\overline{A}\). Indeed, every open neighborhood of \(x\) contains all sufficiently late terms, and hence meets \(A\). The converse requires a countable neighborhood base at \(x\). The following result makes that requirement precise.
Proof. If \(a_n\to x\) with each \(a_n\in A\), every open neighborhood of \(x\) contains a term of the sequence, so it meets \(A\). The neighborhood test for closure gives \(x\in\overline{A}\).
Conversely, suppose \(x\in\overline{A}\), and let \((U_n)_{n\in\mathbb{N}}\) be a countable neighborhood base at \(x\). Define \(V_n=U_1\cap\cdots\cap U_n\). Each \(V_n\) is an open neighborhood of \(x\), and the sets are decreasing: \(V_{n+1}\subseteq V_n\). They are still a neighborhood base, since if \(W\) is any open neighborhood of \(x\), some \(U_k\subseteq W\), and then \(V_k\subseteq U_k\subseteq W\). Since \(x\in\overline{A}\), every \(V_n\) meets \(A\). Choose \(a_n\in V_n\cap A\). Given any open neighborhood \(W\) of \(x\), choose \(k\) with \(V_k\subseteq W\). For every \(n\ge k\), \(a_n\in V_n\subseteq V_k\subseteq W\). Thus \(a_n\to x\). \(\square\)
Worked Example: Closure Without a Convergent Witness Sequence
Let \(X\) be an uncountable set with the cocountable topology: the open sets are \(\varnothing\) and all sets whose complements in \(X\) are countable. This is a topology: a nonempty union contains one of its nonempty open sets, so its complement is countable; a finite intersection of nonempty open sets has complement a finite union of countable sets, hence countable. Fix \(p\in X\) and set \(A=X\setminus\{p\}\).
Every open neighborhood \(U\) of \(p\) meets \(A\). Otherwise \(A\subseteq X\setminus U\), which would make the uncountable set \(A\) a subset of a countable set. Therefore \(p\in\overline{A}\).
Now take any sequence \((a_n)\) in \(A\). Its range \(R=\{a_n:n\in\mathbb{N}\}\) is countable. The set \(U=X\setminus R\) is open, contains \(p\), and contains no term \(a_n\). Thus the sequence does not converge to \(p\). In fact, \(p\) has no countable neighborhood base: if \((U_n)\) were one, each \(X\setminus U_n\) would be countable, so \(\bigcup_n(X\setminus U_n)\) would be countable. Choose \(y\) outside that union and different from \(p\), possible since \(X\) is uncountable. Then \(y\in U_n\) for every \(n\), whereas \(X\setminus\{y\}\) is a neighborhood of \(p\). No \(U_n\) is contained in this neighborhood, contradicting the base property.
Check: the construction rules out every sequence in \(A\), not merely a particular choice of sequence. The missing countable neighborhood base is exactly why the sequence argument cannot be used here.
The converse direction of the theorem is often where a proof hides a gap. To choose one point from each of a countable list of neighborhoods, the list must be a neighborhood base (or be refined to one). Without that property, the resulting sequence need not eventually lie in an arbitrary neighborhood of \(x\).
What Hausdorffness Adds to Compactness
Compactness alone does not imply that a subset is closed. Hausdorffness supplies the separation needed for the familiar compact-subset argument.
Proof. If \(K=\varnothing\), it is closed. Otherwise, fix \(x\in X\setminus K\). For each \(k\in K\), Hausdorffness gives disjoint open neighborhoods \(U_k\) of \(x\) and \(V_k\) of \(k\). The sets \(V_k\), taken with the subspace topology on \(K\), cover \(K\). Compactness gives \(k_1,\ldots,k_m\in K\) such that \(K\subseteq V_{k_1}\cup\cdots\cup V_{k_m}\). The set \(U=U_{k_1}\cap\cdots\cap U_{k_m}\) is an open neighborhood of \(x\). It misses each \(V_{k_i}\), so it misses \(K\), which is covered by those sets. Thus every \(x\notin K\) has an open neighborhood in \(X\setminus K\). Hence \(X\setminus K\) is open and \(K\) is closed. \(\square\)
Worked Example: A Compact Set That Is Not Closed
Let \(X=\{a,b\}\) have the Sierpinski topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This is a topology: it contains \(\varnothing\) and \(X\), and unions and finite intersections of its listed sets remain on the list. The set \(K=\{a\}\) is compact. Indeed, for any open cover of \(K\), one member contains \(a\), and that single member covers \(K\).
But \(K\) is not closed, because \(X\setminus K=\{b\}\) is not open. The space is not Hausdorff: every open neighborhood of \(b\) is \(X\), so no neighborhood of \(a\) can be disjoint from a neighborhood of \(b\). Thus “compact implies closed” is not valid without a separation hypothesis.
Check: compactness here follows directly from the open-cover definition, while closedness fails because the complement is not open. The example tests the two definitions separately.
A related standard result is the compact-to-Hausdorff homeomorphism theorem: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Its proof belongs to the earlier material on the compact-to-Hausdorff homeomorphism argument. The Hausdorff condition on the target cannot simply be deleted.
Worked Example: A Continuous Bijection That Is Not a Homeomorphism
Let the set \(X=\{a,b\}\) have the discrete topology, in which every subset is open, and let \(Y=\{a,b\}\) have the Sierpinski topology \(\{\varnothing,\{a\},Y\}\). Consider the identity bijection \(f:X\to Y\). The finite space \(X\) is compact: from any open cover, choose one member for each point, giving at most two members. The map \(f\) is continuous because the preimage of every open subset of \(Y\) is a subset of the discrete space \(X\), hence open.
However, \(f^{-1}:Y\to X\) is not continuous. The set \(\{b\}\) is open in \(X\), but its preimage under \(f^{-1}\) is \(\{b\}\), which is not open in \(Y\). Therefore \(f\) is not a homeomorphism. The target \(Y\) is not Hausdorff, as verified above, so this does not contradict the compact-to-Hausdorff theorem.
Check: continuity is checked by taking preimages of open sets in the codomain. The inverse fails the same test for the specific open set \(\{b\}\subseteq X\).
Finite Subcovers: Check What Is Being Covered
Compactness says that every open cover of the compact space has a finite subcover. It does not say that every open cover of every subset has a finite subcover, nor does it allow a proof to select finitely many sets without showing that they cover the set under discussion. For a subset \(K\subseteq X\), compactness means compactness in the subspace topology: every cover of \(K\) by sets open in \(K\) has a finite subcover.
Worked Example: An Open Cover With No Finite Subcover
Give \(I=(0,1)\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). For each integer \(n\ge2\), let \(U_n=(1/n,1)\), viewed as an open subset of \(I\). These sets cover \(I\): given \(x\in(0,1)\), choose an integer \(n\ge2\) with \(n>1/x\); then \(1/n<x<1\), so \(x\in U_n\).
Any finite selection has a largest index \(m\), and since \(U_n\subseteq U_m\) for \(n\le m\), its union is \(U_m=(1/m,1)\). This misses \(1/(2m)\), which satisfies \(0<1/(2m)<1/m\). Hence no finite selection covers \(I\), so \(I\) is not compact.
Check: the strict endpoint in \(U_m=(1/m,1)\) matters: the point \(1/m\) is not included. The exhibited point \(1/(2m)\) lies in \(I\) and is missed by the whole finite union.
A common invalid proof says: “For each point choose one set from the cover containing it; therefore finitely many chosen sets cover.” The choices may be one per point, and there may be infinitely many points. The finite conclusion follows only after applying compactness to a genuine open cover of a compact set. In the preceding example, each point has a chosen \(U_n\), but the cover has no finite subcover.
A Practical Audit
When checking a proof, locate the step where a hypothesis is used rather than checking only the conclusion. The following questions expose many common gaps:
- Sequence step: Is there a countable neighborhood base at the point where a sequence is supposed to detect closure?
- Separation step: Does the argument need disjoint neighborhoods, and has Hausdorffness (or another sufficient separation condition) been assumed?
- Compactness step: Is the set being covered compact in the relevant topology, and is the family actually an open cover of that set?
- Finite selection: After selecting finitely many sets, has their union been shown to cover every point required?
- Map step: For a proposed homeomorphism, have both the map and its inverse been shown continuous, or does a correctly hypothesized theorem supply that conclusion?
Common pitfall. “Every compact subset is closed, so \(\{a\}\) is closed in the Sierpinski space.” The first clause omits Hausdorffness. The corrected statement is that compact subsets of Hausdorff spaces are closed; in the Sierpinski space, \(\{a\}\) is compact but not closed because its complement \(\{b\}\) is not open. A counterexample does not merely challenge the conclusion: it identifies which omitted hypothesis the proof needed.
Check Your Understanding
For each question, state which definition or hypothesis controls the argument.
- Why does a sequence in \(A\) converging to \(x\) imply \(x\in\overline{A}\) in any topological space?
- In the cocountable example, why does \(X\setminus\{a_n:n\in\mathbb{N}\}\) contain \(p\) and remain open?
- Which step in the proof that compact subsets of Hausdorff spaces are closed uses compactness?
- Give the open set in the discrete source that demonstrates failure of continuity of the inverse in the finite bijection example.
- Why does every finite selection from the cover \(U_n=(1/n,1)\), \(n\ge2\), fail to cover \(I=(0,1)\)?