Match the Proof to the Structure
Topological problems often admit several possible proof routes, but those routes do not have the same hypotheses. A basis is useful when the question is whether sets are open or a map is continuous. Neighborhoods suggest nets when a point must be detected across an arbitrary topology. An open cover is the right object when compactness is at issue. Sequences can be efficient in metric or first-countable spaces, but they do not test closure in every topological space.
The practical question is not merely “Which method is shortest?” It is “What information does this method require, and does the space provide it?” The examples below use that question to select and justify each proof route.
Use a Basis to Test Continuity
Proof. Suppose first that \(f\) is continuous. Each \(B\in\mathcal{B}\) is open in \(Y\), so \(f^{-1}(B)\) is open in \(X\) by the definition of continuity. Conversely, suppose \(f^{-1}(B)\) is open for every \(B\in\mathcal{B}\). Let \(V\) be any open subset of \(Y\). Since \(\mathcal{B}\) is a basis, \(V\) is a union of basis elements, say \(V=\bigcup_{\alpha}B_\alpha\). Preimages preserve unions, so \(f^{-1}(V)=\bigcup_{\alpha}f^{-1}(B_\alpha)\). Each set in this union is open by assumption, and a union of open sets is open. Thus \(f\) is continuous. \(\square\)
Worked Example: Identity Maps and a Basis Test
Let \(\mathbb{R}\) have its standard topology in the domain and the lower limit topology \(\mathbb{R}_\ell\) in the codomain. The latter has basis consisting of the intervals \([a,b)\), where \(a<b\). Consider the identity map \(f:\mathbb{R}\to\mathbb{R}_\ell\).
Test the basis element \([0,1)\). Its preimage under \(f\) is \([0,1)\). This is not open in the standard topology: every standard open interval containing \(0\) includes negative numbers, so no such interval is contained in \([0,1)\). The basis test therefore shows that \(f\) is not continuous.
Reverse the topologies and consider the identity map \(g:\mathbb{R}_\ell\to\mathbb{R}\). Let \(V\) be standard-open and let \(x\in V\). There is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq V\). Then \(x\in[x,x+\varepsilon)\subseteq V\), and \([x,x+\varepsilon)\) is open in \(\mathbb{R}_\ell\). This gives a lower-limit-open neighborhood inside \(V\) around each \(x\in V\), so \(V\) is lower-limit-open. Thus every open set in the codomain \(\mathbb{R}\) has open preimage, and \(g\) is continuous.
Check: The same set \([0,1)\) is lower-limit-open but not standard-open. The different continuity conclusions follow from the different topologies, not from a change in the underlying identity function.
Use Neighborhoods to Build Nets
For a point \(x\) and a set \(A\) in a topological space \(X\), the neighborhood test says that \(x\in\overline{A}\) exactly when every open neighborhood of \(x\) meets \(A\). A net makes this test into a convergence construction. Index by the open neighborhoods themselves, rather than by pairs that include a selected point; this gives a genuine partial order.
Proof. Suppose \(x\in\overline{A}\). Let \(D\) be the set of open neighborhoods of \(x\), ordered by reverse inclusion: \(U\le V\) means \(V\subseteq U\). This is a partial order, and it is directed because \(U\cap V\) is an open neighborhood of \(x\) with \(U\le U\cap V\) and \(V\le U\cap V\). For each \(U\in D\), the neighborhood test gives \(U\cap A\ne\varnothing\); choose \(a_U\in U\cap A\). This selection uses the axiom of choice for this family of nonempty sets. The net \((a_U)_{U\in D}\) converges to \(x\): for any open neighborhood \(W\) of \(x\), every index \(U\ge W\) satisfies \(U\subseteq W\), and hence \(a_U\in W\).
Conversely, suppose a net \((a_d)_{d\in D}\) has all its terms in \(A\) and converges to \(x\). Every open neighborhood \(W\) of \(x\) contains all sufficiently late terms, so it contains at least one term \(a_d\in A\). Thus \(W\cap A\ne\varnothing\). The neighborhood test gives \(x\in\overline{A}\). \(\square\)
Worked Example: Closure That No Sequence Detects
Let \(X\) be an uncountable set with the cocountable topology: its open sets are \(\varnothing\) and the sets whose complements in \(X\) are countable. Fix \(p\in X\), and let \(A=X\setminus\{p\}\). We show that \(p\in\overline{A}\), although no sequence in \(A\) converges to \(p\).
Take any open neighborhood \(U\) of \(p\). If \(U\cap A\) were empty, then \(U=\{p\}\). But \(X\setminus\{p\}\) is uncountable, so \(\{p\}\) is not open in this topology. Therefore every such \(U\) meets \(A\), and the neighborhood test gives \(p\in\overline{A}\).
Now take any sequence \((a_n)\) in \(A\). Its range \(R=\{a_n:n\in\mathbb{N}\}\) is countable, so \(W=X\setminus R\) is open. Because every \(a_n\ne p\), the point \(p\) belongs to \(W\), but no term of the sequence belongs to \(W\). Thus the sequence is not eventually in this neighborhood of \(p\), and it does not converge to \(p\).
The net construction does work here: index by the open neighborhoods \(U\) of \(p\), ordered by reverse inclusion, and choose \(a_U\in U\cap A\). Each choice is possible by the argument above, and the resulting net converges to \(p\) by the theorem.
When Sequences Are Enough
A sequence is a net whose index set is \(\mathbb{N}\) with its usual order. In a first-countable space, each point has a countable neighborhood base, and closure can be tested by sequences: if \(x\in\overline{A}\), choose a decreasing countable neighborhood base \(U_1\supseteq U_2\supseteq\cdots\) at \(x\), and choose \(a_n\in U_n\cap A\). Then \(a_n\to x\). Conversely, a sequence in \(A\) converging to \(x\) forces every neighborhood of \(x\) to meet \(A\). The construction relies on the countable neighborhood base; the preceding example shows why the sequence conclusion cannot be used in an arbitrary space.
The same limitation matters when testing continuity. In general spaces, continuity implies that convergent nets map to convergent nets, and the converse holds too.
Proof. If \(f\) is continuous and \(V\) is an open neighborhood of \(f(x)\), then \(f^{-1}(V)\) is an open neighborhood of \(x\). Since \(x_d\to x\), all sufficiently late \(x_d\) lie in \(f^{-1}(V)\), so all sufficiently late \(f(x_d)\) lie in \(V\). This proves convergence of the image net.
For the converse, suppose \(f\) is not continuous. Then there is an open set \(V\subseteq Y\) for which \(f^{-1}(V)\) is not open. Choose \(x\in f^{-1}(V)\) that is not an interior point of \(f^{-1}(V)\). Every open neighborhood \(U\) of \(x\) then contains some \(y_U\notin f^{-1}(V)\). Index by the open neighborhoods \(U\) of \(x\), again ordered by reverse inclusion, and choose one such \(y_U\) for each \(U\). The net \(y_U\) converges to \(x\), by the same neighborhood argument as before, but every \(f(y_U)\) lies outside \(V\). Since \(V\) is a neighborhood of \(f(x)\), the image net does not converge to \(f(x)\). This contradicts the assumed net property, so \(f\) is continuous. \(\square\)
Use Open Covers to Prove or Disprove Compactness
Compactness is defined in terms of covers, so an open-cover argument is the direct method. A space is compact if every cover by open sets has a finite subcover. To prove compactness, start with an arbitrary open cover and extract a finite subcover. To disprove compactness, construct one open cover and show that each finite selection misses a point.
Worked Example: Every Open Cover of [0,1] Has a Finite Subcover
Let \(\mathcal{U}\) be any cover of \([0,1]\) by sets open in the subspace topology. Define \(S\) to be the set of \(t\in[0,1]\) such that \([0,t]\) is covered by finitely many members of \(\mathcal{U}\). The set \(S\) is nonempty: choose \(U\in\mathcal{U}\) containing \(0\). Since \(U\) is relatively open, some \(\delta>0\) satisfies \([0,\delta)\cap[0,1]\subseteq U\). Hence a positive \(t\le1\) with \(t<\delta\) belongs to \(S\).
Let \(c=\sup S\), which exists because \(S\) is nonempty and bounded above by \(1\). Choose \(U\in\mathcal{U}\) containing \(c\). Relative openness gives \(\delta>0\) such that \((c-\delta,c+\delta)\cap[0,1]\subseteq U\). Since \(c>0\), the definition of supremum provides \(s\in S\) with \(c-\delta<s\le c\). A finite collection from \(\mathcal{U}\) covers \([0,s]\), and \(U\) covers \([s,c]\), so a finite collection covers \([0,c]\). Thus \(c\in S\).
If \(c<1\), choose \(r>0\) such that \(c+r\le1\) and \(r<\delta\). The same \(U\) covers \([c,c+r]\). Together with the finite cover of \([0,c]\), this gives a finite cover of \([0,c+r]\), so \(c+r\in S\), contradicting \(c=\sup S\). Therefore \(c=1\), and \([0,1]\) has a finite subcover. As \(\mathcal{U}\) was arbitrary, \([0,1]\) is compact.
Check: The argument began with an arbitrary open cover and produced a finite subcover; it did not assume a particular form for the cover or use a sequence criterion.
Worked Example: An Open Cover With No Finite Subcover
Consider \(X=(0,1)\) with the subspace topology from \(\mathbb{R}\), and for each \(n\in\mathbb{N}\) with \(n\ge2\), let \(U_n=(1/n,1)\). Each \(U_n\) is open in \(X\), since \(U_n=X\cap(1/n,2)\).
These sets cover \(X\). Indeed, if \(x\in(0,1)\), choose an integer \(n\ge2\) with \(n>1/x\). Then \(1/n<x<1\), so \(x\in U_n\). But the sets are nested: if \(2\le n\le N\), then \(U_n\subseteq U_N\). Therefore any nonempty finite selection has union \(U_N\) for its largest index \(N\). This union misses \(x=1/(2N)\), since \(0<1/(2N)<1/N\). The empty finite selection covers no points either. Thus this open cover has no finite subcover, and \(X\) is not compact.
A Common Mistake: Using the Wrong Test
A frequent error is to conclude that a point is not in a closure simply because no sequence from the set converges to it. In the cocountable example, that calculation would assert \(p\notin\overline{A}\) because every sequence in \(A\) has a neighborhood of \(p\) that contains none of its terms. But the conclusion is false: every open neighborhood of \(p\) meets \(A\), so \(p\in\overline{A}\). The correction is to check the hypothesis behind the sequence test. Without a countable neighborhood base, use the neighborhood test or construct a net.
Likewise, a continuity argument that checks preimages of a few convenient open sets is incomplete unless those sets form a basis (or another sufficient test family has been justified). For compactness, checking one particular cover cannot prove compactness: the definition quantifies over every open cover. The proof route should match the quantifier in the property being tested.
Check Your Understanding
For each question, decide which proof route fits and state any hypothesis it needs.
- Explain why testing the preimages of basis elements is enough to prove continuity.
- Describe the directed set used to construct a net converging to a point in \(\overline{A}\).
- Why does the cocountable example show that sequences cannot test closure in every topological space?
- For the cover \(U_n=(1/n,1)\) of \((0,1)\), identify a point missed by a given finite subcollection.
- In the proof that \([0,1]\) is compact, where is completeness of the real numbers used?