When Convergent Sequences Reveal Too Little
In a metric space, continuity can be tested by checking what happens to every convergent sequence. That test may fail in a general topological space: a space can have so few convergent sequences that preserving them gives very little information about a map. We will see this happen on an uncountable set with the cocountable topology. There, every convergent sequence is eventually constant, so even a discontinuous map can preserve all convergent sequences.
Let \(X\) be an uncountable set. Its cocountable topology is the collection \(\mathcal{T}\) consisting of \(\varnothing\) and all subsets \(U\subseteq X\) for which \(X\setminus U\) is countable. A sequence \((x_n)\) converges to \(x\) if for every open neighborhood \(U\) of \(x\), there is \(N\in\mathbb{N}\) such that \(x_n\in U\) whenever \(n\ge N\).
To check that this is a topology, note first that \(X\) and \(\varnothing\) belong to \(\mathcal{T}\). For any union of open sets, if every set in the union is empty, the union is empty; otherwise its complement is contained in the countable complement of one nonempty member, so the union is open. For a finite intersection of nonempty open sets, the complement is a finite union of countable sets and is therefore countable. If one of the sets being intersected is empty, the intersection is empty. Thus \(\mathcal{T}\) is a topology.
Proof. Suppose first that \(x_n=x\) for every \(n\ge N\). Every open neighborhood \(U\) of \(x\) contains \(x\), so it contains \(x_n\) for every \(n\ge N\). Hence \(x_n\to x\).
Conversely, suppose \(x_n\to x\), and let \(A=\{x_n:n\in\mathbb{N},\ x_n\ne x\}\), the set of values of the sequence that differ from \(x\). The set \(A\) is at most countable. Therefore \(U=X\setminus A\) is open, and \(x\in U\). Convergence gives an \(N\) such that \(x_n\in U\) for all \(n\ge N\). But every term \(x_n\ne x\) belongs to \(A\) and hence lies outside \(U\). Thus \(x_n=x\) for every \(n\ge N\), as required. \(\square\)
Examples of Convergence and Its Absence
Worked Example: A Sequence That Converges
Choose points \(a,b,x\in X\), with no requirement that they be distinct. Define \(x_1=a\), \(x_2=b\), and \(x_n=x\) for every \(n\ge3\). Then this sequence converges to \(x\): given any open neighborhood \(U\) of \(x\), the terms \(x_n\) belong to \(U\) for all \(n\ge3\), since those terms equal \(x\).
The theorem also shows that the sequence cannot converge to any \(y\ne x\). If it converged to \(y\), it would have to be eventually equal to \(y\), but all its terms from index \(3\) onward equal \(x\). Since \(x\ne y\), that is impossible.
Check: the conclusion does not depend on whether \(a\) or \(b\) equals \(x\). In every case, \(x_n=x\) for all \(n\ge3\), so the sequence is eventually equal to \(x\) and converges to \(x\).
Worked Example: A Sequence of Distinct Points Has No Limit
Choose distinct points \(b_1,b_2,\dots\) in \(X\), and let \(B=\{b_n:n\in\mathbb{N}\}\). The sequence \((b_n)\) has no limit. To verify this, consider any proposed limit \(y\in X\).
If \(y\notin B\), then \(U=X\setminus B\) is an open neighborhood of \(y\). It contains none of the terms \(b_n\), so the sequence is not eventually in \(U\).
If \(y=b_k\) for some \(k\in\mathbb{N}\), then \(U=X\setminus(B\setminus\{b_k\})\) is open and contains \(y\). Among all the terms \(b_n\), only \(b_k\) belongs to \(U\), because the points \(b_n\) are distinct. Thus \(b_n\in U\) for just the single index \(n=k\), not for all sufficiently large \(n\). The sequence does not converge to \(y\) in this case either.
Since both possibilities for \(y\) have been ruled out, the sequence has no limit. This agrees with the theorem: a sequence of distinct points cannot be eventually equal to any one point.
Sequential Continuity and Continuity
A map \(f:X\to Y\) between topological spaces is sequentially continuous if \(x_n\to x\) in \(X\) always implies \(f(x_n)\to f(x)\) in \(Y\). The target \(Y\) can have any topology. A continuous map is always sequentially continuous: if \(V\) is an open neighborhood of \(f(x)\), continuity makes \(f^{-1}(V)\) an open neighborhood of \(x\); convergence of \(x_n\) then ensures that \(x_n\) is eventually in \(f^{-1}(V)\), so \(f(x_n)\) is eventually in \(V\).
Proof. Let \(f:X\to Y\) be any map, where \(Y\) has any topology, and suppose \(x_n\to x\) in \(X\). By the convergence theorem, there is \(N\in\mathbb{N}\) such that \(x_n=x\) for every \(n\ge N\). Applying \(f\) gives \(f(x_n)=f(x)\) for every \(n\ge N\). A sequence eventually equal to \(f(x)\) converges to \(f(x)\) in every topology on \(Y\), because every open neighborhood of \(f(x)\) contains \(f(x)\). Hence \(f(x_n)\to f(x)\), proving sequential continuity. \(\square\)
Worked Example: A Discontinuous Map to a Discrete Space
Fix \(p\in X\), and let \(Y=\{0,1\}\) have the discrete topology \(\mathcal{P}(Y)\), in which every subset is open. Define \(f:X\to Y\) by \(f(p)=1\) and \(f(x)=0\) for every \(x\ne p\).
The map is not continuous. The singleton \(\{1\}\) is open in \(Y\), but \(f^{-1}(\{1\})=\{p\}\) is not open in \(X\). Indeed, \(\{p\}\) is nonempty and its complement \(X\setminus\{p\}\) is uncountable, so \(\{p\}\) does not have countable complement. It therefore does not belong to the cocountable topology.
Nevertheless, \(f\) is sequentially continuous. If \(x_n\to x\), then \(x_n=x\) for every sufficiently large \(n\). For those indices, \(f(x_n)=f(x)\), so the image sequence is eventually equal to \(f(x)\) and converges to it in the discrete topology.
Check: take any \(x\ne p\). The open neighborhood \(X\setminus\{p\}\) of \(x\) maps entirely to \(0\). At \(p\), however, continuity would require the open neighborhood \(f^{-1}(\{1\})=\{p\}\) of \(p\), which is not open. Thus the failure of continuity occurs at \(p\), while sequential continuity holds for every convergent sequence.
Why the Space Is Not First-Countable
The example does not contradict the sequence test for continuity in first-countable spaces. Recall that a countable local base at \(x\) is a countable collection of open neighborhoods of \(x\) such that every open neighborhood of \(x\) contains at least one member of the collection. A space is first-countable if each point has a countable local base.
Proof. Fix \(x\in X\), and suppose, for contradiction, that \(U_1,U_2,\dots\) is a countable local base at \(x\). For each \(n\), the complement \(C_n=X\setminus U_n\) is countable. The union \(\bigcup_{n\in\mathbb{N}}C_n\) is countable, so uncountability of \(X\) allows us to choose \(y\in X\setminus\bigl(\{x\}\cup\bigcup_{n\in\mathbb{N}}C_n\bigr)\). In particular, \(y\ne x\), and \(y\in U_n\) for every \(n\).
The set \(V=X\setminus\{y\}\) is an open neighborhood of \(x\). Since the \(U_n\) form a local base, some \(U_n\) must satisfy \(U_n\subseteq V\). But \(y\in U_n\) and \(y\notin V\), contradicting \(U_n\subseteq V\). Thus no countable local base exists at \(x\). Since \(x\) was arbitrary, the space is not first-countable at any point. \(\square\)
A Common Mistake
Common pitfall: To try to detect the discontinuity of the map in the worked example, one might claim that there is a sequence of points different from \(p\) converging to \(p\). For instance, choose distinct \(b_n\ne p\) and propose \(b_n\to p\). This calculation fails: \(B=\{b_n:n\in\mathbb{N}\}\) is countable, so \(U=X\setminus B\) is an open neighborhood of \(p\), but \(b_n\notin U\) for every \(n\). The proposed sequence does not converge to \(p\).
The correction is to check convergence against every open neighborhood, not just to note that the terms differ from the proposed limit by some informal notion of closeness. In this topology, the neighborhood \(X\setminus B\) rules out that sequence. More generally, the theorem shows that no sequence can approach \(p\) while differing from \(p\) infinitely often. This is why the discontinuity of \(f\) cannot be detected by convergent sequences.
The key distinction is therefore between continuity, which tests all open neighborhoods, and sequential continuity, which tests only convergent sequences. In the cocountable topology on an uncountable set, the latter condition is satisfied by every map, even though many maps are not continuous.
Check Your Understanding
Use the neighborhood definition of convergence and the stated cocountable topology to justify your answers.
- Let \(X\) be uncountable with the cocountable topology. Prove that a sequence eventually equal to \(x\) converges to \(x\).
- Why is \(X\setminus A\) an open neighborhood of \(x\) when \(A\) is the countable set of values \(x_n\ne x\) and \(x_n\to x\)?
- Show that a sequence of distinct points in \(X\) has no limit.
- Give a nonconstant map from \(X\) to a two-point discrete space that is sequentially continuous but not continuous, and verify both claims.
- In the proof that \(X\) is not first-countable at \(x\), explain why the point \(y\) can be chosen outside \(\{x\}\cup\bigcup_{n\in\mathbb{N}}(X\setminus U_n)\).