When Do Sequences Detect Hausdorffness?
A Hausdorff space separates any two distinct points by disjoint open neighborhoods. This separation prevents a sequence from converging to both points. The converse is subtler: if two points cannot be separated, can we build a sequence converging to both? First countability supplies the countable neighborhood lists needed to do so.
Throughout, convergence means topological convergence: a sequence \((x_n)\) in a space \(X\) converges to \(x\in X\) if, for every open neighborhood \(U\) of \(x\), there is \(N\in\mathbb{N}\) such that \(x_n\in U\) whenever \(n\ge N\).
A finite neighborhood base can be written as a sequence by repeating its members. Thus, in a first-countable space, we may write a neighborhood base at each point as a sequence \(U_1,U_2,\ldots\). The listed neighborhoods need not be nested. We will make them nested by taking finite intersections.
Hausdorffness Always Implies Unique Sequence Limits
Proof. Suppose a sequence \((x_n)\) converges to both \(x\) and \(y\), and suppose \(x\ne y\). By the Hausdorff property, there are disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Convergence to \(x\) gives an index \(N_U\) such that \(x_n\in U\) whenever \(n\ge N_U\). Convergence to \(y\) gives an index \(N_V\) such that \(x_n\in V\) whenever \(n\ge N_V\). For \(n\ge\max(N_U,N_V)\), the term \(x_n\) belongs to \(U\cap V\), contradicting \(U\cap V=\varnothing\). Therefore \(x=y\), as required. \(\square\)
This direction needs no countability assumption. The converse below uses first countability: it turns the failure to separate two points into a sequence that visits increasingly small neighborhoods of both.
The First-Countable Characterization
Proof. The forward implication is the proposition just proved. For the reverse implication, we prove the contrapositive. Suppose \(X\) is first countable but not Hausdorff. Then there are distinct points \(x,y\in X\) for which no open neighborhood of \(x\) is disjoint from an open neighborhood of \(y\). Choose countable neighborhood bases \((U_n)\) at \(x\) and \((V_n)\) at \(y\). Define the open neighborhoods \(W_n=\bigcap_{i=1}^{n}U_i\) of \(x\) and \(Z_n=\bigcap_{i=1}^{n}V_i\) of \(y\). They are open neighborhoods because each is a finite intersection of open neighborhoods of the relevant point. Since \(x,y\) cannot be separated, \(W_n\cap Z_n\ne\varnothing\) for every \(n\). Choose \(z_n\in W_n\cap Z_n\) for each \(n\); this countable sequence of choices is available in the usual set-theoretic framework with countable choice.
To check convergence to \(x\), let \(U\) be any open neighborhood of \(x\). Since \((U_n)\) is a neighborhood base, \(U_k\subseteq U\) for some \(k\). Whenever \(n\ge k\), the definition of \(W_n\) gives \(z_n\in W_n\subseteq U_k\subseteq U\). Thus \(z_n\to x\). For any open neighborhood \(V\) of \(y\), choose \(j\) with \(V_j\subseteq V\). For every \(n\ge j\), \(z_n\in Z_n\subseteq V_j\subseteq V\), so \(z_n\to y\) as well. The sequence has two distinct limits, proving the contrapositive and hence the reverse implication. \(\square\)
The finite intersections matter: a neighborhood base need not be nested, so \(z_n\in U_n\) alone would not ensure that the sequence eventually remains in an earlier base neighborhood \(U_k\). By using \(W_n\), each later choice lies in every earlier listed neighborhood. The same reasoning applies to \(Z_n\).
Worked Examples
Worked Example: The Standard Real Line
Let \(\mathbb{R}\) have its standard topology. Verify first countability and the sequence-based Hausdorff test.
At each \(x\in\mathbb{R}\), the open intervals \(U_n=(x-1/n,x+1/n)\), for \(n\in\mathbb{N}\), form a neighborhood base. Indeed, if \(U\) is any open neighborhood of \(x\), it contains an interval \((x-r,x+r)\) for some \(r>0\). Choose \(n\) with \(1/n<r\). Then \(U_n\subseteq(x-r,x+r)\subseteq U\). Thus \(\mathbb{R}\) is first countable.
It is Hausdorff: given distinct \(x,y\), put \(d=|x-y|>0\). The intervals \((x-d/3,x+d/3)\) and \((y-d/3,y+d/3)\) are disjoint, since their centers are distance \(d\) apart and their radii sum to \(2d/3<d\). By the theorem, every sequence in \(\mathbb{R}\) has at most one limit. For a direct check, if a sequence converged to both \(x\) and \(y\), it would eventually belong to both of these disjoint neighborhoods, which is impossible.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Determine whether the space is first countable and Hausdorff, and exhibit a sequence with two limits.
At \(a\), the single neighborhood \(\{a\}\) is a neighborhood base: it is contained in every open neighborhood of \(a\). At \(b\), the only open neighborhood is \(X\), so \(\{X\}\) is a neighborhood base. Both bases are finite, hence \(X\) is first countable.
The space is not Hausdorff. The only open neighborhood of \(b\) is \(X\), and every open neighborhood of \(a\) contains \(a\); consequently, any neighborhood of \(a\) intersects \(X\). In particular, no neighborhoods of \(a\) and \(b\) are disjoint. Consider the constant sequence \(x_n=a\) for every \(n\). It converges to \(a\), since every neighborhood of \(a\) contains \(a\). It also converges to \(b\), since the only neighborhood of \(b\) is \(X\), which contains every term. Thus the two distinct limits \(a\) and \(b\) demonstrate the theorem’s contrapositive in this example.
Why First Countability Cannot Be Dropped
Worked Example: A Non-Hausdorff Space with Unique Sequence Limits
Let \(X\) be an uncountable set with the countable-complement topology: the open sets are \(\varnothing\) and all subsets whose complements in \(X\) are countable. This is a topology: arbitrary unions of nonempty open sets have countable complement, since their complement is contained in the complement of any one of those sets; finite intersections have countable complement because their complements are finite unions of countable sets.
This space is not Hausdorff. If \(U,V\) are nonempty open sets, then \(X\setminus U\) and \(X\setminus V\) are countable, so \(X\setminus(U\cap V)\) is countable. Since \(X\) is uncountable, \(U\cap V\ne\varnothing\). In particular, neighborhoods of two distinct points cannot be disjoint.
Nevertheless, every convergent sequence has exactly one limit. Let \((x_n)\) be any sequence and let \(R=\{x_n:n\in\mathbb{N}\}\) be its range, which is countable. If the sequence converges to \(x\), the set \(U=X\setminus(R\setminus\{x\})\) is open and contains \(x\): its complement is countable, and \(x\notin R\setminus\{x\}\). Convergence means that the sequence is eventually in \(U\). Every term of the sequence belongs to \(R\), and the only point of \(R\) that lies in \(U\) is \(x\) (if \(x\notin R\), there are no such points). Hence the sequence must eventually equal \(x\). Conversely, a sequence eventually equal to \(x\) converges to \(x\), because every neighborhood of \(x\) contains \(x\). If a sequence is eventually equal to \(x\), it cannot also be eventually equal to a distinct \(y\): a term sufficiently far along would have to equal both. Therefore no sequence has two limits.
Finally, \(X\) is not first countable. Suppose \((U_n)\) were a countable neighborhood base at some \(x\). Each complement \(X\setminus U_n\) is countable, so the set \(\{x\}\cup\bigcup_{n\in\mathbb{N}}(X\setminus U_n)\) is countable. Choose \(y\) outside this set, possible because \(X\) is uncountable. The open neighborhood \(X\setminus\{y\}\) of \(x\) would have to contain some base member \(U_n\). But \(y\notin X\setminus U_n\), so \(y\in U_n\), contradicting \(U_n\subseteq X\setminus\{y\}\). No countable neighborhood base at \(x\) exists. This example shows exactly why the first-countability hypothesis is needed for the converse.
A Common Mistake
Pitfall: It is incorrect to claim that a space is Hausdorff whenever every sequence has at most one limit. The tempting argument is to start with two points that cannot be separated and assume the countable construction from the theorem is available. The countable-complement example shows the gap: it is non-Hausdorff, yet its convergent sequences are eventually constant and therefore have unique limits. It is not first countable, so the theorem does not apply.
The safe conclusion is conditional: Hausdorffness always implies uniqueness of sequence limits, while the reverse implication is guaranteed here when the space is first countable. In spaces without that countability assumption, sequences may not detect a failure of neighborhood separation.
Check Your Understanding
For each question, identify where first countability is needed and use the definition of convergence to justify your answer.
- In the theorem’s proof, why are the neighborhoods \(W_n\) defined using finite intersections rather than simply setting \(W_n=U_n\)?
- Show that a constant sequence with value \(x\) converges to \(x\) in every topological space.
- In the Sierpinski space above, which open neighborhoods of \(b\) must a sequence eventually enter in order to converge to \(b\)?
- Explain why two disjoint open neighborhoods rule out a sequence converging to both of their respective points.
- In the countable-complement example, why does eventual equality to \(x\) prevent convergence to any \(y\ne x\)?