Three Separation Tests That Must Not Be Confused
When checking how well a topology separates points, it is easy to replace a required condition with a weaker one. A closed singleton gives a useful test for \(T_1\), but it does not provide disjoint neighborhoods. Likewise, unique limits for convergent sequences are necessary in Hausdorff spaces, but they do not guarantee Hausdorffness in every space. The reliable approach is to return to the precise neighborhood condition each axiom asks for.
Let \((X,\mathcal{T})\) be a topological space. We use the following definitions. They quantify over distinct points, so the conditions are vacuously satisfied if \(X\) has fewer than two points.
These conditions ask for different things. \(T_0\) requires at least one open set that distinguishes the points; \(T_1\) requires such neighborhoods in both directions; Hausdorffness requires a pair of neighborhoods that are disjoint. In particular, merely finding two neighborhoods that exclude the other point does not prove that those neighborhoods are disjoint.
Closed Singletons Are a \(T_1\) Test
Proof. Suppose \(X\) is \(T_1\). Fix \(x\in X\). For each \(y\in X\setminus\{x\}\), the \(T_1\) condition gives an open neighborhood \(U_y\) of \(y\) that does not contain \(x\). Thus \(U_y\subseteq X\setminus\{x\}\), and every point of \(X\setminus\{x\}\) belongs to its corresponding \(U_y\). Therefore \(X\setminus\{x\}=\bigcup_{y\ne x}U_y\) is open, so \(\{x\}\) is closed. If there are no such \(y\), the complement is empty and is open as well.
Conversely, suppose every singleton is closed. If \(x\ne y\), then \(X\setminus\{y\}\) is open, contains \(x\), and excludes \(y\); also \(X\setminus\{x\}\) is open, contains \(y\), and excludes \(x\). These are exactly the two neighborhoods required by \(T_1\). \(\square\)
This theorem does not say that the two neighborhoods it produces are disjoint. They are complements of different singletons, and their intersection can contain many points. Hausdorffness asks for the stronger, separate condition of disjoint neighborhoods.
Worked Example: Closed Singletons Without Disjoint Neighborhoods
Let \(X=\mathbb{N}\) with the cofinite topology: the open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite. We check the singleton test, then check directly whether distinct points have disjoint open neighborhoods.
For each \(n\in\mathbb{N}\), the complement \(X\setminus\{n\}\) has one point removed, so it is open. Hence \(\{n\}\) is closed. The theorem also verifies that this space is \(T_1\).
Now let \(U,V\) be any two nonempty open sets. Their complements \(X\setminus U\) and \(X\setminus V\) are finite, so their union is finite. Since \(X=\mathbb{N}\) is infinite, there is a natural number outside that finite union. That number belongs to both \(U\) and \(V\), so \(U\cap V\ne\varnothing\). In particular, no two nonempty open neighborhoods of distinct points can be disjoint. The space is not Hausdorff.
Check: for distinct \(m,n\), the natural candidate neighborhoods \(U=X\setminus\{n\}\) and \(V=X\setminus\{m\}\) do contain \(m\) and \(n\), respectively, while excluding the other point. But \(U\cap V=X\setminus\{m,n\}\), which is nonempty because \(\mathbb{N}\) has points other than \(m,n\). The neighborhoods establish \(T_1\), not Hausdorffness.
Common mistake. “Every singleton is closed, so for \(x\ne y\), the open sets \(X\setminus\{y\}\) and \(X\setminus\{x\}\) separate \(x\) and \(y\).” The first part of this calculation is correct: the sets contain \(x\) and \(y\), respectively, and exclude the other point. The conclusion is not: “separate” here does not mean “disjoint.” In the example, their intersection is \(X\setminus\{x,y\}\), which is nonempty. To prove Hausdorffness, one must exhibit neighborhoods \(U,V\) and verify the equation \(U\cap V=\varnothing\).
\(T_0\) Does Not Require Both Directions
Worked Example: A \(T_0\) Space That Is Not \(T_1\)
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This is a topology: it contains \(\varnothing,X\), and unions and finite intersections of its members remain in the collection.
The open set \(\{a\}\) contains \(a\) and not \(b\), so it distinguishes the two points. Thus \(X\) is \(T_0\). But there is no open neighborhood of \(b\) that excludes \(a\): the only open set containing \(b\) is \(X\), and \(X\) contains \(a\). The \(T_1\) requirement fails. Equivalently, \(\{a\}\) is not closed, because its complement \(\{b\}\) is not open. This example shows why checking only one direction is enough for \(T_0\) but not for \(T_1\).
Check: the open sets that contain \(a\) are \(\{a\}\) and \(X\); at least one excludes \(b\). The only open set that contains \(b\) is \(X\); none excludes \(a\). So the one-direction test succeeds and the two-direction test fails.
Unique Sequence Limits Do Not Always Give Hausdorffness
In a Hausdorff space, a sequence cannot converge to two distinct points: disjoint neighborhoods of the two proposed limits would force the sequence eventually into both, which is impossible. The converse is not valid for arbitrary spaces. The following example verifies unique sequence limits directly while also verifying failure of Hausdorffness.
Worked Example: Unique Sequence Limits in the Cocountable Topology
Give \(X=\mathbb{R}\) the cocountable topology: its open sets are \(\varnothing\) and the sets \(U\) for which \(\mathbb{R}\setminus U\) is countable. First consider two nonempty open sets \(U,V\). Their complements are countable, so \((\mathbb{R}\setminus U)\cup(\mathbb{R}\setminus V)\) is countable. Since \(\mathbb{R}\) is uncountable, this union cannot be all of \(\mathbb{R}\). A point outside the union belongs to \(U\cap V\). Thus every two nonempty open sets intersect, and distinct points cannot have disjoint open neighborhoods. This space is not Hausdorff.
Now let \((x_n)\) be a sequence converging to \(p\in\mathbb{R}\), and let \(S=\{x_n:n\in\mathbb{N}\}\) be its range. The set \(S\setminus\{p\}\) is countable and does not contain \(p\), so \(U=\mathbb{R}\setminus(S\setminus\{p\})\) is an open neighborhood of \(p\). Convergence means that there is \(N\in\mathbb{N}\) such that \(x_n\in U\) for every \(n\ge N\). Since every term lies in \(S\), the only point of \(S\) that can lie in \(U\) is \(p\). Hence \(x_n=p\) for every \(n\ge N\): every convergent sequence is eventually constant at its limit.
If the same sequence converged to both \(p\) and \(q\), it would therefore be eventually equal to \(p\) and eventually equal to \(q\). Taking an index greater than or equal to both eventual-constancy indices gives \(p=x_n=q\), so \(p=q\). Thus convergent sequences have unique limits in this space, even though the space is not Hausdorff.
Check: the argument uses the open set with complement \(S\setminus\{p\}\), not the set with complement \(S\). If \(p\in S\), the latter set would omit \(p\) and would not be a neighborhood of the proposed limit. Removing \(p\) from the countable complement is essential to the convergence test.
A Reliable Checklist
When a separation argument is uncertain, write down exactly what must be exhibited and check it point by point. The following distinctions prevent the most common substitutions:
| Claim to verify | What must be shown | What is not enough |
|---|---|---|
| \(T_0\) | For each distinct pair, one open set contains exactly one of the two points. | Nothing weaker than distinguishing the pair. |
| \(T_1\) | Each point has an open neighborhood excluding the other. | Finding a neighborhood in only one direction. |
| Hausdorff | There are open neighborhoods \(U,V\) with \(U\cap V=\varnothing\). | Showing only that each neighborhood excludes the other point. |
| Unique sequence limits | No sequence converges to two distinct points. | Using this as a Hausdorff test without an additional hypothesis. |
Two useful implications can be proved directly. A Hausdorff space is \(T_1\): for fixed \(x\), each \(y\ne x\) has a Hausdorff neighborhood excluding \(x\), and the union of these neighborhoods is \(X\setminus\{x\}\), which is open. Also, a \(T_1\) space is \(T_0\), because its neighborhood condition in both directions supplies, in particular, a neighborhood distinguishing the pair. The examples above show why the reverse implications cannot simply be assumed: the Sierpinski space is \(T_0\) but not \(T_1\), and the infinite cofinite space is \(T_1\) but not Hausdorff.
The practical lesson is to match the evidence to the axiom. Closed singletons test \(T_1\); disjoint neighborhoods test Hausdorffness; and sequence behavior must be interpreted with care. A counterexample is especially effective when its topology is explicit and each claimed property is checked from the definition.
Check Your Understanding
For each question, identify the exact definition being tested before deciding what must be proved.
- In the two-point topology \(\{\varnothing,\{a\},\{b\},X\}\), which of \(T_0\), \(T_1\), and Hausdorffness hold? Justify each answer from the open sets.
- In an infinite set with the cofinite topology, why must any two nonempty open sets intersect?
- Explain why \(X\setminus\{y\}\) and \(X\setminus\{x\}\) need not be disjoint when \(x\ne y\).
- In the cocountable topology on \(\mathbb{R}\), construct an open neighborhood of a proposed sequence limit \(p\) that forces a convergent sequence to be eventually \(p\).
- Which implication is proved by closed singletons: \(T_0\), \(T_1\), or Hausdorffness? State the exact equivalence.