Separating Points by Neighborhoods
Separation axioms measure how effectively a topology lets open sets distinguish different points. The conditions considered here are local: for each pair of distinct points, we ask what open neighborhoods they have. The order of the requirements matters. A neighborhood that distinguishes two points in one direction need not distinguish them in the reverse direction, and two distinguishing neighborhoods need not be disjoint.
Throughout, \((X,\mathcal{T})\) is a topological space. An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) with \(x\in U\). In each definition below, the points being tested are distinct. If \(X\) has fewer than two points, there are no such pairs, so each condition holds vacuously.
In the \(T_0\) condition, either point may be the one whose neighborhood excludes the other. In the \(T_1\) condition, both directions must be witnessed, possibly by different open sets. The Hausdorff condition asks for neighborhoods that do not intersect at all; consequently, they exclude the other point as well. These definitions do not say that every open neighborhood of a point separates it from another point. They require only that suitable neighborhoods exist for each pair.
Discrete and Indiscrete Topologies
Two basic topologies make the definitions especially direct to check. For a set \(X\), the discrete topology is \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. The indiscrete topology is \(\mathcal{T}=\{\varnothing,X\}\), so the only nonempty open set is \(X\).
Worked Example: The Discrete Topology
Let \(X\) be any set with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). Test the three separation conditions.
Take distinct \(x,y\in X\). The singleton sets \(\{x\}\) and \(\{y\}\) are open, since every subset of \(X\) is open. The set \(\{x\}\) is a neighborhood of \(x\) and does not contain \(y\), while \(\{y\}\) is a neighborhood of \(y\) and does not contain \(x\). This gives the two witnesses required for \(T_1\), and in particular provides a witness for \(T_0\). Also, \(\{x\}\cap\{y\}=\varnothing\), so these neighborhoods establish the Hausdorff condition.
If \(X\) is empty or has one point, there is no distinct pair to test, and each condition holds vacuously. Thus every discrete space, including these degenerate cases, is \(T_0\), \(T_1\), and Hausdorff.
Check: for every distinct pair, the actual neighborhoods used are \(\{x\}\) and \(\{y\}\); each contains its named point, excludes the other, and their intersection is empty. This verifies all three conditions directly.
Worked Example: The Indiscrete Topology
Let \(X\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). Determine when it satisfies each separation condition.
If \(X\) contains distinct points \(x,y\), any open neighborhood of \(x\) must be nonempty, so it can only be \(X\). This neighborhood contains \(y\). The same reasoning applies to every open neighborhood of \(y\): it must be \(X\), which contains \(x\). Therefore no neighborhood of either point excludes the other, and the \(T_0\) condition fails. The \(T_1\) condition fails because it already requires a neighborhood in each direction that excludes the other point. The Hausdorff condition fails because the only possible neighborhoods of the pair are both \(X\), and \(X\cap X=X\ne\varnothing\).
If \(X\) has no points or exactly one point, there is no distinct pair, and all three conditions hold vacuously. Hence an indiscrete space is \(T_0\), \(T_1\), and Hausdorff exactly when its underlying set has at most one point.
Check: for a space with at least two points, the only open neighborhood of either member of a distinct pair is \(X\). It cannot exclude the other point, and two copies of \(X\) are not disjoint. So each failure follows from the defining neighborhood test.
The contrast is not about the number of open sets alone: it is about whether the available open sets can separate a chosen pair. The discrete topology supplies singleton neighborhoods, while the indiscrete topology supplies no proper nonempty open set at all.
Every Topology on Two Points
A two-point set is small enough to classify all its topologies. This gives examples where the one-sided \(T_0\) requirement holds but the stronger requirements fail.
Proof. Every topology on \(X=\{a,b\}\) must contain \(\varnothing\) and \(X\). The only other subsets that could be open are \(\{a\}\) and \(\{b\}\). There are four possibilities for which of these two sets are included: neither, only \(\{a\}\), only \(\{b\}\), or both. Each possibility gives a topology. In fact, the listed collections contain \(\varnothing\) and \(X\), and unions and finite intersections of their sets remain in the same collection. There can be no other topology because there are no other subsets to consider.
For \(\{\varnothing,\{a\},X\}\), the open neighborhood \(\{a\}\) contains \(a\) and excludes \(b\), so the space is \(T_0\). But every open neighborhood of \(b\) is \(X\), so there is no neighborhood of \(b\) excluding \(a\). Thus it is not \(T_1\). Any neighborhood of \(b\) is \(X\), which intersects every neighborhood of \(a\), so the space is not Hausdorff. For \(\{\varnothing,\{b\},X\}\), the neighborhood \(\{b\}\) witnesses \(T_0\); every neighborhood of \(a\) is \(X\), so the same tests show it is neither \(T_1\) nor Hausdorff. In the indiscrete case, the only neighborhoods of either point are \(X\), which neither exclude the other point nor form a disjoint pair. In the discrete case, \(\{a\}\) and \(\{b\}\) are open, exclude one another, and are disjoint. These checks establish the stated properties for all four topologies. \(\square\)
Worked Example: One Open Singleton on a Two-Point Set
Take \(X=\{a,b\}\) with \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Test each condition by listing the neighborhoods of the pair.
The open neighborhoods of \(a\) are \(\{a\}\) and \(X\), and the only open neighborhood of \(b\) is \(X\). For \(T_0\), use \(\{a\}\): it contains \(a\), and \(b\notin\{a\}\). Thus the pair is distinguished in at least one direction. For \(T_1\), a neighborhood of \(b\) excluding \(a\) would be needed, but the only such neighborhood candidate containing \(b\) is \(X\), and \(a\in X\). Therefore the required two-sided witnesses do not exist.
For Hausdorffness, every neighborhood of \(b\) is \(X\). If \(U\) is any neighborhood of \(a\), then \(U\) is either \(\{a\}\) or \(X\); in both cases \(U\cap X=U\ne\varnothing\). Hence no neighborhoods of \(a\) and \(b\) are disjoint.
Check: \(\{a\}\) is the \(T_0\) witness, while the complete list of neighborhoods of \(b\) is just \(\{X\}\). This confirms that the missing reverse witness prevents \(T_1\), and the intersection computation prevents Hausdorffness.
What the Tests Do—and Do Not—Say
The definitions can be compared by the kind of witness they demand. A \(T_0\) witness is one open set that contains one point and excludes the other. For \(T_1\), one needs a witness in each direction. For Hausdorffness, one needs a pair of neighborhoods whose intersection is empty. Disjointness is stronger than merely requiring each neighborhood to exclude the other point: two neighborhoods may both exclude the opposite point and still share some third point.
| Condition | Requirement for each distinct \(x,y\) | Two-point topology with only \(\{a\}\) open |
|---|---|---|
| \(T_0\) | Some open neighborhood of one point excludes the other | Satisfied by \(\{a\}\) |
| \(T_1\) | A neighborhood of each point excludes the other | Fails: no neighborhood of \(b\) excludes \(a\) |
| Hausdorff (\(T_2\)) | The points have disjoint open neighborhoods | Fails: every neighborhood of \(b\) is \(X\) |
Common pitfall. A common incorrect argument is: “The set \(\{a\}\) separates \(a\) from \(b\), so the space \((\{a,b\},\{\varnothing,\{a\},X\})\) is \(T_1\).” This verifies only one direction. For \(T_1\), one must also find an open neighborhood of \(b\) that excludes \(a\). The only open set containing \(b\) is \(X\), and \(a\in X\); that required neighborhood does not exist. The correction is to check each direction separately for \(T_1\), and to check the intersection of the chosen neighborhoods when testing Hausdorffness.
These tests are useful whenever a problem asks whether a space has a particular separation property: fix an arbitrary distinct pair and either construct the required neighborhoods or show why no such neighborhoods can exist. In a discrete space the singleton sets provide the neighborhoods immediately. In an indiscrete space with at least two points, the absence of proper nonempty open sets rules them out. On a two-point space, checking which singletons are open completes the classification.
Check Your Understanding
For each question, identify the required neighborhood witness before deciding whether the condition holds.
- In the topology \(\{\varnothing,\{b\},\{a,b\}\}\) on \(\{a,b\}\), which open set witnesses \(T_0\)?
- For an indiscrete space with exactly one point, why do the \(T_0\), \(T_1\), and Hausdorff conditions all hold?
- In the two-point topology \(\{\varnothing,\{a\},\{b\},\{a,b\}\}\), give disjoint open neighborhoods for the two distinct points.
- Can a space with two points and exactly one open singleton be \(T_1\)? Check the required neighborhood in the direction from the point without an open singleton.
- State the difference between the neighborhood requirement for \(T_1\) and the neighborhood requirement for Hausdorffness.