Three Separation Conditions
The separation axioms describe how well a topology can distinguish distinct points. Their definitions differ in what a neighborhood must do: \(T_0\) requires a one-sided distinction, \(T_1\) requires both one-sided distinctions, and Hausdorffness requires disjoint neighborhoods. These differences are visible in small finite examples and in the contrast between the real line and an infinite space with the cofinite topology.
Let \((X,\mathcal{T})\) be a topological space. An open neighborhood of \(x\in X\) is an open set containing \(x\). The conditions below concern pairs of distinct points; if there are no such pairs, each condition is satisfied vacuously.
These conditions form a hierarchy. The implications themselves are useful when classifying a space: a failure of \(T_0\) rules out both stronger conditions, and a failure of \(T_1\) rules out Hausdorffness.
Proof. Suppose \(X\) is Hausdorff and \(x\ne y\). Choose disjoint open neighborhoods \(U\) of \(x\) and \(V\) of \(y\). Since \(V\) contains \(y\) and is disjoint from \(U\), the set \(U\) does not contain \(y\); likewise \(V\) does not contain \(x\). This is the \(T_1\) condition. Now suppose \(X\) is \(T_1\). For distinct \(x,y\), the \(T_1\) condition gives an open neighborhood of \(x\) that excludes \(y\), so an open set contains one of the points and not the other. Thus \(X\) is \(T_0\). \(\square\)
Classifying Familiar Spaces
Worked Example: The Real Line
Give \(\mathbb{R}\) its standard topology. If \(x\ne y\), let \(r=|x-y|/3\), which is positive. The open intervals \((x-r,x+r)\) and \((y-r,y+r)\) contain \(x\) and \(y\), respectively. Their centers are \(|x-y|=3r\) apart, while the sum of their radii is \(2r\), so the intervals are disjoint. Therefore \(\mathbb{R}\) is Hausdorff, and the proposition shows it is also \(T_1\) and \(T_0\).
Check: if \(x<y\), the right endpoint of the interval around \(x\) is \(x+(y-x)/3\), while the left endpoint of the interval around \(y\) is \(y-(y-x)/3\). The difference between these endpoints is \((y-x)/3>0\), confirming the intervals do not overlap.
Worked Example: Discrete and Indiscrete Topologies
Let \(X\) be any set with the discrete topology, in which every subset is open. For distinct \(x,y\in X\), the neighborhoods \(\{x\}\) and \(\{y\}\) are open and disjoint. Hence \(X\) is Hausdorff, and therefore \(T_1\) and \(T_0\). This also covers the empty set: there are no distinct pairs to test.
Now give \(X\) the indiscrete topology \(\{\varnothing,X\}\). If \(X\) has at least two points, every nonempty open set is \(X\). No open set can contain one of a distinct pair while excluding the other, so the space is not \(T_0\), and consequently is neither \(T_1\) nor Hausdorff. If \(X\) is empty or has one point, all three conditions hold vacuously. In the one-point case the discrete and indiscrete topologies coincide.
Check: for a distinct pair in a discrete space, \(\{x\}\cap\{y\}=\varnothing\). In an indiscrete space with at least two points, the only open neighborhood of either point is \(X\), and \(X\cap X=X\ne\varnothing\).
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). The open set \(\{a\}\) contains \(a\) and excludes \(b\), so the space is \(T_0\). But there is no open neighborhood of \(b\) that excludes \(a\): the only open set containing \(b\) is \(X\). Thus the space is not \(T_1\), and it is not Hausdorff.
The failure of Hausdorffness can also be checked directly. Any open neighborhood of \(b\) is \(X\), and every open neighborhood of \(a\) intersects \(X\). Thus no pair of disjoint open neighborhoods can separate \(a\) and \(b\).
Check: the sole open neighborhood of \(b\) is \(X=\{a,b\}\), which contains \(a\); the open neighborhood \(\{a\}\) does not contain \(b\). The two directions required for \(T_1\) do not both hold.
Finite Spaces and Their Open Neighborhoods
For finite spaces, it is possible to turn the \(T_0\) and \(T_1\) tests into a concise description. For each point \(x\) in a finite space, define \(U_x\) to be the intersection of all open sets containing \(x\). There are only finitely many subsets of \(X\), so this is a finite intersection of open sets and is open. It contains \(x\), and it is contained in every open neighborhood of \(x\).
Proof. First suppose \(X\) is \(T_0\), and take distinct \(x,y\). Some open set contains one of them but not the other. If it contains \(x\) and not \(y\), then \(U_x\), being contained in every open neighborhood of \(x\), does not contain \(y\). But \(y\in U_y\), so \(U_x\ne U_y\). If instead the open set contains \(y\) and not \(x\), the same reasoning gives \(U_y\ne U_x\). Conversely, suppose \(U_x\ne U_y\). If \(y\notin U_x\), then the open set \(U_x\) contains \(x\) and excludes \(y\). If \(y\in U_x\), the inequality of the sets implies \(x\notin U_y\): otherwise \(U_x\) and \(U_y\) would each contain the other point, which would mean every open neighborhood of either point contains the other, and hence the two collections of open neighborhoods coincide, giving \(U_x=U_y\). Thus \(U_y\) contains \(y\) and excludes \(x\). In either case the pair is \(T_0\)-distinguishable.
Now suppose the finite space is \(T_1\). For each \(x\in X\) and each \(y\ne x\), choose an open neighborhood of \(x\) excluding \(y\). Intersect these finitely many neighborhoods, one for each \(y\ne x\). The intersection is open, contains \(x\), and contains no other point, so \(\{x\}\) is open. If \(X\) has only one point, the intersection over no other points is taken to be \(X=\{x\}\), which is open as well. Therefore every singleton is open, and every subset is a union of open singletons, so \(X\) is discrete. Conversely, a discrete space is \(T_1\), as shown above. Finally, a Hausdorff space is \(T_1\) by the proposition, so a finite Hausdorff space is discrete. A discrete space is Hausdorff, proving the last equivalence. \(\square\)
Worked Example: A Three-Point Topology
Let \(X=\{a,b,c\}\) and let \(\mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}\). These sets are nested, so unions and finite intersections remain in the collection; this is a topology. The smallest open neighborhoods are \(U_a=\{a\}\), \(U_b=\{a,b\}\), and \(U_c=X\). They are pairwise distinct, so the finite-space criterion gives \(T_0\).
The space is not \(T_1\): there is no open neighborhood of \(b\) excluding \(a\), since the open sets containing \(b\) are \(\{a,b\}\) and \(X\). By the hierarchy it is not Hausdorff. This example shows that a finite \(T_0\) space need not be discrete; the theorem says that the stronger \(T_1\) condition forces discreteness in the finite case.
Check: the listed \(U_x\) are intersections of the open neighborhoods of each point: those of \(a\) are \(\{a\},\{a,b\},X\); those of \(b\) are \(\{a,b\},X\); and the only one of \(c\) is \(X\).
Infinite Cofinite Spaces
On an infinite set \(X\), the cofinite topology consists of \(\varnothing\) and all subsets whose complement in \(X\) is finite. It gives a particularly useful separation example: points can be distinguished in the \(T_1\) sense, but distinct points cannot be separated by disjoint open sets.
Worked Example: An Infinite Cofinite Space
Let \(X\) be infinite with the cofinite topology, and take distinct \(x,y\in X\). The set \(X\setminus\{y\}\) is open because its complement is the finite set \(\{y\}\); it contains \(x\) and excludes \(y\). Likewise \(X\setminus\{x\}\) is open, contains \(y\), and excludes \(x\). Thus \(X\) is \(T_1\), and consequently \(T_0\).
To test Hausdorffness, let \(U\) and \(V\) be any nonempty open sets. The sets \(X\setminus U\) and \(X\setminus V\) are finite. If \(U\cap V\) were empty, then every point of \(X\) would lie outside \(U\) or outside \(V\), giving \(X=(X\setminus U)\cup(X\setminus V)\). The right side is a union of two finite sets and is finite, contradicting that \(X\) is infinite. Therefore any two nonempty open sets intersect. In particular, no two distinct points have disjoint open neighborhoods, so \(X\) is not Hausdorff.
Check: the two open neighborhoods chosen for \(x\) and \(y\) have intersection \(X\setminus\{x,y\}\). Since \(X\) is infinite, removing two points leaves a nonempty set. The neighborhoods verify \(T_1\), but they do not verify Hausdorffness.
A Common Mistake and a Comparison
Common mistake. A tempting but incorrect argument says that an infinite cofinite space is Hausdorff because \(X\setminus\{y\}\) and \(X\setminus\{x\}\) are open neighborhoods of \(x\) and \(y\). Those sets are open, but they are not disjoint: their intersection is \(X\setminus\{x,y\}\), which is nonempty when \(X\) is infinite. The \(T_1\) test asks for neighborhoods that exclude the other point; the Hausdorff test asks for neighborhoods whose entire intersection is empty. These are different requirements.
| Space | \(T_0\) | \(T_1\) | Hausdorff |
|---|---|---|---|
| \(\mathbb{R}\) with its standard topology | Yes | Yes | Yes |
| Any set with the discrete topology | Yes | Yes | Yes |
| Indiscrete space with at least two points | No | No | No |
| Sierpinski space \(\{\varnothing,\{a\},X\}\) | Yes | No | No |
| Finite space | Depends on its open sets | Exactly when discrete | Exactly when discrete |
| Infinite cofinite space | Yes | Yes | No |
When checking a new example, begin with a pair of distinct points and apply the definitions literally. For \(T_0\), find one open set that distinguishes the pair. For \(T_1\), find an excluding neighborhood in each direction. For Hausdorffness, find two neighborhoods and verify that their intersection is empty. Keeping these tests separate prevents a successful \(T_1\) calculation from being mistaken for a Hausdorff proof.
Check Your Understanding
For each question, use the open sets of the stated topology and the definitions of the separation conditions.
- Why is every Hausdorff space \(T_1\), and why is every \(T_1\) space \(T_0\)?
- Classify the empty set and a one-point set with their indiscrete topologies.
- In the Sierpinski space, which direction of the \(T_1\) condition fails for the pair \(a,b\)?
- Explain why every finite \(T_1\) space is discrete.
- For an infinite cofinite space, why must any two nonempty open sets intersect?