The T1 Condition and a Closed-Set Test
The T1 axiom describes how neighborhoods distinguish distinct points. Its definition refers to pairs of points and open sets, but there is an equivalent test that refers only to closed sets: every one-point set must be closed. This equivalence is useful because it turns a separation condition into a direct calculation with complements.
Let \((X,\mathcal{T})\) be a topological space. An open neighborhood of \(x\in X\) is an open set containing \(x\). A set \(C\subseteq X\) is closed when \(X\setminus C\) is open. For \(y\in X\), the singleton \(\{y\}\) is the set whose only element is \(y\).
The neighborhoods used for different pairs need not be the same. The T1 condition asks only that one point at a time can be excluded from a neighborhood of another point. It does not require a neighborhood to exclude every other point simultaneously.
Proof. Suppose first that \((X,\mathcal{T})\) is T1. Fix \(y\in X\). For each \(x\in X\setminus\{y\}\), the T1 condition gives an open set \(U_x\) containing \(x\) and not containing \(y\). Thus \(x\in U_x\subseteq X\setminus\{y\}\), and therefore \(X\setminus\{y\}=\bigcup_{x\in X\setminus\{y\}}U_x\). A union of open sets is open, so \(X\setminus\{y\}\) is open and \(\{y\}\) is closed. If \(X\setminus\{y\}\) is empty, the union is the empty set, which is open, so the argument includes the one-point space.
Conversely, suppose every singleton is closed. Let \(x,y\in X\) with \(x\ne y\). Since \(\{y\}\) is closed, \(X\setminus\{y\}\) is open. This open set contains \(x\) and does not contain \(y\). Thus the required neighborhood exists for each ordered pair of distinct points, and \((X,\mathcal{T})\) is T1. If \(X\) is empty, there are no such pairs and there are no singletons, so both conditions hold vacuously. \(\square\)
Applying the Characterization
For a concrete topology, the closed-singleton test gives a short route: write down the complement of each singleton and decide whether that complement is open. The test is also a reminder to check closedness, not openness. These are separate properties; a set can be open, closed, both, or neither.
Worked Example: The Discrete Topology
Let \(X\) be any set with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), in which every subset of \(X\) is open. For each \(y\in X\), the complement \(X\setminus\{y\}\) is a subset of \(X\), hence is open. Therefore \(\{y\}\) is closed. The closed-singleton characterization shows that \(X\) is T1.
This topology does satisfy the axioms for a topology: it contains \(\varnothing\) and \(X\); a union of subsets of \(X\) is a subset of \(X\); and a finite intersection of subsets of \(X\) is a subset of \(X\). Each of these sets belongs to \(\mathcal{P}(X)\). If \(X=\varnothing\), there are no singletons and the T1 condition holds vacuously; otherwise the complement argument applies to every point.
Check: for distinct \(x,y\in X\), \(X\setminus\{y\}\) is an open neighborhood of \(x\) that excludes \(y\). This verifies the original neighborhood definition as well as the closed-singleton test.
Worked Example: A Cofinite Topology
Let \(X=\mathbb{Z}\), and give it the cofinite topology: \(\mathcal{T}\) consists of \(\varnothing\) and all subsets \(U\subseteq\mathbb{Z}\) whose complement \(\mathbb{Z}\setminus U\) is finite. First, this is a topology. It contains \(\varnothing\) and \(\mathbb{Z}\). For any family of open sets, if every member is empty, its union is empty; otherwise choose a nonempty member \(U\). The complement of the union is contained in the finite set \(\mathbb{Z}\setminus U\), so the union is open. For a finite intersection, an empty member makes the intersection empty; otherwise the complement of the intersection is a finite union of finite complements, hence finite. The intersection is therefore open.
Fix \(n\in\mathbb{Z}\). The complement of \(\{n\}\) is open because its complement in \(\mathbb{Z}\) is the finite set \(\{n\}\). Thus every singleton is closed and \((\mathbb{Z},\mathcal{T})\) is T1.
However, \(\{n\}\) is not open: its complement \(\mathbb{Z}\setminus\{n\}\) is infinite, so \(\{n\}\) does not have finite complement. This does not contradict T1, which requires singletons to be closed, not open. More generally, the closed sets here are \(\mathbb{Z}\) and the finite subsets: taking complements in the definition of \(\mathcal{T}\) gives exactly these possibilities.
Check: the set \(2\mathbb{Z}\) of even integers is infinite and is not all of \(\mathbb{Z}\), so it is not closed by the description just obtained. Its complement, the odd integers, is infinite, so \(2\mathbb{Z}\) is not open either. Thus even in this T1 space, not every subset is closed or open.
Worked Example: The Indiscrete Topology
Let \(X=\{a,b,c\}\) with the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). Its closed sets are also just \(\varnothing\) and \(X\), since the complements of these two open sets are \(X\) and \(\varnothing\). Each singleton, such as \(\{a\}\), is nonempty and is not \(X\). Hence no singleton is closed, and the space is not T1.
The neighborhood definition confirms the failure. The only open set containing \(a\) is \(X\), and it contains \(b\) as well. There is no open neighborhood of \(a\) excluding \(b\). In fact, for any two distinct points in this space, the only neighborhood of either point is \(X\), which contains both.
The size of the underlying set matters in the degenerate cases. If \(X=\varnothing\), there are no singletons and no distinct pairs, so the space is T1 vacuously. If \(X=\{a\}\) with the indiscrete topology \(\{\varnothing,X\}\), then \(\{a\}=X\) is closed, and there are no distinct pairs to test. This one-point space is T1. The failure occurs for the three-point example because it has distinct points that cannot be separated by the required neighborhoods.
Finite Sets and a Common Pitfall
Proof. Let \(F\) be a finite subset of a T1 space. If \(F=\varnothing\), it is closed because its complement \(X\) is open. Otherwise write \(F=\{x_1,\ldots,x_k\}\), where \(k\) is a positive integer and the \(x_i\) are its elements. By the closed-singleton characterization, each \(\{x_i\}\) is closed. The complement of \(F\) is \(X\setminus F=\bigcap_{i=1}^{k}(X\setminus\{x_i\})\). Each set in this finite intersection is open, so the intersection is open. Hence \(F\) is closed. \(\square\)
The proposition concerns finite subsets only. The cofinite example above shows that an infinite proper subset need not be closed in a T1 space. In that example \(2\mathbb{Z}\) is infinite and proper, so it is not closed.
Common pitfall: confusing closed singletons with open singletons. A mistaken test might say, “The space is T1 only if each singleton is open.” In the cofinite topology on \(\mathbb{Z}\), this would incorrectly rule out T1: for example, \(\{0\}\) is not open because \(\mathbb{Z}\setminus\{0\}\) is infinite. The correct calculation is that \(\mathbb{Z}\setminus\{0\}\) is open, since its complement is the finite set \(\{0\}\). Therefore \(\{0\}\) is closed, as required. The test is about the complement of the singleton being open.
| Topology | Complement of a singleton | T1? |
|---|---|---|
| Discrete topology on \(X\) | Open, because every subset is open | Yes |
| Cofinite topology on \(\mathbb{Z}\) | Open, because its complement is finite | Yes |
| Indiscrete topology on \(\{a,b,c\}\) | Not open, because only \(\varnothing\) and \(X\) are open | No |
Check Your Understanding
For each question, distinguish the statement that singletons are closed from the stronger-looking but different statement that singletons are open.
- In the cofinite topology on an infinite set \(X\), use complements to show that every singleton is closed.
- In the indiscrete topology on a set with at least two points, identify two distinct points for which the T1 neighborhood condition fails.
- Why is the empty space T1 under both the neighborhood definition and the closed-singleton characterization?
- Give a reason that the cofinite topology on \(\mathbb{Z}\) is not discrete.
- Prove that a finite union of closed sets is closed, and use this to recover the proposition about finite subsets of a T1 space.