Why Finiteness Changes the T1 Condition
The T1 axiom says that individual points can be separated by open sets. A useful equivalent formulation is that every singleton is closed. In an infinite space, closed singletons do not generally have to be open. In a finite space, however, the complement of a singleton is a finite union of closed singletons. That finite-union step forces every singleton to be open, and then every subset is open.
Throughout, \((X,\mathcal{T})\) is a topological space. A space is T1 if for any distinct points \(x,y\in X\), there is an open set containing \(x\) but not \(y\), and an open set containing \(y\) but not \(x\). We use the equivalent characterization from the earlier material on T1 spaces: \(X\) is T1 if and only if \(\{x\}\) is closed for every \(x\in X\). A space is discrete if every subset of \(X\) is open.
Proof. First suppose \(X\) is finite and T1. If \(X=\varnothing\), its only subset is \(\varnothing\), which is open, so \(X\) is discrete. Now suppose \(X\ne\varnothing\), and fix \(x\in X\). The set \(X\setminus\{x\}\) is the union of the singletons \(\{y\}\) over all \(y\in X\) with \(y\ne x\). There are finitely many such points. Each \(\{y\}\) is closed by the T1 condition, and a finite union of closed sets is closed. If there are no such points, then \(X\setminus\{x\}=\varnothing\), which is closed as well. Thus \(X\setminus\{x\}\) is closed in every case, so its complement \(\{x\}\) is open. Since this holds for every \(x\in X\), every subset \(A\subseteq X\) is a union of open singletons and is therefore open. Hence \(X\) is discrete.
Conversely, suppose \(X\) is discrete. For every \(x\in X\), the set \(X\setminus\{x\}\) is open because every subset is open. Therefore \(\{x\}\) is closed. The closed-singleton characterization gives that \(X\) is T1. This also covers \(X=\varnothing\), for which the T1 condition has no pairs of distinct points to check. \(\square\)
Unpacking the Finite-Union Argument
The key step is not that a singleton is closed: T1 already gives that. The key is that, when \(X\) is finite, the complement of a point is built from only finitely many other singletons. A finite union of closed sets is closed, so the complement of that union is open. Once each singleton is open, arbitrary subsets are open because each is a union of some of those singletons.
For every \(y\in X\), the singleton \(\{y\}\) is closed.
Write \(X\setminus\{x\}=\bigcup_{y\in X,\ y\ne x}\{y\}\). Finiteness of \(X\) makes this a finite union of closed sets, including the possibility of the empty union.
The set \(X\setminus\{x\}\) is closed, so \(\{x\}\) is open.
For \(A\subseteq X\), \(A=\bigcup_{x\in A}\{x\}\). This is a union of open sets, so \(A\) is open, even if \(A\) is empty.
Worked Example: Three Points in the Real Line
Let \(X=\{-2,1,4\}\) have the subspace topology inherited from \(\mathbb{R}\) with its standard topology. We check directly that this finite subspace is discrete, and then verify the T1 condition.
For the point \(-2\), the open interval \((-3,-1)\) intersects \(X\) in \(\{-2\}\): it contains \(-2\), while \(1\) and \(4\) are not in the interval. For \(1\), the interval \((0,2)\) intersects \(X\) in \(\{1\}\). For \(4\), the interval \((3,5)\) intersects \(X\) in \(\{4\}\). By the definition of the subspace topology, these intersections are open in \(X\). Thus every singleton is open, and every subset of \(X\), being a union of singletons, is open.
In particular, the complements of the singletons are open: \(X\setminus\{-2\}=\{1,4\}\), \(X\setminus\{1\}=\{-2,4\}\), and \(X\setminus\{4\}=\{-2,1\}\). So each singleton is also closed, which verifies that \(X\) is T1.
Check: the eight subsets of \(X\) are \(\varnothing\), the three singletons, the three two-point subsets, and \(X\). The empty set is open by the topology axioms; the singletons are open as shown; the two-point subsets are unions of two open singletons; and \(X\) is open by the topology axioms. Hence the subspace topology is exactly the discrete topology.
Worked Example: A Finite Discrete Metric
Let \(X=\{a,b,c\}\), and define \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). Give \(X\) the topology generated by the open balls of this metric. We compute small balls to see the topology.
For each \(x\in X\), the ball \(B_d(x,1/2)\) contains \(x\), since \(d(x,x)=0<1/2\). It contains no other point \(y\), since \(d(x,y)=1\) when \(y\ne x\), and \(1\not<1/2\). Therefore \(B_d(a,1/2)=\{a\}\), \(B_d(b,1/2)=\{b\}\), and \(B_d(c,1/2)=\{c\}\). Each singleton is open. Every subset is a union of these balls, so every subset is open.
The T1 check can also be made using closed singletons. The complements \(X\setminus\{a\}=\{b,c\}\), \(X\setminus\{b\}=\{a,c\}\), and \(X\setminus\{c\}=\{a,b\}\) are open as unions of open balls. Thus each singleton is closed and \(X\) is T1.
Check: for each \(x\in X\), the open ball of radius \(1/2\) is exactly \(\{x\}\), because the distance to \(x\) is \(0<1/2\) and the distance to either other point is \(1\), which is not less than \(1/2\). This confirms the singleton calculations and the discrete topology.
A Finite Ordered Example
Worked Example: The Order Topology on Four Points
Let \(X=\{0,1,2,3\}\) with the order topology from the usual order. We exhibit an open singleton around each point. In the order topology, open intervals and the corresponding open rays at endpoints are basic open sets.
The ray \((-\infty,1)\cap X\) is \(\{0\}\), since \(0<1\) and none of \(1,2,3\) is less than \(1\). The interval \((0,2)\cap X\) is \(\{1\}\), since the only element of \(X\) strictly between \(0\) and \(2\) is \(1\). Likewise, \((1,3)\cap X=\{2\}\), and the ray \((2,\infty)\cap X=\{3\}\). These four basic open sets are the four singletons, so every subset of \(X\) is open.
For an explicit T1 check, the complements are \(X\setminus\{0\}=\{1,2,3\}\), \(X\setminus\{1\}=\{0,2,3\}\), \(X\setminus\{2\}=\{0,1,3\}\), and \(X\setminus\{3\}=\{0,1,2\}\). Each is a union of open singletons, hence open. The four singletons are therefore closed as well.
Check: each displayed ray or interval includes exactly the stated point: the endpoints are excluded from open intervals, and the rays use strict inequalities. Thus the open-set calculation shows that the order topology is discrete, in agreement with the theorem.
Why the Hypothesis Matters
Finiteness is used precisely when we conclude that \(X\setminus\{x\}\) is a finite union of closed singletons. It is not enough to know that every singleton is closed: infinite unions of closed sets need not be closed. For example, in \(\mathbb{R}\) with its standard topology, every singleton is closed, but \(\{0\}\) is not open. Indeed, every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\), and that interval contains points other than \(0\). So \(\mathbb{R}\) is T1 but not discrete.
Common pitfall. A mistaken argument says: “Every \(\{y\}\) is closed, so \(X\setminus\{0\}=\bigcup_{y\ne 0}\{y\}\) is closed; therefore \(\{0\}\) is open.” In \(\mathbb{R}\), the displayed union has infinitely many terms, and the rule that finite unions of closed sets are closed does not apply. In fact, \(\mathbb{R}\setminus\{0\}\) is not closed: its complement \(\{0\}\) is not open, as every neighborhood of \(0\) contains other real numbers. The corrected finite-space argument explicitly counts on there being only finitely many terms in the union.
A useful consequence is that every subspace of a finite discrete space is discrete. If \(A\subseteq X\) and \(B\subseteq A\), then \(B\) is open in \(X\), and \(B=B\cap A\) is therefore open in the subspace topology on \(A\). In particular, once a finite T1 space has been shown discrete, all of its subspaces inherit the discrete topology.
Check Your Understanding
Use the closed-singleton argument and keep track of where finiteness is needed.
- For a finite T1 space \(X\) and \(x\in X\), write \(X\setminus\{x\}\) as a union of closed singletons and explain why that union is closed.
- How does the proof handle a space with exactly one point? What is the union representing the complement of its point?
- In the four-point ordered example, why is \((0,2)\cap X\) equal to \(\{1\}\), rather than including either endpoint?
- Give the specific invalid step in the argument that an arbitrary union of closed singletons must be closed.
- Show directly from the subspace topology definition that every subspace of a discrete space is discrete.