Separating a Point from a Closed Set
Hausdorff separation concerns two distinct points. Regularity asks for a stronger kind of separation: given a point and a closed set that does not contain it, can open sets separate them? This property is useful because closed sets can contain many points, and separation must hold for the whole set at once. A convention matters here: some authors build the \(T_1\) condition into the word “regular,” while others do not.
Throughout, \((X,\mathcal{T})\) is a topological space. A set is closed if its complement is open. Two sets are separated by disjoint open sets if there are open sets containing them, respectively, whose intersection is empty.
The \(T_1\) axiom means that every singleton \(\{x\}\) is closed. We use the convention that \(T_3\) means regular and \(T_1\). Consequently, in this tutorial “regular” alone does not assert \(T_1\). Under the alternative convention in which “regular” already includes \(T_1\), the separation property defined above is sometimes called “regular without \(T_1\).” Stating the convention prevents an apparent disagreement between classifications.
Examples in the Real Line and Discrete Spaces
Worked Example: Separating a Closed Set in the Real Line
Let \(\mathbb{R}\) have its standard topology. We verify directly that it is regular. Let \(F\subseteq\mathbb{R}\) be closed and \(x\notin F\). Since \(\mathbb{R}\setminus F\) is open and contains \(x\), there is \(r>0\) such that \((x-r,x+r)\subseteq\mathbb{R}\setminus F\). Set
The interval \(U\) is open and contains \(x\). The interval \([x-r/2,x+r/2]\) is closed in \(\mathbb{R}\), so \(V\) is open. Also, \(U\cap V=\varnothing\). Because \([x-r/2,x+r/2]\subseteq(x-r,x+r)\subseteq\mathbb{R}\setminus F\), no point of \(F\) lies in the closed interval removed to form \(V\); hence \(F\subseteq V\). These sets satisfy the regularity condition, including when \(F=\varnothing\).
Check: the chosen open interval \(U\) and open set \(V\) are disjoint by construction, \(x\in U\), and \(F\subseteq V\). Thus the separation works for an arbitrary closed \(F\), not only for a single point. \(\mathbb{R}\) is also \(T_1\): for each \(x\), the complement of \(\{x\}\) is open, so \(\mathbb{R}\) is \(T_3\).
The same argument is a special case of a standard result: every metric space is regular and \(T_1\). The general metric-ball separation proof is treated in the later material on metric balls as separators. For the \(T_1\) part, the reason is direct: if \(d\) is a metric and \(y\ne x\), the ball \(B_d(y,d(x,y)/2)\) misses \(x\). Such balls show that \(X\setminus\{x\}\) is open.
Worked Example: The Discrete Topology
Let \(X\) be any set with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset of \(X\) is open. It is also closed, because its complement is open. In particular, every singleton is closed and \(X\) is \(T_1\).
To check regularity, take any closed \(F\subseteq X\) and any \(x\in X\setminus F\). Choose \(U=\{x\}\) and \(V=F\). Both sets are open in the discrete topology, \(x\in U\), and \(F\subseteq V\). Since \(x\notin F\), \(U\cap V=\varnothing\). This proves regularity, so every discrete space is \(T_3\), including the empty space: in that case there are no choices of \(x\) and \(F\) to check, and the separation condition holds vacuously.
Check: if \(F=\varnothing\), the same choices give \(U=\{x\}\) and \(V=\varnothing\), which are open and disjoint. No nonemptiness assumption on \(F\) is needed.
These examples illustrate how to verify regularity from its definition: start with an arbitrary closed set and a point outside it, then exhibit the two required open sets. It is not enough to separate selected pairs of points unless that argument also handles every closed set in the condition.
Cofinite Topologies: The Role of \(T_1\)
For a set \(X\), the cofinite topology consists of \(\varnothing\) and all subsets whose complement in \(X\) is finite. A space in this topology is \(T_1\): the complement \(X\setminus\{x\}\) of any singleton is open, since its complement is finite. Whether the space is regular depends on whether \(X\) is finite or infinite.
Worked Example: Cofinite Topology on an Infinite Set
Let \(X\) be infinite with the cofinite topology. Its closed sets are \(X\) and the finite subsets of \(X\): taking complements in the definition of open sets gives precisely these possibilities. We show that \(X\) is not regular.
Choose distinct \(x,y\in X\), and let \(F=\{y\}\). This is a closed set and \(x\notin F\). Every open neighborhood of \(x\) is nonempty, and every open set containing \(y\) is nonempty. In fact, any two nonempty open sets \(U,V\) intersect. To see this, \(X\setminus U\) and \(X\setminus V\) are finite, so their union is finite. An infinite set cannot be contained in that finite union; choose \(z\in X\setminus((X\setminus U)\cup(X\setminus V))\). Then \(z\in U\cap V\).
Therefore no open neighborhood of \(x\) can be disjoint from an open set containing \(F=\{y\}\). The point \(x\) and the closed set \(F\) cannot be separated as required, so \(X\) is not regular. Since \(X\) is \(T_1\), it is not \(T_3\) either.
Check: the obstruction applies to every candidate pair \(U,V\), not merely to a particular choice: both must be nonempty, and any two nonempty cofinite-open sets intersect.
Worked Example: Cofinite Topology on a Finite Set
Now let \(X\) be finite and give it the cofinite topology. Every subset \(A\subseteq X\) has finite complement, so every subset is open; the topology is discrete. By the discrete-space argument above, \(X\) is regular and \(T_1\), hence \(T_3\).
For a concrete check, take \(X=\{a,b,c\}\), \(F=\{b\}\), and \(x=a\). The sets \(U=\{a\}\) and \(V=\{b\}\) are open in the cofinite topology because their complements \(\{b,c\}\) and \(\{a,c\}\) are finite. They are disjoint, contain \(x\) and \(F\), respectively, and so separate the point from the closed set. The general argument works for every \(x\notin F\), not just these choices.
Check: there is no distinction between the finite cofinite topology and the discrete topology: all subsets are open in both. This also covers \(X=\varnothing\), for which the topology is \(\{\varnothing\}\) and the separation condition is vacuous.
Regularity Does Not Imply \(T_1\)
The \(T_1\) requirement in \(T_3\) is not redundant. In an indiscrete topology, whose only open sets are \(\varnothing\) and \(X\), the closed sets are also just \(\varnothing\) and \(X\). For any point \(x\) and closed set \(F\) with \(x\notin F\), the only possibility is \(F=\varnothing\). Taking \(U=X\) and \(V=\varnothing\) separates \(x\) from \(F\). Thus every indiscrete space is regular according to the separation definition here. If \(X\) has at least two points, however, a singleton is not closed: its complement is a nonempty proper subset, hence is not open. Such a space is not \(T_1\), and therefore is not \(T_3\).
Common pitfall. The claim “regular spaces are automatically \(T_1\)” is false under the convention used here. For a concrete wrong argument, let \(X=\{a,b\}\) have the indiscrete topology. It is wrong to infer from the successful separation test that \(\{a\}\) must be closed: the test only considers closed sets \(F\) and points outside \(F\). The only closed set omitting a point is \(\varnothing\), and the test succeeds by taking \(U=X\), \(V=\varnothing\). But \(\{a\}\) is not closed because \(X\setminus\{a\}=\{b\}\) is not open. The correction is to keep the two conditions separate: regularity is the point-versus-closed-set separation property, while \(T_1\) requires every singleton to be closed. \(T_3\) requires both.
| Space | Regular? | \(T_1\)? | \(T_3\)? |
|---|---|---|---|
| \(\mathbb{R}\) with its standard topology | Yes; direct separation above | Yes | Yes |
| Any discrete space | Yes | Yes | Yes |
| Infinite set with the cofinite topology | No | Yes | No |
| Finite set with the cofinite topology | Yes; it is discrete | Yes | Yes |
| Indiscrete space with at least two points | Yes | No | No |
Check Your Understanding
For each question, keep the regularity test and the \(T_1\) condition distinct.
- In an infinite cofinite space, why can no two nonempty open sets be disjoint?
- Show directly that a discrete space is \(T_1\), as well as regular.
- For an indiscrete space with at least two points, identify a singleton that is not closed and explain why regularity still holds.
- Is the cofinite topology on a finite set \(T_3\)? Justify your answer by identifying the topology.
- In the convention used here, which two properties must a space satisfy to be \(T_3\)?