Separating Closed Sets
Hausdorffness asks whether distinct points can be separated by open neighborhoods. Normality asks a stronger, set-based question: can two disjoint closed sets be placed inside disjoint open sets? The distinction matters because a closed set may contain many points, and separating each pair of points separately does not itself give one pair of open sets that separates the entire closed sets.
Throughout, a neighborhood means an open neighborhood. The sets used in the definition below are closed in the given space \((X,\mathcal{T})\), and the separating sets are open in that same topology.
In this course, the T4 axiom means normality together with the T1 axiom. Recall that T1 means every singleton \(\{x\}\) is closed. Thus a T4 space must meet both requirements: it must separate every pair of disjoint closed sets by disjoint open sets, and its points must be closed sets.
Some texts build T1 into the word “normal.” Stating the convention avoids ambiguity: here the two conditions are kept separate. For any space, normality handles the separation of closed sets; the extra T1 condition rules out spaces in which points fail to be closed.
Discrete Spaces
In the discrete topology on \(X\), every subset of \(X\) is open. Since the complement of any subset is also open, every subset is closed as well. These facts make the separation test direct, even when \(X\) is infinite.
Proof. Let \(C,F\subseteq X\) be disjoint closed sets in a discrete space. Set \(U=C\) and \(V=F\). Both sets are open by discreteness, they contain \(C\) and \(F\), respectively, and \(U\cap V=C\cap F=\varnothing\). This proves normality. For every \(x\in X\), the complement \(X\setminus\{x\}\) is open, so \(\{x\}\) is closed. Thus the space is T1 as well, and hence T4. If \(X=\varnothing\), there are no points to check, and the only closed set is empty; choosing \(U=V=\varnothing\) verifies normality. \(\square\)
Worked Example: Separating Sets in a Discrete Space
Let \(X=\{a,b,c,d\}\) have the discrete topology, and take \(C=\{a,c\}\) and \(F=\{b\}\). The sets are disjoint because neither \(a\) nor \(c\) equals \(b\). Choose \(U=C=\{a,c\}\) and \(V=F=\{b\}\). Both are open in the discrete topology, and \(U\cap V=\varnothing\). They contain the required closed sets, so this pair is separated.
In fact, no special choice of neighborhoods is needed for any pair in a discrete space: the closed sets themselves are open. This verifies the definition for all disjoint closed pairs, not merely for the particular pair in this calculation. Also, each singleton is closed, so this space is T4.
A Metric-Space Separation Test
The definition also applies when a topology comes from a metric. In the next example, \(\mathbb{R}\) has its standard topology, induced by the usual metric \(d(x,y)=|x-y|\). We separate one specified pair of closed sets. This checks the definition for that pair; it is not, by itself, a proof of normality of the whole real line, which requires checking every disjoint closed pair.
Worked Example: Two Closed Sets in the Real Line
Take \(C=\{0\}\) and \(F=[2,3]\) in \(\mathbb{R}\). The set \(C\) is closed because its complement \((-\infty,0)\cup(0,\infty)\) is open. The set \(F\) is closed because its complement \((-\infty,2)\cup(3,\infty)\) is open. They are disjoint since \(0\notin[2,3]\).
Choose \(U=(-1,1)\) and \(V=(3/2,7/2)\). These are open intervals, and \(0\in U\), so \(C\subseteq U\). Since \(2>3/2\) and \(3<7/2\), every point of \([2,3]\) lies in \(V\), so \(F\subseteq V\). Finally, every point of \(U\) is less than \(1\), whereas every point of \(V\) is greater than \(3/2\). Since \(1<3/2\), \(U\cap V=\varnothing\). Thus these two closed sets have disjoint open neighborhoods, as required by the normality test.
The example illustrates what must be shown for a chosen pair: establish that the sets are closed and disjoint, then exhibit open supersets and verify that those supersets do not intersect. It does not allow us to infer that arbitrary disjoint closed sets in an arbitrary metric space can be separated. A general result about metric spaces is treated later.
The Cofinite Topology
For a set \(X\), the cofinite topology consists of \(\varnothing\) and all subsets whose complements in \(X\) are finite. On an infinite set, this topology offers a useful contrast with the discrete topology: points are closed, but distinct nonempty open sets cannot be disjoint.
Proof. For each \(x\in X\), the complement \(X\setminus\{x\}\) has one point removed and is therefore open in the cofinite topology. Hence \(\{x\}\) is closed, proving T1. To show the space is not normal, choose distinct points \(p,q\in X\). The sets \(\{p\}\) and \(\{q\}\) are disjoint closed sets. Suppose \(U,V\) were disjoint open sets with \(p\in U\) and \(q\in V\). Both are nonempty, so both have finite complements. Then \(X\setminus(U\cap V)=(X\setminus U)\cup(X\setminus V)\) is finite, because it is a union of two finite sets. Since \(X\) is infinite, this implies \(U\cap V\ne\varnothing\), contradicting the supposition that they are disjoint. Therefore this pair of closed sets cannot be separated, and the space is not normal. Since T4 requires normality, the space is not T4. \(\square\)
Worked Example: Why Cofinite Neighborhoods Cannot Separate Two Points
Let \(X=\mathbb{N}=\{1,2,3,\ldots\}\) with the cofinite topology, and take \(C=\{1\}\), \(F=\{2\}\). These are disjoint and closed: each singleton has an open complement. Consider any open set \(U\) containing \(1\) and any open set \(V\) containing \(2\). Since \(U\) is nonempty and open, \(X\setminus U\) is finite; since \(V\) is nonempty and open, \(X\setminus V\) is finite.
If \(U\cap V\) were empty, then every natural number would lie outside \(U\) or outside \(V\), giving \(X=(X\setminus U)\cup(X\setminus V)\). The right side is finite, being a union of two finite sets, but \(X=\mathbb{N}\) is infinite. This is impossible. Therefore \(U\cap V\ne\varnothing\) for every such pair \(U,V\), so no disjoint open neighborhoods separate \(C\) and \(F\). This is a direct failure of normality, despite the space being T1.
For comparison, if \(X\) is finite, every subset has finite complement. The cofinite topology is then the discrete topology: every subset is open. By the discrete-space result, a finite set with the cofinite topology is normal and T4. The infinitude assumption in the proposition is therefore essential.
Normality Does Not Automatically Give T4
A frequent error is to treat normality and T4 as synonyms without checking the convention. Under the convention used here, the T1 condition is an additional requirement. The indiscrete topology supplies a concrete example showing why it cannot simply be omitted.
Worked Example: A Normal Space That Is Not T4
Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\). Its closed sets, obtained by taking complements of open sets, are also just \(\varnothing\) and \(X\). Among these closed sets, the only disjoint pairs are \((\varnothing,\varnothing)\), \((\varnothing,X)\), and \((X,\varnothing)\).
For the pair \((\varnothing,\varnothing)\), choose the disjoint open sets \(U=V=\varnothing\). For \((\varnothing,X)\), choose \(U=\varnothing\) and \(V=X\). For \((X,\varnothing)\), choose \(U=X\) and \(V=\varnothing\). Each pair contains the corresponding closed sets and is disjoint. These are all disjoint closed pairs, so the space is normal.
However, \(\{a\}\) is not closed: its complement \(\{b\}\) is not open in the indiscrete topology. Thus the space is not T1, and consequently it is not T4. This example shows that checking only the closed-set separation clause does not establish T4.
Common pitfall. In an infinite cofinite space, it is wrong to assume that a singleton is an open neighborhood of its point and use two different singletons as disjoint neighborhoods. For example, in the cofinite topology on \(\mathbb{N}\), \(\{1\}\) is closed, not open: its complement is infinite, so it is not cofinite. The open set \(\mathbb{N}\setminus\{2\}\) does contain \(1\), but it also intersects every nonempty cofinite open neighborhood of \(2\). The correction is to check openness in the stated topology, rather than importing the open sets of the discrete topology.
The main distinction can be kept in view with two questions. First, can every pair of disjoint closed sets be placed in disjoint open supersets? If so, the space is normal. Second, is every singleton closed? If so, it is T1. Both answers must be yes for the T4 axiom used here.
Check Your Understanding
For each question, use the definitions of normality and T1 directly.
- Why does the discrete topology on any set make every pair of disjoint closed sets separable by open sets?
- In the cofinite topology on an infinite set, prove that two distinct singleton sets are closed but cannot have disjoint open neighborhoods.
- Is the indiscrete topology on a set with exactly one point T1? Is it normal? Justify both answers.
- What changes about the cofinite topology when the underlying set is finite?
- Explain why separating one particular pair of closed sets in \(\mathbb{R}\) does not, on its own, prove that \(\mathbb{R}\) is normal.