Why the \(T_3\) and \(T_4\) Labels Need Care
Separation axioms are often described by asking whether certain sets can be placed inside disjoint open sets. Regularity asks about a point and a closed set not containing it; normality asks about two disjoint closed sets. The distinction matters: a point need not be a closed set. Thus normality does not automatically give the point–closed-set separation used in regularity unless an additional condition is present.
There are two common ways to package that additional condition. One convention defines “regular” and “normal” without requiring \(T_1\), then defines \(T_3\) and \(T_4\) by adding \(T_1\). Another convention builds \(T_1\) into the words “regular” and “normal,” so that those terms already mean \(T_3\) and \(T_4\), respectively. The underlying separation statements are the same; what changes is where the \(T_1\) requirement is stated.
Empty sets are allowed among the closed sets in these definitions. For example, if \(F=\varnothing\), the weak regularity condition can be met with \(U=X\) and \(V=\varnothing\). If one of \(C,F\) is empty in the weak normality condition, the same choice of open sets separates them. Thus these cases do not require an extra hypothesis.
The \(T_1\) Condition and Singletons
The key fact is that \(T_1\) turns the point in a point–closed-set pair into a closed set. More precisely, \(X\) is \(T_1\) exactly when every singleton \(\{x\}\) is closed. To see the relation with the usual point-separation definition of \(T_1\), suppose every singleton is closed. For distinct \(x,y\), the set \(X\setminus\{y\}\) is open, contains \(x\), and does not contain \(y\); the set \(X\setminus\{x\}\) is open, contains \(y\), and does not contain \(x\). Conversely, if for each distinct \(x,y\) there is an open set containing \(x\) but not \(y\), then for each fixed \(y\), the union of all such open sets as \(x\) ranges over \(X\setminus\{y\}\) is \(X\setminus\{y\}\). This union is open, so \(\{y\}\) is closed.
Proof. Assume \(X\) is weakly normal and \(T_1\). To prove weak regularity, take any closed set \(F\) and any point \(x\in X\setminus F\). Since \(X\) is \(T_1\), the singleton \(\{x\}\) is closed. The sets \(\{x\}\) and \(F\) are disjoint because \(x\notin F\). Weak normality therefore gives disjoint open sets \(U,V\) such that \(\{x\}\subseteq U\) and \(F\subseteq V\). The first containment says \(x\in U\), so these are precisely the open neighborhoods required by weak regularity. This works also when \(F=\varnothing\), since weak normality applies to the disjoint closed sets \(\{x\}\) and \(\varnothing\). Thus \(X\) is weakly regular. The \(T_1\) condition is already part of the assumption, so \(X\) is \(T_3\) under the stated convention. \(\square\)
The proof identifies exactly where \(T_1\) is used: it makes \(\{x\}\) closed, so weak normality can be applied to \(\{x\}\) and \(F\). Without that condition, the point may not be a closed set, and the defining hypothesis of normality may not apply to the pair in question.
Worked Example: A Normal Space That Is Not Regular
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Its closed sets, obtained by taking complements of the open sets, are \(\varnothing,\{b\},X\).
Check weak normality. The only disjoint pairs of closed sets are pairs in which at least one set is empty: the two nonempty closed sets \(\{b\}\) and \(X\) are not disjoint, and \(\{b\}\) cannot be paired disjointly with itself. For any disjoint closed pair with one member empty, take the open set \(X\) around the nonempty member and \(\varnothing\) around the empty member; if both are empty, take \(\varnothing\) for both. The chosen open sets are disjoint and contain the required closed sets. Hence \(X\) is weakly normal.
Check failure of weak regularity. The point \(a\) is not in the closed set \(\{b\}\). The open neighborhoods of \(a\) are \(\{a\}\) and \(X\), while the only open set containing \(\{b\}\) is \(X\). Every open neighborhood of \(\{b\}\) therefore intersects every open neighborhood of \(a\): the intersection with \(\{a\}\) is \(\{a\}\), and the intersection with \(X\) is \(X\). There are no disjoint open neighborhoods of \(a\) and \(\{b\}\). Thus \(X\) is not weakly regular.
Finally, \(\{a\}\) is not closed, since its complement \(\{b\}\) is not open. The example is not \(T_1\), so it does not contradict the theorem. It shows that weak normality alone does not imply weak regularity.
How the Conventions Translate
It is useful to keep the separation condition and the \(T_1\) condition distinct when reading a definition. The following table records the two common packages. “Point–closed separation” and “closed-set separation” refer exactly to the conditions in the definition box above.
| Label | Separation condition | Additional condition |
|---|---|---|
| Weak regularity | Separate a point from a closed set not containing it | None included in the label |
| \(T_3\), separate-condition convention | Point–closed separation | \(T_1\) |
| Regularity, bundled convention | Point–closed separation | \(T_1\) is included in “regular” |
| Weak normality | Separate two disjoint closed sets | None included in the label |
| \(T_4\), separate-condition convention | Closed-set separation | \(T_1\) |
| Normality, bundled convention | Closed-set separation | \(T_1\) is included in “normal” |
Under the bundled convention, the theorem can be stated briefly as “normal implies regular,” because the \(T_1\) assumption is already part of the word “normal.” Under the separate-condition convention, the precise statement is “weakly normal plus \(T_1\) implies weakly regular.” The difference is terminological, not a difference in the proof.
Worked Example: The Indiscrete Topology
Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). The closed sets are also \(\varnothing\) and \(X\), because each is the complement of an open set.
Check weak normality. Two disjoint closed sets cannot both be nonempty: the only nonempty closed set is \(X\), and \(X\cap X=X\ne\varnothing\). Thus every disjoint pair has an empty member. If \(C=\varnothing\) and \(F=X\), choose \(U=\varnothing\) and \(V=X\); if \(C=X\) and \(F=\varnothing\), choose \(U=X\) and \(V=\varnothing\); and if both are empty, choose \(U=V=\varnothing\). Each pair of chosen open sets is disjoint and contains the corresponding closed sets. So the space is weakly normal.
Check weak regularity. The only closed set \(F\) for which there is a point \(x\notin F\) is \(F=\varnothing\). For any \(x\in X\), choose \(U=X\) and \(V=\varnothing\); these are disjoint open sets with \(x\in U\) and \(F\subseteq V\). Hence the space is weakly regular.
Nevertheless, \(X\) is not \(T_1\): the singleton \(\{a\}\) is not closed, since its complement \(\{b\}\) is not open. Under the convention that requires \(T_1\) for \(T_3\) and \(T_4\), this space is neither \(T_3\) nor \(T_4\). This example shows why the weak separation properties and the \(T_3,T_4\) labels must not be conflated.
Checking Both Axioms in a \(T_1\) Example
Worked Example: A Two-Point Discrete Space
Let \(X=\{a,b\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)=\{\varnothing,\{a\},\{b\},X\}\). Every subset is open, and every subset is closed because its complement is also open. In particular, both singletons are closed, so \(X\) is \(T_1\).
Check weak regularity. Let \(F\) be closed and let \(x\notin F\). Choose \(U=\{x\}\) and \(V=F\). Both sets are open, they are disjoint because \(x\notin F\), and they contain \(x\) and \(F\), respectively. This also works when \(F=\varnothing\), since then \(V=\varnothing\).
Check weak normality. Let \(C,F\) be disjoint closed sets. Choose \(U=C\) and \(V=F\). Both are open, their intersection is \(C\cap F=\varnothing\), and each contains the corresponding closed set. This also covers an empty member of the pair. Therefore \(X\) is weakly normal and weakly regular; since it is \(T_1\), it is both \(T_3\) and \(T_4\) under the separate-condition convention.
This example is a direct check of the definitions, not a claim that \(T_3\) and \(T_4\) are equivalent in general. The implication established above goes from \(T_4\) to \(T_3\).
A Common Pitfall
Wrong argument: “The Sierpinski space is normal, so for the point \(a\) and the closed set \(\{b\}\), normality gives disjoint open neighborhoods.” This calculation silently treats \(\{a\}\) as a closed set. In the topology \(\{\varnothing,\{a\},\{a,b\}\}\), the complement of \(\{a\}\) is \(\{b\}\), which is not open; hence \(\{a\}\) is not closed. Normality only guarantees separation for pairs of disjoint closed sets, so it cannot be applied to \(\{a\}\) and \(\{b\}\).
Correction: first check whether the point singleton is closed. If the space is \(T_1\), it is, and the proof that normality implies regularity applies. If the space is not known to be \(T_1\), weak normality alone does not supply the required point–closed-set neighborhoods; the Sierpinski example verifies that the desired separation can fail.
The practical rule is to expand the labels before using an implication: write \(T_4\) as closed-set separation plus \(T_1\), and write \(T_3\) as point–closed-set separation plus \(T_1\). Then the proof is a single valid application of normality to the closed sets \(\{x\}\) and \(F\), with \(T_1\) justifying that \(\{x\}\) is closed.
Check Your Understanding
In each question, state whether you are using weak regularity or regularity with \(T_1\) included.
- In the Sierpinski topology \(\{\varnothing,\{a\},\{a,b\}\}\), which singleton is closed? Use the open neighborhoods to test weak regularity.
- Explain the exact line in the proof that \(T_4\) implies \(T_3\) where \(T_1\) is required.
- In the indiscrete topology on \(\{a,b\}\), verify that every disjoint pair of closed sets has an empty member. Why does this not make the space \(T_4\) under the separate-condition convention?
- Show directly from the definitions that a discrete space is weakly regular and weakly normal.
- Suppose a text defines “normal” to include \(T_1\). Translate “normal implies regular” into the convention that states \(T_1\) separately.