Two Different Separation Questions
Separation axioms describe ways to keep points and sets apart using open sets or continuous functions. Two of the central questions are these: can a point be separated from a closed set by a continuous function, and can two disjoint closed sets be separated by disjoint open sets? The first is complete regularity; the second is normality. They are related, but they are not the same condition.
Throughout, a topological space is written \((X,\mathcal{T})\). A space is \(T_1\) if every singleton is closed. A continuous function to \([0,1]\) means a continuous map to that interval with its usual subspace topology from \(\mathbb{R}\).
These definitions use different kinds of separation data. Complete regularity asks for a function for a point and a closed set; normality asks for open neighborhoods for two closed sets. The \(T_1\) requirement is included explicitly in our definition of complete regularity and therefore in Tychonoffness. Normality, as defined here, does not include \(T_1\).
Normal \(T_1\) Spaces Are Tychonoff
The key connection is Urysohn’s lemma. In the earlier material on Urysohn’s lemma, the result was proved: if \(X\) is normal and \(F,G\) are disjoint closed subsets of \(X\), then there is a continuous \(f:X\to[0,1]\) with \(f=0\) on \(F\) and \(f=1\) on \(G\). We apply that result to a singleton and a closed set.
Proof. Let \(X\) be normal and \(T_1\), and let \(F\subseteq X\) be closed. If \(x\in X\setminus F\), then \(\{x\}\) and \(F\) are disjoint closed sets: \(\{x\}\) is closed by the \(T_1\) hypothesis, and disjointness follows from \(x\notin F\). Urysohn’s lemma gives a continuous \(f:X\to[0,1]\) with \(f(x)=0\) and \(f(y)=1\) for all \(y\in F\). This is exactly the required point–closed-set separation. The \(T_1\) condition is already part of the assumptions, so \(X\) is completely regular. If \(X\) has no points, the point–closed-set condition is vacuous and the conclusion still holds. \(\square\)
The role of \(T_1\) is specific: it ensures that the singleton used as one of Urysohn’s two closed sets really is closed. Normality alone, under the definition above, does not provide that fact.
Worked Example: A Normal Space That Is Not Tychonoff
Let \(X=\{a,b\}\) have the Sierpinski topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Its closed sets, obtained by taking complements of its open sets, are \(\varnothing,\{b\},X\).
Normality. Two disjoint closed sets cannot both be nonempty: the only nonempty proper closed set is \(\{b\}\), and it intersects itself; \(X\) intersects every nonempty set. Thus, in any disjoint pair, at least one set is empty. If \(F=\varnothing\), take \(U=\varnothing\) and \(V=X\); these are disjoint open sets containing \(F\) and the other closed set. If \(G=\varnothing\), take \(U=X\) and \(V=\varnothing\). So \(X\) is normal by the definition.
Failure of \(T_1\). The singleton \(\{a\}\) is not closed, since its complement \(\{b\}\) is not open. Therefore \(X\) is not \(T_1\), and it is not Tychonoff under our definition. This example shows why normality, without a \(T_1\) assumption, does not imply Tychonoffness.
Check. The example does not contradict the theorem: the theorem assumes both normality and \(T_1\), and this space fails the latter condition.
Complete Regularity Implies Regularity
Complete regularity also gives a familiar open-set separation property. A space is regular here if it is \(T_1\) and, whenever \(x\notin F\) for a closed set \(F\), there are disjoint open sets containing \(x\) and \(F\), respectively.
Proof. Let \(X\) be completely regular, let \(F\subseteq X\) be closed, and let \(x\notin F\). Choose a continuous \(f:X\to[0,1]\) with \(f(x)=0\) and \(f(y)=1\) for every \(y\in F\). Put \(U=f^{-1}([0,1/2))\) and \(V=f^{-1}((1/2,1])\). Both intervals are open in the subspace \([0,1]\), so \(U\) and \(V\) are open in \(X\). Also \(x\in U\), \(F\subseteq V\), and \(U\cap V=\varnothing\), since no value of \(f\) can be both less than and greater than \(1/2\). The space is \(T_1\) by complete regularity, so it is regular. \(\square\)
Thus, with the stated conventions, normal \(T_1\) spaces are completely regular, and completely regular spaces are regular. The first implication uses Urysohn’s lemma; the second follows directly from the separating function.
Worked Example: A Finite Discrete Space
Let \(X=\{p,q,r\}\) have the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open and closed. We verify both separation properties and construct the point–closed-set function explicitly.
For disjoint closed sets \(F,G\subseteq X\), choose \(U=F\) and \(V=G\). Since all subsets are open, these are open; they are disjoint by assumption and contain the respective closed sets. Hence \(X\) is normal. Every singleton is closed, so \(X\) is \(T_1\).
Now take any closed \(F\subseteq X\) and any \(x\notin F\). Define \(f:X\to[0,1]\) by \(f(x)=0\) and \(f(y)=1\) for all \(y\ne x\). For any open \(O\subseteq[0,1]\), the preimage \(f^{-1}(O)\) is one of \(\varnothing,\{x\},X\setminus\{x\},X\), depending on whether \(O\) contains 0, 1, both, or neither. Each is open in the discrete topology, so \(f\) is continuous. Since \(x\notin F\), every \(y\in F\) differs from \(x\), and therefore \(f(y)=1\). This verifies complete regularity.
Check. The same \(f\) separates \(x\) from \(F\), including when \(F=\varnothing\). If \(F=\varnothing\), the requirement \(f(y)=1\) for every \(y\in F\) has no points to check, while \(f(x)=0\) still holds.
Complete Regularity Does Not Imply Normality
The converse to the theorem fails: a Tychonoff space need not be normal. A standard example is the Sorgenfrey plane, the product \(\mathbb{R}_{\ell}\times\mathbb{R}_{\ell}\) of two lower-limit lines. In \(\mathbb{R}_{\ell}\), basic open sets are intervals \([a,b)\) with \(a<b\); the plane has the product topology, whose basis consists of rectangles \([a,b)\times[c,d)\).
Worked Example: The Sorgenfrey Plane Is Tychonoff but Not Normal
First verify complete regularity directly. Each interval \([a,b)\) is open by the definition of the lower-limit topology. It is also closed: its complement is \((-\infty,a)\cup[b,\infty)\), where \((-\infty,a)=\bigcup_{t<a}[t,a)\) and \([b,\infty)=\bigcup_{t\ge b}[t,t+1)\); both pieces are open. Consequently every basic rectangle in the Sorgenfrey plane is clopen, meaning both closed and open.
The plane is \(T_1\). Given distinct points \((a,b)\) and \((c,d)\), at least one coordinate differs. If \(a\ne c\), choose a basic interval \([a,a+\delta)\) that excludes \(c\): when \(c>a\), choose \(0<\delta<c-a\), and when \(c<a\), every such interval excludes \(c\). A basic rectangle around \((a,b)\) using this interval in the first coordinate excludes \((c,d)\). If \(a=c\), use the same construction in the second coordinate, since then \(b\ne d\). Thus for any two distinct points there is an open neighborhood of the first missing the second, which implies that every singleton is closed.
Now let \(F\) be closed and let \(x\notin F\). The open set \(X\setminus F\) contains \(x\), so the rectangle basis gives a basic rectangle \(R\) with \(x\in R\subseteq X\setminus F\). This \(R\) is clopen. Define \(f:X\to[0,1]\) by \(f=0\) on \(R\) and \(f=1\) on \(X\setminus R\). It is continuous: for any open \(O\subseteq[0,1]\), its preimage is one of \(\varnothing,R,X\setminus R,X\), all open. Also \(f(x)=0\) and \(f=1\) on \(F\). This proves complete regularity.
The Sorgenfrey plane is not normal; this is the nonnormality result for that space established in the earlier material on the Sorgenfrey plane. Together, that result and the verification above give a Tychonoff space that is not normal. Hence complete regularity and normality are not interchangeable, even though normality together with \(T_1\) implies complete regularity.
A Common Mistake
Common pitfall. One might argue, “A completely regular space separates points from closed sets, so it separates any two disjoint closed sets; therefore it is normal.” The conclusion does not follow from the stated separation condition. That condition supplies a function for each chosen point and a closed set, not a single pair of disjoint open neighborhoods for two arbitrary closed sets. For example, in the Sorgenfrey plane the point–closed-set condition holds by the clopen-rectangle construction above, but the plane is not normal. The mistaken step is replacing a point by an entire closed set without a justification that produces disjoint open neighborhoods for the two sets.
Keep the hypotheses and tasks distinct:
| Property | What must be separated? | Separation supplied |
|---|---|---|
| Regular | A point and a closed set not containing it | Disjoint open neighborhoods |
| Completely regular | A point and a closed set not containing it | A continuous function to \([0,1]\), with values 0 and 1 |
| Normal | Two disjoint closed sets | Disjoint open neighborhoods |
| Tychonoff | A point and a closed set, with \(T_1\) | Complete regularity, including \(T_1\) |
Check Your Understanding
For each question, identify which definition or implication supplies the relevant separation.
- Why does the proof that normal \(T_1\) spaces are completely regular require the singleton \(\{x\}\) to be closed?
- In the Sierpinski example, which closed sets can form a disjoint pair, and how does that verify normality?
- Explain why the preimage of an open subset of \([0,1]\) under the finite discrete example’s function is open.
- In the proof that complete regularity implies regularity, why are \(f^{-1}([0,1/2))\) and \(f^{-1}((1/2,1])\) disjoint open sets?
- Which implication fails in the Sorgenfrey-plane example, and which earlier result supplies the needed nonnormality claim?