Complete Regularity via Point-Closed Set Separation

Separating a Point from a Closed Set

Disjoint open neighborhoods separate sets topologically; a continuous real-valued function can separate them numerically. For a point \(x\) and a closed set \(F\) not containing it, the useful goal is a function that takes one value at \(x\) and a different value everywhere on \(F\). Complete regularity expresses the existence of such functions for every point and closed set. Urysohn’s lemma supplies them when the space is normal and \(T_1\).

Throughout, \(X\) has topology \(\mathcal{T}\). A space is \(T_1\) if every singleton is closed. It is normal if every two disjoint closed sets have disjoint open sets containing them, respectively. Here normality does not include \(T_1\) as part of its definition.

Definition: The space \(X\) is completely regular if, whenever \(F\subseteq X\) is closed and \(x\in X\setminus F\), there is a continuous function \(f:X\to[0,1]\) such that \(f(x)=0\) and \(f(y)=1\) for every \(y\in F\). The requirement is vacuous when there is no such pair \(x,F\).

The prescribed range and values matter: the function is zero at the point and one on the entire closed set, not merely at one selected point of that set. The \(T_1\) axiom is not included in this definition. When complete regularity is paired with \(T_1\), the resulting separation property is often called the Tychonoff property.

A Local Formulation

Proposition (Local function criterion): A space \(X\) is completely regular if and only if, for every open set \(U\) and every \(x\in U\), there is a continuous \(f:X\to[0,1]\) with \(f(x)=0\) and \(f(y)=1\) for every \(y\in X\setminus U\).

Proof. Suppose first that \(X\) is completely regular. For an open set \(U\), its complement \(X\setminus U\) is closed. Since \(x\in U\), we have \(x\notin X\setminus U\). Apply complete regularity to \(x\) and \(F=X\setminus U\) to obtain the required function. Conversely, suppose the stated local condition holds. Given a closed set \(F\) and \(x\in X\setminus F\), the set \(U=X\setminus F\) is open and contains \(x\). The local condition gives a continuous \(f\) with \(f(x)=0\) and \(f=1\) on \(X\setminus U=F\). This is exactly complete regularity. \(\square\)

This formulation shows how to use complete regularity around an open neighborhood: the function can be made to take its larger value everywhere outside that neighborhood. A further consequence is that the point has an open neighborhood whose closure remains inside \(U\).

Corollary: If \(X\) is completely regular, \(U\) is open, and \(x\in U\), then there is an open set \(V\) such that \(x\in V\) and \(\overline{V}\subseteq U\).

Proof. Choose \(f\) from the local function criterion and put \(V=f^{-1}([0,1/2))\). The interval \([0,1/2)\) is open in the subspace \([0,1]\), because it is the intersection of \([0,1]\) with the open interval \((-1,1/2)\) in \(\mathbb{R}\). Continuity of \(f\) therefore makes \(V\) open, and \(f(x)=0\) gives \(x\in V\). Also \(V\subseteq f^{-1}([0,1/2])\). The latter set is closed in \(X\), since \([0,1/2]\) is closed in \([0,1]\) and \(f\) is continuous. Thus \(\overline{V}\subseteq f^{-1}([0,1/2])\). On \(X\setminus U\), the function equals \(1\), so \(f^{-1}([0,1/2])\subseteq U\). Combining these inclusions gives \(\overline{V}\subseteq U\). \(\square\)

Normal \(T_1\) Spaces Are Completely Regular

The key input is Urysohn’s lemma, in the form that applies to two disjoint closed sets.

Urysohn’s lemma: If \(X\) is normal and \(A,B\subseteq X\) are disjoint closed sets, then there is a continuous function \(u:X\to[0,1]\) such that \(u(a)=0\) for every \(a\in A\) and \(u(b)=1\) for every \(b\in B\). The construction and continuity proof are part of the earlier material on Urysohn’s lemma.

Theorem: Every normal \(T_1\) space is completely regular.

Proof. Let \(F\) be closed in \(X\), and let \(x\in X\setminus F\). Since \(X\) is \(T_1\), the singleton \(\{x\}\) is closed. It is disjoint from \(F\), because \(x\notin F\). Normality allows Urysohn’s lemma to be applied to the disjoint closed sets \(A=\{x\}\) and \(B=F\). It gives a continuous \(u:X\to[0,1]\) with \(u(x)=0\) and \(u(y)=1\) for every \(y\in F\). This is the function required by the definition of complete regularity. The argument also covers \(F=\varnothing\): the two closed sets remain disjoint, and the condition on values on \(F\) is vacuous. \(\square\)

Each hypothesis has a distinct role. Normality supplies the continuous function for two disjoint closed sets; \(T_1\) ensures that the point itself is a closed set to which Urysohn’s lemma applies. Without \(T_1\), normality alone does not guarantee complete regularity.

Worked Applications

Worked Example: Separating a Point from an Interval in \(\mathbb{R}\)

Give an explicit continuous function that separates \(x=0\) from the closed set \(F=[2,3]\) in \(\mathbb{R}\) with its standard topology.

Define \(f:\mathbb{R}\to[0,1]\) by \(f(t)=\min(1,|t|)\). The absolute-value function is continuous, and taking the minimum with the constant function \(1\) preserves continuity. Equivalently, \(\min(a,b)=(a+b-|a-b|)/2\), so the formula uses only continuous operations. The values lie in \([0,1]\) because \(|t|\geq 0\), and the minimum is at most \(1\).

At the chosen point, \(f(0)=\min(1,|0|)=\min(1,0)=0\). If \(t\in F=[2,3]\), then \(2\leq t\leq3\), hence \(|t|=t\geq2>1\). Therefore \(f(t)=\min(1,|t|)=1\) for every \(t\in F\). The function has the required values on the point and the whole closed set.

Check: At the endpoint \(t=2\), the formula gives \(f(2)=\min(1,2)=1\); at \(t=3\), it gives \(f(3)=\min(1,3)=1\). At \(t=0\), it gives \(0\), as required.

Worked Example: Separation in a Finite Discrete Space

Let \(X=\{a,b,c,d\}\) have the discrete topology, in which every subset is open. Separate \(x=b\) from \(F=\{a,c\}\).

The set \(F\) is closed because its complement \(\{b,d\}\) is open. Define \(f:X\to[0,1]\) by \(f(a)=1\), \(f(b)=0\), \(f(c)=1\), and \(f(d)=1/2\). These values all lie in \([0,1]\), and \(f(b)=0\) while \(f(a)=f(c)=1\). Thus \(f\) equals \(1\) at every point of \(F\).

To check continuity, take any open set \(O\subseteq[0,1]\). Its preimage \(f^{-1}(O)\) is some subset of \(X\), and every subset of \(X\) is open in the discrete topology. Hence \(f\) is continuous. This verifies the separation directly from the definition rather than relying only on the general theorem.

Check: For example, \(f^{-1}(\{1\})=\{a,c\}=F\), \(f^{-1}(\{0\})=\{b\}\), and \(f^{-1}(\{1/2\})=\{d\}\); all three sets are open in \(X\).

Worked Example: Why the \(T_1\) Hypothesis Matters

Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). This space is normal but not \(T_1\), and it is not completely regular.

The closed sets are the complements of the open sets, namely \(\varnothing,\{b\},X\). Any two disjoint closed sets must include \(\varnothing\), since the only two nonempty closed sets are \(\{b\}\) and \(X\), which are not disjoint. For a pair \(\varnothing,C\) of disjoint closed sets, the open sets \(\varnothing\) and \(X\) are disjoint neighborhoods of \(\varnothing\) and \(C\), respectively. The same choice with the roles reversed handles \(C,\varnothing\). Thus \(X\) is normal. It is not \(T_1\), since \(\{a\}\) is not closed: its complement \(\{b\}\) is not open.

Now take the closed set \(F=\{b\}\) and the point \(x=a\notin F\). Suppose there were a continuous \(f:X\to[0,1]\) with \(f(a)=0\) and \(f(b)=1\). The open set \((1/2,3/2)\cap[0,1]\) in \([0,1]\) contains \(f(b)=1\) but not \(f(a)=0\). Its preimage under \(f\) would be \(\{b\}\). Continuity would require this preimage to be open in \(X\), but \(\{b\}\notin\mathcal{T}\). This contradiction shows that no such function exists, so \(X\) is not completely regular.

Check: The failure is not that \(\{b\}\) is unsuitable as a closed set—it is closed. Rather, the singleton \(\{a\}\) is not closed, so the normal \(T_1\) argument cannot apply Urysohn’s lemma to \(\{a\}\) and \(F\).

A Common Pitfall

A tempting but invalid argument is: “Given \(x\notin F\), the sets \(\{x\}\) and \(F\) are disjoint and closed, so Urysohn’s lemma gives a separating function.” Disjointness is not enough: both sets must be closed. The example above has \(x=a\) and \(F=\{b\}\), with \(F\) closed but \(\{a\}\) not closed. Normality can still hold because its condition only concerns pairs that are already closed. The \(T_1\) hypothesis repairs precisely this gap by making every singleton closed.

Keep the logical order explicit: start with a point \(x\) outside a closed set \(F\); use \(T_1\) to establish that \(\{x\}\) is closed; check that \(\{x\}\cap F=\varnothing\); and only then apply Urysohn’s lemma. Reversing or omitting one of these checks can turn a valid separation proof into an unsupported claim.

Check Your Understanding

For each question, identify the hypotheses needed before choosing a separating function.

State the definition of complete regularity, including the range and prescribed values of the function.
Explain why \(X\setminus U\) is the appropriate closed set to use in the local function criterion when \(x\in U\).
In the proof that a normal \(T_1\) space is completely regular, where is each of normality and \(T_1\) used?
For the discrete space in the worked example, verify continuity by describing the preimage of an arbitrary open subset of \([0,1]\).
In the two-point example, which preimage contradicts continuity for a function with \(f(a)=0\) and \(f(b)=1\)?