From Separation to a Smaller Neighborhood
Regularity is usually expressed as a separation property: a point and a closed set not containing it can be placed in disjoint open sets. The closure-shrinking criterion gives a local version of the same idea. Instead of choosing two separating open sets at once, it asks whether each point inside an open set has a smaller open neighborhood whose closure still fits inside the original set.
Throughout, \((X,\mathcal{T})\) is a topological space. A set is closed if its complement is open. The closure \(\overline{A}\) of \(A\subseteq X\) is the intersection of all closed subsets of \(X\) containing \(A\). Thus \(\overline{A}\) is closed, contains \(A\), and is contained in every closed set that contains \(A\).
One elementary consequence of the closure definition will be used in both directions. If \(U\) and \(W\) are disjoint open sets, then \(X\setminus W\) is closed and contains \(U\). Therefore \(\overline{U}\subseteq X\setminus W\). In particular, \(\overline{U}\cap W=\varnothing\). This explains how separation can force a closure to stay away from a specified open set.
The Closure-Shrinking Criterion
Proof. First suppose \(X\) is regular. Let \(V\) be open and \(x\in V\). The set \(F=X\setminus V\) is closed, and \(x\notin F\). By regularity, there are disjoint open sets \(U,W\) such that \(x\in U\) and \(F\subseteq W\). Since \(U\cap W=\varnothing\), the closure observation gives \(\overline{U}\subseteq X\setminus W\). The inclusion \(F\subseteq W\) implies \(X\setminus W\subseteq X\setminus F=V\). Also \(U\subseteq\overline{U}\), so \(x\in U\subseteq\overline{U}\subseteq V\).
Conversely, suppose the stated shrinking condition holds. Let \(F\) be closed and \(x\notin F\), and set \(V=X\setminus F\), which is open and contains \(x\). Choose an open \(U\) with \(x\in U\) and \(\overline{U}\subseteq V\). Set \(W=X\setminus\overline{U}\). Because \(\overline{U}\) is closed, \(W\) is open. The inclusion \(\overline{U}\subseteq X\setminus F\) implies \(F\subseteq W\), while \(U\subseteq\overline{U}\) implies \(U\cap W=\varnothing\). Thus \(U,W\) separate \(x\) and \(F\), proving regularity. If \(X\) is empty, both conditions hold vacuously, so the equivalence also covers that case. \(\square\)
The two directions use complementary constructions. For the forward direction, regularity separates \(x\) from the closed complement of \(V\); the open neighborhood around \(x\) cannot have closure meeting the open set around that complement. For the reverse direction, the complement of the smaller neighborhood’s closure is the open set that separates the entire closed set from \(x\).
Worked Examples
Worked Example: Shrinking an Interval in the Real Line
Give a concrete closure-shrinking neighborhood for \(x=2\) inside \(V=(1,5)\) in \(\mathbb{R}\) with its standard topology.
Take \(U=(1.5,2.5)\). It is open and contains \(2\), since \(1.5<2<2.5\). Its closure is \([1.5,2.5]\): the interval is closed and contains \(U\), and every open interval about either endpoint meets \(U\), so no closed set containing \(U\) can omit either endpoint. Finally, \(1<1.5\le y\le2.5<5\) for every \(y\in[1.5,2.5]\). Therefore $$ 2\in(1.5,2.5)\subseteq[1.5,2.5]\subseteq(1,5). $$ Check: the left and right endpoints of the closure are \(1.5\) and \(2.5\); both lie strictly between \(1\) and \(5\), as required.
More generally, if \(x\in V\) and \(V\) is open in the standard topology, there is an \(r>0\) such that \((x-r,x+r)\subseteq V\). The open interval \(U=(x-r/2,x+r/2)\) has closure \([x-r/2,x+r/2]\), which is contained in \((x-r,x+r)\). Thus the criterion holds throughout the real line. This is an instance of the criterion, not a proof that every metric space is regular.
Worked Example: The Discrete Topology
Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open. For any \(V\subseteq X\) and \(x\in V\), choose \(U=\{x\}\). Every subset is also closed, so \(\overline{U}=\{x\}\). Hence \(x\in U=\overline{U}\subseteq V\), including when \(V=\{x\}\).
For a direct separation check, let \(F\subseteq X\) be closed and \(x\notin F\). The open sets \(\{x\}\) and \(F\) are disjoint, contain \(x\) and \(F\), respectively, and meet the definition of regularity. If \(F=\varnothing\), the choice \(W=\varnothing\) works. Thus the criterion and regularity agree in this example, and \(X\) is also \(T_1\) because each singleton is closed.
Worked Example: An Infinite Cofinite Topology
Let \(X=\mathbb{Z}\) with the cofinite topology: the open sets are \(\varnothing\) and the subsets whose complements in \(X\) are finite. The closed sets are \(X\) and the finite subsets. Every singleton is closed, so this space is \(T_1\).
Take \(x=0\) and \(V=X\setminus\{1\}\). This \(V\) is open and contains \(0\). Any open \(U\) containing \(0\) is nonempty and cofinite, hence infinite. A closed set containing \(U\) cannot be finite, so the only closed set containing \(U\) is \(X\). By the definition of closure, \(\overline{U}=X\). Consequently no such \(U\) can satisfy \(\overline{U}\subseteq V\), since \(1\in X\) but \(1\notin V\). The criterion fails.
In terms of separation, the same point and closed set cannot be separated: use \(F=\{1\}\). Every open neighborhood of \(0\) is cofinite, and every open neighborhood of \(1\) is cofinite. Two cofinite subsets of the infinite set \(\mathbb{Z}\) intersect, since their union of finite complements is finite and cannot exhaust \(\mathbb{Z}\). So disjoint separating neighborhoods do not exist. This example shows that \(T_1\) alone does not guarantee regularity.
Worked Example: The Indiscrete Topology and the \(T_1\) Condition
Let \(X=\{a,b\}\) with the indiscrete topology \(\{\varnothing,X\}\). Its only closed sets are also \(\varnothing\) and \(X\). The only nonempty open set \(V\) is \(X\); for either \(x\in X\), choose \(U=X\). Then \(\overline{U}=X\), so \(x\in U\subseteq\overline{U}\subseteq V\). The shrinking condition holds.
This space is not \(T_1\), because \(\{a\}\) is not closed: its complement \(\{b\}\) is not open. In the point–closed-set definition of regularity, the only closed set not containing \(x\) is \(\varnothing\), and it can be separated from \(x\) using \(U=X\) and \(W=\varnothing\). Thus the space is regular under that definition, but it is not \(T_3\). The example shows why the criterion characterizes regularity but does not, by itself, supply the extra \(T_1\) condition in \(T_3\).
A Common Pitfall: Forgetting the Separation Convention
A frequent error is to say that the closure-shrinking condition is equivalent to \(T_3\) without checking the convention for regularity. The indiscrete example gives a concrete correction: it satisfies closure shrinking, but fails \(T_1\). Therefore the precise statement is that closure shrinking is equivalent to point–closed-set regularity. If \(T_3\) means regular plus \(T_1\), then \(T_3\) is equivalent to closure shrinking together with \(T_1\).
Another useful check is that the inclusion must concern the closure, not just the smaller open set. In the cofinite example, \(U=V=X\setminus\{1\}\) is an open neighborhood of \(0\) contained in \(V\), but \(\overline{U}=X\), which is not contained in \(V\). The condition \(U\subseteq V\) alone does not express the needed separation: it is \(\overline{U}\subseteq V\) that ensures the complementary open set \(X\setminus\overline{U}\) can separate \(U\) from the closed set.
The criterion is useful because it converts a global separation question into a local neighborhood choice. It is especially convenient when open sets come with a concrete way to produce smaller ones. The argument itself requires no metric, countability condition, or \(T_1\) assumption; those hypotheses matter only when a particular application or a convention such as \(T_3\) requires them.
Check Your Understanding
For each question, identify whether it concerns regularity, \(T_1\), or both.
- In the theorem’s forward direction, which closed set is separated from \(x\) when \(x\in V\)?
- Explain why disjoint open sets \(U,W\) imply \(\overline{U}\subseteq X\setminus W\).
- In the cofinite topology on \(\mathbb{Z}\), why does every open neighborhood of \(0\) have closure \(X\)?
- Does the indiscrete topology on \(\{a,b\}\) satisfy the closure-shrinking criterion? Is it \(T_3\)? Give the reason for each answer.
- State the closure-shrinking characterization of \(T_3\) using the convention that \(T_3\) means regular and \(T_1\).