Nested Open Neighborhoods for Urysohn’s Lemma

Why Build a Nested Family?

To separate two disjoint closed sets by neighborhoods, it is not enough to choose one open set around each set. In Urysohn’s construction, one needs a whole ordered family of open sets: as the index increases, each set must fit inside the next, with its closure still inside that next set. Normality supplies exactly the shrinking step needed to build this family.

Let \((X,\mathcal{T})\) be a topological space. A set is closed when its complement is open. The closure \(\overline{C}\) of a set \(C\subseteq X\) is the smallest closed set containing \(C\). A space is normal if every pair of disjoint closed sets has disjoint open neighborhoods. We will construct open sets indexed by the dyadic rationals in \([0,1]\), namely numbers of the form \(k/2^n\) in that interval.

The endpoint choices encode the separation: \(U_0\) contains \(A\), while \(U_1\) omits \(B\). The closure condition is stronger than merely requiring \(U_r\subseteq U_s\). It leaves room to insert a new open set between the closure of one set and a later set.

The Shrinking Step

Proof. The sets \(F\) and \(X\setminus G\) are closed and disjoint: \(F\subseteq G\) implies \(F\cap(X\setminus G)=\varnothing\). By normality, there are disjoint open sets \(V,W\) with \(F\subseteq V\) and \(X\setminus G\subseteq W\). Since \(V\cap W=\varnothing\), we have \(V\subseteq X\setminus W\). The set \(X\setminus W\) is closed and contains \(V\), so the definition of closure gives \(\overline{V}\subseteq X\setminus W\). Also \(X\setminus W\subseteq G\), because \(X\setminus G\subseteq W\). Thus \(F\subseteq V\subseteq\overline{V}\subseteq G\). \(\square\)

This argument also covers \(F=\varnothing\) or \(X\setminus G=\varnothing\): normality applies to disjoint closed sets even when one is empty. The lemma is the mechanism for every midpoint insertion.

Constructing the Dyadic Neighborhoods

Proof. We first construct the sets at each finite dyadic level and ensure that consecutive indices \(r<s\) at that level satisfy \(\overline{U_r}\subseteq U_s\). At level zero the indices are \(0\) and \(1\). Apply the shrinking lemma with \(F=A\) and \(G=X\setminus B\). These are, respectively, closed and open, and \(A\subseteq X\setminus B\) because \(A\cap B=\varnothing\). Set \(U_0=V\) from the lemma and set \(U_1=X\setminus B\). Then \(A\subseteq U_0\subseteq\overline{U_0}\subseteq U_1\).

Suppose the sets have been constructed at level \(n\), where the indices are \(k/2^n\) for \(0\leq k\leq 2^n\), and consecutive indices satisfy the required closure inclusion. For each consecutive pair \(r<s\), insert the midpoint \(t=(r+s)/2\). The set \(F=\overline{U_r}\) is closed and is contained in the open set \(G=U_s\) by the level-\(n\) condition. Apply the shrinking lemma to obtain an open \(U_t\) with \(\overline{U_r}\subseteq U_t\subseteq\overline{U_t}\subseteq U_s\). This proves the required inclusion for both new consecutive pairs, \(r<t\) and \(t<s\). Doing this for every consecutive pair constructs level \(n+1\) without changing the sets already chosen.

Every dyadic index occurs at some finite level. To check the condition for any dyadics \(r<s\), choose a level containing both. At that level, list the consecutive indices from \(r\) to \(s\). The closure of the set at the first index is contained in the next set, which is contained in each later set in the list. Hence \(\overline{U_r}\subseteq U_s\). In particular \(U_r\subseteq U_s\), since every set is contained in its closure. As \(U_s\subseteq U_1=X\setminus B\), all the sets omit \(B\); and \(A\subseteq U_0\). The construction has all the stated properties. \(\square\)

The argument uses only finitely many shrinking steps to construct any one finite level. It does not require choosing a new set by an infinite limiting procedure. The dyadic family as a whole is obtained by carrying out the finite construction successively at levels \(0,1,2,\ldots\).

Why Normality Matters—and a Common Pitfall

The shrinking lemma depends on normality. Here is a finite example where the desired nesting can fail. Let \(X=\{a,b,c\}\) have topology \(\{\varnothing,\{c\},\{a,c\},\{b,c\},X\}\). The closed sets, obtained by taking complements, are \(X,\{a,b\},\{a\},\{b\},\varnothing\). Thus \(A=\{a\}\) and \(B=\{b\}\) are disjoint closed sets. The open neighborhoods of \(a\) are \(\{a,c\}\) and \(X\); those of \(b\) are \(\{b,c\}\) and \(X\). Every such pair intersects, so this space is not normal.

In fact, no dyadic family for these \(A,B\) is possible. Its endpoint would have to be \(U_1=X\setminus B=\{a,c\}\). An open \(U_0\) containing \(a\) is either \(\{a,c\}\) or \(X\). The first has closure \(X\), since the only closed set containing \(c\) is \(X\); the second also has closure \(X\). Neither closure is contained in \(U_1\), contrary to the required condition \(\overline{U_0}\subseteq U_1\). This verifies concretely what the normality hypothesis provides.

Common pitfall. For \(A=(-\infty,0]\) and \(B=[1,\infty)\), one might try \(U_0=(-\infty,0)\) because it is open and avoids \(B\). But \(0\in A\) and \(0\notin U_0\), so it is not a neighborhood of all of \(A\). The corrected family above has \(U_0=(-\infty,1/2)\), which does contain \(0\), while its closure \((-\infty,1/2]\) remains inside \(U_1=(-\infty,1)\). Checking the endpoint containment is as important as checking the nesting.