From Nested Open Sets to a Function
Urysohn’s construction turns a sequence of increasingly large open neighborhoods into a real-valued function. The indexing matters: dyadic rationals provide arbitrarily fine levels between zero and one, while the nesting ensures that membership in an earlier open set implies membership in every later one. Here we define the function and verify its prescribed values on the two closed sets. The continuity argument is a separate step.
Let \((X,\mathcal{T})\) be a topological space, and let \(A,B\subseteq X\) be disjoint closed sets. Write \(\mathcal{D}\) for the dyadic rationals in \([0,1]\): \[ \mathcal{D}=\left\{\frac{k}{2^n}: n\geq 0,\ 0\leq k\leq 2^n\right\}. \] Suppose we have open sets \(U_r\in\mathcal{T}\), one for each \(r\in\mathcal{D}\), satisfying \[ A\subseteq U_0,\qquad U_1=X\setminus B,\qquad \overline{U_r}\subseteq U_s\quad\text{whenever }r<s. \] In particular, \(U_r\subseteq U_s\) if \(r<s\), since every set is contained in its closure. The earlier construction of nested open neighborhoods in a normal space supplies such a family; for the definition and value calculation below, these displayed properties are the assumptions we need.
When \(S_x\) is nonempty, it is a subset of \([0,1]\), so it has an infimum in \(\mathbb{R}\); that infimum belongs to \([0,1]\). The special convention for the empty set is needed because its usual extended-real infimum is not a real number in \([0,1]\). It is also consistent with the intended value on \(B\): points of \(B\) belong to none of the open sets in the family.
Why the Dyadic Infimum Has the Required Values
Proof. First take \(a\in A\). Since \(A\subseteq U_0\), we have \(0\in S_a\). Every element of \(S_a\) is nonnegative, so \(0\) is a lower bound for \(S_a\); because \(0\in S_a\), its infimum is exactly \(0\). Thus \(f(a)=0\).
Now take \(b\in B\). The identity \(U_1=X\setminus B\) gives \(b\notin U_1\). For any \(r\in\mathcal{D}\) with \(r<1\), nesting gives \(U_r\subseteq U_1\), so \(b\notin U_r\) as well. Therefore \(S_b=\varnothing\), and the definition gives \(f(b)=1\). Finally, if \(S_x\ne\varnothing\), its infimum lies between zero and one because \(S_x\subseteq[0,1]\); if \(S_x=\varnothing\), then \(f(x)=1\). This proves the range assertion in both cases. \(\square\)
The nesting hypothesis does more than ensure the endpoint values. It orders the sets by their indices: once a point belongs to some \(U_r\), it belongs to every \(U_s\) with \(s>r\), because \(U_r\subseteq\overline{U_r}\subseteq U_s\). The index set \(S_x\) is therefore upward-closed among the dyadics. Its infimum records the threshold where membership begins, whether or not the point belongs to the open set at that threshold.
We will use the following elementary property when computing such thresholds. Dyadic rationals occur arbitrarily close above any number in \([0,1)\). Consequently, for \(0\leq t<1\), \[ \inf\{r\in\mathcal{D}:r>t\}=t. \]
Proof. Every member of the set on the left is greater than \(t\), so \(t\) is a lower bound. To show that no number greater than \(t\) is a lower bound, let \(\varepsilon>0\). Choose \(n\) so large that \(2^{-n}<\varepsilon\) and \(2^{-n}<1-t\). Let \(k\) be the least integer strictly greater than \(2^nt\). Then \(r=k/2^n\) is dyadic, \(t<r\), and \(r\leq t+2^{-n}<t+\varepsilon\). The choice \(2^{-n}<1-t\) ensures \(k\leq 2^n\), so \(r\leq1\). Thus the set contains an element less than \(t+\varepsilon\) for every \(\varepsilon>0\), proving its infimum is \(t\). \(\square\)
Worked Example: A Point at a Dyadic Threshold
Take \(X=[0,1]\) with its usual subspace topology from \(\mathbb{R}\), \(A=\{0\}\), and \(B=\{1\}\). Define \[ U_0=[0,1/4),\qquad U_r=[0,1/4+3r/4)\ \ (0<r<1),\qquad U_1=[0,1). \] These sets are open in \([0,1]\): each is the intersection of \([0,1]\) with an open interval in \(\mathbb{R}\). Also \(A\subseteq U_0\) and \(U_1=[0,1]\setminus\{1\}=X\setminus B\).
For \(r<1\), the closure of \(U_r\) in \([0,1]\) is the interval with the same left endpoint and the included right endpoint. If \(0<r<s<1\), then \(1/4+3r/4<1/4+3s/4\), so \(\overline{U_r}\subseteq U_s\). The same inclusion holds from \(U_0\) to any \(U_s\) with \(s>0\), since \(1/4<1/4+3s/4\); and for \(s=1\), each \(\overline{U_r}\) with \(r<1\) lies in \([0,1)\). Thus the required nesting holds.
Compute \(f(5/8)\). The point \(5/8\) is not in \(U_0\). For \(0<r<1\), it belongs to \(U_r\) exactly when \[ \frac{5}{8}<\frac{1}{4}+\frac{3r}{4} \quad\Longleftrightarrow\quad \frac{3}{8}<\frac{3r}{4} \quad\Longleftrightarrow\quad \frac{1}{2}<r. \] It belongs to \(U_1\) as well. Hence \(S_{5/8}=\{r\in\mathcal{D}:r>1/2\}\), whose infimum is \(1/2\). Therefore \(f(5/8)=1/2\), even though \(5/8\notin U_{1/2}\). The endpoint checks are \(f(0)=0\), since \(0\in U_0\), and \(f(1)=1\), since \(1\notin U_r\) for every \(r\in\mathcal{D}\). These agree with the theorem.
Worked Example: A Finite Discrete Space
Let \(X=\{a,p,q,b\}\) have the discrete topology, so every subset is open and closed. Set \(A=\{a\}\) and \(B=\{b\}\). For each dyadic \(r\in[0,1]\), define \[ U_r=\{a\}\cup \begin{cases}\{p\},&r>1/4,\\ \varnothing,&r\leq1/4,\end{cases} \cup \begin{cases}\{q\},&r>3/4,\\ \varnothing,&r\leq3/4.\end{cases} \] The set \(b\) is never included. In particular \(U_0=\{a\}\) and \(U_1=\{a,p,q\}=X\setminus B\). These sets are open, and \(U_r\subseteq U_s\) whenever \(r<s\). Since every set is closed in the discrete topology, \(\overline{U_r}=U_r\), so \(\overline{U_r}\subseteq U_s\) as required.
The index set for \(a\) is all of \(\mathcal{D}\), including \(0\), so \(f(a)=0\). The point \(p\) belongs exactly when \(r>1/4\); hence \(S_p=\{r\in\mathcal{D}:r>1/4\}\), and its infimum is \(1/4\). Likewise \(q\) belongs exactly when \(r>3/4\), so \(f(q)=3/4\). The point \(b\) belongs to no \(U_r\), giving \(S_b=\varnothing\) and \(f(b)=1\).
For an explicit check at the cutoffs, \(p\notin U_{1/4}\) but \(p\in U_{1/2}\), while \(q\notin U_{3/4}\) but \(q\in U_1\). These membership tests do not make \(1/2\) or \(1\) the corresponding infimum: the dyadic indices approach the relevant cutoff from above. The computed values satisfy \(f(A)=\{0\}\) and \(f(B)=\{1\}\).
Worked Example: The Empty Index Set
Use the same space \(X=[0,1]\), with its usual topology, and the same sets \(A=\{0\}\), \(B=\{1\}\), but take the simpler family \(U_r=[0,1)\) for every \(r\in\mathcal{D}\). Each \(U_r\) is open in \(X\), \(A\subseteq U_0\), and \(U_1=X\setminus B\). For every pair \(r<s\), \(\overline{U_r}=[0,1]\) is not a subset of \(U_s=[0,1)\), so this family does not satisfy the closure-nesting hypothesis. It cannot be used for the full construction as stated.
This example isolates why the value calculation on \(B\) requires care: regardless of nesting, \(1\notin U_r\) for every \(r\), so \(S_1=\varnothing\) and the convention gives \(f(1)=1\). But the failed closure condition means that no general conclusions that rely on the nested family, especially the later continuity proof, follow from this example. To see a valid instance with an empty index set, return to the first worked example: \(U_1=[0,1)\) and every earlier \(U_r\subseteq[0,1)\), so the point \(1\) has \(S_1=\varnothing\), while the closure inclusions hold. Its assigned value is precisely \(1\).
A Common Pitfall: Infimum Is Not Minimum
In the first worked example, the index set for \(5/8\) is all dyadics strictly greater than \(1/2\). A tempting but incorrect calculation is to inspect a few indices and declare \(5/8\) the “first” such dyadic, hence the minimum. But \(9/16\) is also dyadic, and \[ \frac{1}{2}<\frac{9}{16}<\frac{5}{8}. \] More generally, for each \(n\geq2\), the dyadic number \(1/2+1/2^n\) is greater than \(1/2\) and approaches \(1/2\) as \(n\) increases. There is no least dyadic strictly above \(1/2\); the infimum is \(1/2\), which is not itself in the index set. The function must use the infimum, not a presumed minimum.
There is a second convention to keep distinct. If \(x\) is in no \(U_r\), the infimum of \(S_x\) is not an ordinary real number. Assigning \(f(x)=1\) in this case is part of the definition, not an application of the nonempty-set infimum rule. Under the stated hypotheses, every \(b\in B\) is exactly such a point. By contrast, every \(x\notin B\) belongs to \(U_1=X\setminus B\), so its index set is nonempty.
The construction therefore assigns a number in \([0,1]\) to every point and separates the two closed sets at the level of values: zero on \(A\), one on \(B\). The nested sets encode thresholds; the infimum captures those thresholds even when no set at the limiting index contains the point. These facts establish the definition and endpoint values, but not continuity. Continuity requires a further argument about how the nested open sets control the inverse images of intervals.
Check Your Understanding
Use the definition of \(S_x\), the nesting assumptions, and the infimum convention to answer the following.
- Why does \(b\in B\) imply \(b\notin U_r\) for every dyadic \(r<1\), as well as for \(r=1\)?
- In the first worked example, compute the index set \(S_{1/2}\) and determine \(f(1/2)\).
- In a finite discrete space, if a point belongs to exactly the dyadic sets with indices \(r>3/8\), what is its function value? Is that value attained as an index of a set containing the point?
- Explain why the convention \(f(x)=1\) when \(S_x=\varnothing\) is needed to make the function take values in \([0,1]\).
- Give the nesting step that shows \(U_r\subseteq U_s\) whenever \(r<s\).