Proving Continuity of the Urysohn Function

From Dyadic Infima to Continuity

The dyadic-infimum construction defines a function using membership in a family of nested open sets. The infimum is useful precisely because it records how far through the family a point first appears. But an infimum need not be attained, so continuity cannot be justified by treating it as a minimum. The key is to identify the strict sublevel and superlevel sets of the function and show that both are open.

Let $(X,\mathcal{T})$ be a normal topological space, meaning that any two disjoint closed subsets of $X$ have disjoint open neighborhoods. Let $A,B\subseteq X$ be disjoint closed sets. Write $D=\{m/2^n:n\in\{0,1,2,\ldots\},\ 0\le m\le 2^n\}$ for the dyadic rationals in $[0,1]$. The nested-open-set construction for Urysohn’s lemma gives open sets $U_r$, for $r\in D$, with $A\subseteq U_0$, $U_1=X\setminus B$, and $\overline{U_r}\subseteq U_s$ whenever $r<s$ in $D$. This construction is the earlier nested-neighborhood result; the continuity proof below uses exactly these properties.

Definition: For $x\in X$, define $S_x=\{r\in D:x\in U_r\}\cup\{1\}$, and set $f(x)=\inf S_x$. The added value $1$ ensures that $S_x$ is nonempty even when $x$ belongs to none of the $U_r$. Thus $f(x)$ is defined and lies in $[0,1]$ for every $x\in X$.

The nesting condition has two useful consequences. First, $U_r\subseteq U_s$ whenever $r<s$, since $U_r\subseteq\overline{U_r}\subseteq U_s$. Second, points that lie in $\overline{U_r}$ lie in every later $U_s$. These facts will translate comparisons involving $f(x)$ into membership and nonmembership in open sets.

The Strict Sublevel and Superlevel Sets

Theorem 1 (Level-set formulas): For $0<a\le 1$ and $0\le a<1$, respectively, $$ \{x\in X:f(x)<a\}=\bigcup_{\substack{r\in D\\r<a}}U_r, \qquad \{x\in X:f(x)>a\}=\bigcup_{\substack{r\in D\\a<r<1}}\bigl(X\setminus\overline{U_r}\bigr). $$ For $a\le 0$, the strict sublevel set is empty; for $a>1$, it is $X$. For $a<0$, the strict superlevel set is $X$; for $a\ge 1$, it is empty.

Proof. Suppose $0<a\le1$. If $f(x)<a$, then $\inf S_x<a$. By the defining property of an infimum, some element $t\in S_x$ satisfies $t<a$; otherwise every element of $S_x$ would be at least $a$, forcing $\inf S_x\ge a$. Since $a\le1$, this $t$ cannot be the added value $1$. Hence $t\in D$, $t<a$, and $x\in U_t$. Conversely, if $x\in U_r$ for some $r\in D$ with $r<a$, then $r\in S_x$, so $f(x)=\inf S_x\le r<a$. This proves the first formula.

Now suppose $0\le a<1$. If $f(x)>a$, density of the dyadic rationals gives $r,s\in D$ with $a<r<s<f(x)$. If $x\in\overline{U_r}$, nesting would give $x\in U_s$, and then $f(x)\le s<f(x)$, a contradiction. Thus $x\in X\setminus\overline{U_r}$, so $x$ belongs to the union on the right.

Conversely, suppose $a<r<1$ and $x\notin\overline{U_r}$. For each $t\in D$ with $t\le r$, we have $U_t\subseteq\overline{U_r}$: for $t=r$ this follows from $U_r\subseteq\overline{U_r}$, and for $t<r$ it follows from $U_t\subseteq\overline{U_t}\subseteq U_r\subseteq\overline{U_r}$. Therefore $x$ belongs to no $U_t$ with $t\le r$. Every element of $S_x$ is consequently greater than $r$, except possibly the added value $1$, which is also greater than $r$. It follows that $f(x)=\inf S_x\ge r>a$. This proves the second formula. The remaining claims follow from $0\le f(x)\le1$. $\square$

Each union in the formulas is a union of open sets: $U_r$ is open, and $X\setminus\overline{U_r}$ is open because $\overline{U_r}$ is closed. The endpoint restrictions matter. For example, when $a=1$, points with $f(x)=1$ do not belong to the strict sublevel set; when $a=0$, no point belongs to the strict superlevel set only if its value is at most zero, so the formula for $f(x)>0$ is still needed.

Continuity and Separation of the Closed Sets

Theorem 2 (Continuity of the dyadic-infimum function): Under the nested-open-set hypotheses above, $f:X\to[0,1]$ is continuous.

Proof. Regard $f$ first as a real-valued function. For every real $a$, the strict sublevel set $\{x:f(x)<a\}$ is open: the first level-set formula proves this when $0<a\le1$, and the remaining cases are either $\varnothing$ or $X$. Likewise, $\{x:f(x)>a\}$ is open for every real $a$, by the second formula and its stated endpoint cases.

Let $(u,v)$ be any open interval in $\mathbb{R}$. Its inverse image is $f^{-1}((u,v))=\{x:f(x)>u\}\cap\{x:f(x)<v\}$, an intersection of two open sets, hence open. Every open subset of $\mathbb{R}$ is a union of open intervals, so its inverse image under $f$ is open as well. This is the definition of continuity. Since the values of $f$ lie in $[0,1]$, the same function is continuous as a map into $[0,1]$ with its subspace topology. $\square$

Theorem 3 (Urysohn’s lemma, values zero and one): If $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous function $f:X\to[0,1]$ such that $f(x)=0$ for every $x\in A$ and $f(x)=1$ for every $x\in B$.

Proof. If $X$ is empty, its unique map to $[0,1]$ is continuous and the conditions are vacuous. If $A=\varnothing$, the constant function with value $1$ works; if $B=\varnothing$, the constant function with value $0$ works. In the remaining case, use the nested-open-set construction for the disjoint closed sets $A,B$, and define $f$ by the dyadic infimum. Theorem 2 gives continuity. For $x\in A$, the inclusion $A\subseteq U_0$ gives $0\in S_x$; since $f(x)\ge0$, it follows that $f(x)=0$. For $x\in B$, we have $x\notin U_1=X\setminus B$. Also, $U_r\subseteq U_1$ whenever $r<1$, so $x\notin U_r$ for every $r\in D$. Thus $S_x=\{1\}$, and $f(x)=1$. $\square$

Worked Examples

Worked Example: An Infimum That Is Not a Minimum

Use the nested open sets in the unit interval $I=[0,1]$, with its usual subspace topology, given for $r<1$ by $U_r=[0,(1+r)/2)$, and let $U_1=[0,1)$. Consider the point $x=5/8$.

For $r<1$, membership means $5/8<(1+r)/2$. Multiplying by $2$ and subtracting $1$ gives $r>1/4$. In particular, $r=1/4$ does not work: its endpoint is $(1+1/4)/2=5/8$, and the interval is open at that endpoint. But $r=3/8$ does work, since $(1+3/8)/2=11/16$ and $5/8=10/16<11/16$. Dyadic values greater than $1/4$ can be chosen arbitrarily close to $1/4$, so the infimum of the membership indices is $1/4$, even though $x\notin U_{1/4}$. Including the added value $1$ does not change this infimum. Therefore $f(5/8)=1/4$.

Check: the membership condition is exactly $r>1/4$, not $r\ge1/4$. For example, $r=1/8$ gives endpoint $9/16$, and $5/8=10/16$ is not less than $9/16$; $r=3/8$ gives endpoint $11/16$, which is greater than $10/16$. This confirms both the strict inequality and the infimum value.

Worked Example: Strict Level Sets on the Unit Interval

For the same family on $I$, compute the strict sublevel and superlevel sets at $a=3/4$. The function is $f(x)=0$ for $0\le x\le1/2$, $f(x)=2x-1$ for $1/2<x<1$, and $f(1)=1$. Indeed, for $x<1$, membership in $U_r$ is equivalent to $r>2x-1$, with $r=0$ already giving membership when $x<1/2$; at $x=1/2$, every positive dyadic $r$ works, so the infimum is $0$. The point $1$ belongs to no $U_r$ and has value $1$.

The inequality $f(x)<3/4$ holds exactly when $x<7/8$, including all $x\le1/2$. Hence $\{x:f(x)<3/4\}=[0,7/8)$. This also follows from the union of $U_r$ over dyadic $r<3/4$: their right endpoints $(1+r)/2$ approach $7/8$ from below, and the endpoint $7/8$ is not included.

The inequality $f(x)>3/4$ holds exactly when $x>7/8$, so $\{x:f(x)>3/4\}=(7/8,1]$. In the superlevel formula, each $r>3/4$ contributes $I\setminus\overline{U_r}=((1+r)/2,1]$; as such dyadic $r$ decrease toward $3/4$, these sets together contain exactly the points greater than $7/8$.

Check: at $x=7/8$, the function value is $2(7/8)-1=7/4-1=3/4$. Thus this point belongs to neither strict level set, as required.

Worked Example: A Finite Discrete Space

Let $X=\{p,q,r\}$ have the discrete topology, in which every subset is open, and take $A=\{p\}$, $B=\{q\}$. Set $U_t=\{p\}$ for every dyadic $t<1$ and $U_1=X\setminus B=\{p,r\}$. These sets are open. For $s<1$, $\overline{U_t}=U_t=\{p\}\subseteq U_s$ whenever $t<s$; for $s=1$, $\{p\}\subseteq\{p,r\}$. Thus the required nesting holds.

For $p$, every dyadic index belongs to $S_p$, including $0$, so $f(p)=0$. For $q$, no $U_t$ contains it, and the added value gives $S_q=\{1\}$, so $f(q)=1$. For $r$, no $U_t$ with $t<1$ contains it, so $S_r=\{1\}$ and $f(r)=1$. The strict sublevel set at $a=1/2$ is therefore $\{p\}$, while the strict superlevel set at $a=1/2$ is $\{q,r\}$. Both are open in the discrete topology, consistent with continuity.

Check: the function takes only the values $0$ and $1$, so the displayed sets follow directly from the strict comparisons $0<1/2$ and $1>1/2$. In particular, $p$ is not in the strict superlevel set, and $q,r$ are not in the strict sublevel set.

A Common Pitfall: Replacing the Infimum by a Minimum

In the first worked example, $f(5/8)=1/4$, but $5/8\notin U_{1/4}$. A tempting but incorrect calculation is to write $\{x:f(x)<a\}=\bigcup_{r\le a}U_r$. At $a=1/4$, this would put $5/8$ in the sublevel set because it includes $U_{1/4}$, even though $f(5/8)=1/4$ and $1/4<1/4$ is false. The correct formula uses $r<a$, not $r\le a$. The infimum describes the greatest lower bound of the membership indices; it need not itself be a membership index.

The companion superlevel formula uses complements of closures, not merely complements of the open sets. Closure is what lets nesting rule out membership at every earlier dyadic index. Together, the two strict formulas expose the mechanism behind continuity: both kinds of strict level sets are open, and their intersections give inverse images of open intervals.

Check Your Understanding

Use the strict inequalities in the definitions and formulas when answering.

Why is the added value $1$ included in $S_x$, and what is $f(x)$ when $x\in B$?
In the unit-interval example, determine $f(3/4)$ and state whether $3/4$ belongs to $U_{1/2}$.
Write the level-set formulas for $a=0$ and $a=1$, identifying any empty sets.
Explain why $x\in\overline{U_r}$ implies $f(x)\le s$ for every dyadic $s>r$.
Why do open strict sublevel and superlevel sets imply that the inverse image of every open interval is open?