4.4The Urysohn Lemma

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The Ursyohn lemma states that for any normal space $X$ and any pair of closed subsets $A, B \subseteq X$ , there exists a continuous $f: X \to [a, b]$ (for some $a, b \in \bR$ ) such that $f|_A \equiv a$ and $f|_B \equiv b$ .

[4.4.1]Theorem(The Urysohn Lemma)#

Let $X$ be a normal space, and let $A, B$ be disjoint, closed subsets of $X$ . Then there is a continuous map

f: X \to [0, 1]

for which $f|_A \equiv 0$ and $f|_B \equiv 1$ .

Before we prove this lemma, we consider a class of numbers known as dyadic rationals, a set of fractions whose denominator is a power of two. Namely, we consider the dyadic rationals contained in $[0, 1]$ . Let $D_n$ be the set of dyadic rationals of order $n$ contained in $[0, 1]$ (i.e., pure fractions whose denominator is of the form $2^n$ ). For example,

D_0 = \{0, 1\}, \quad D_1 = \left\{0, \frac{1}{2}, 1 \right\}, \quad D_2 = \left\{0, \frac14, \frac12, \frac34, 1 \right\}, \quad \cdots.

Suppose we bisect the interval $[0, 1]$ $n$ times. Then the endpoints of bisection correspond exactly with the values in $D_n$ , motivating them geometrically. If we bisect the interval a countably infinite number of times (like we do in the proof of, say, Bolzano-Weierstrass), the endpoints yielded correspond exactly with the values in $D = \bigcup_{n \in \bZ_+} D_n$ . Crucially, $D$ is dense in $[0, 1]$ , so the open sets indexed by $D$ are fine enough to bring about a function that satisfies the hypotheses.

We will bisect $[0, 1]$ a countably infinite number of times, use the shrinking lemma to force an open set each time we bisect, and index these open sets with the dyadic rationals. Then we will define a map using these open sets that satisfy the conditions of the hypothesis, completing the proof.

Proof.

We start by creating our family of open sets indexed by $D$ . We start with $D_0$ : consider the set $U_1 = X \setminus B$ . Then, since $A \cap B = 0$ , we have that $A \subseteq U_1$ . We have that $A$ is closed and $U_1$ is open, and so by the shrinking lemma, there is an open set $U_0$ such that

A \subseteq U_0 \subseteq \bar{U_0} \subseteq U_1.

Now we move to $D_1$ . Note that $\bar{U_0}$ is a closed subset of the open set $U_1$ , so we apply the shrinking lemma again, yielding a set $U_{1/2}$ such that

\bar{U_0} \subseteq U_{1/2} \subseteq \bar{U_{1/2}} \subseteq U_1.

Now we move to $D_2$ . Note that $\bar{U_0}$ is a closed subset of the open set $U_{1/2}$ , and that $\bar{U_{1/2}}$ is a closed subset of the open set $U_1$ . Then we have open sets $U_{1/4}, U_{3/4}$ such that

U_0 \subseteq \overline{U_0} \subseteq U_{1/4} \subseteq \overline{U_{1/4}} \subseteq U_{1/2} \subseteq \overline{U_{1/2}} \subseteq U_{3/4} \subseteq \overline{U_{3/4}} \subseteq U_1.

Repeat this process inductively, yielding the family of open sets $\{U_r\}_{r \in D}$ . We are now ready to define the continuous function.

Define $f: X \to [0, 1]$ by the function

f(x) = \inf\{r \in D: x \in U_r\} \quad (\inf(\emptyset) = 1).

For any $x \in X$ , we identify the smallest dyadic rational $r$ for which the open set $U_r$ contains $x$ . If no such $U_r$ exists, then $f$ emits $1$ . Immediately, we note that $f|_A \equiv 0$ , since $A \subseteq U_0$ and so every element of $A$ is contained in $U_0$ . Conversely, we note that $f|_B \equiv 1$ , since $U_1 \subseteq B$ and so no element of $B$ is contained in $\{U_r\}$ .

Finally, we prove that $f$ is continuous, completing the proof. We show continuity on the subbasis elements $[0, a)$ and $(a, 1]$ for any $a \in (0, 1)$ . Note that if $f(x) < a$ if and only if $x \in U_r$ for dyadic rationals $r$ such that $r < a$ . Thus

f^{-1}([0, a)) = \bigcup_{r \in D, r < a} U_r.

Conversely, suppose $f(x) > a$ . By density of $D$ in $[0, 1]$ , we can pick two dyadic rationals $r, s$ such that $a < r < s < f(x)$ . Since $s < f(x)$ , we note that $x \notin U_s$ . Since $\bar{U_r} \subseteq U_s$ , note that $x \notin \bar{U_r}$ as well. Thus

f^{-1}((a, 1]) = \bigcup_{r \in D, r > a} (X \setminus \bar{U_r}).

Thus the preimage of all subbasis elements are open, meaning $f$ is continuous. We have completed the proof.

[4.4.2]Remark(Urysohn's Lemma Applies to Any Closed Interval)#

Note the argument involved is invariant under translation and scaling of the codomain. Thus Urysohn's lemma guarantees any continuous map $f: X \to [a, b]$ for which $f|_A \equiv a$ and $f|_B \equiv b$ .

[4.4.3]Definition(Separated by a Continuous Function)#

Suppose $X$ is a topological space and $A, B \subseteq X$ are subsets. If there exists a continuous function $f: X \to [0, 1]$ such that $f|_A \equiv 0$ and $f|_B \equiv 1$ , then we say that $A$ and $B$ can be separated by a continuous function.

[4.4.4]Theorem(Normal iff. Disjoint, Closed Sets Separable by Continuous Function)#

A space $X$ is normal if and only if, any pair of disjoint, closed sets can be separated by a continuous function.

Proof.

$(\Longrightarrow)$ : This is the statement of Urysohn's Lemma.

$(\Longleftarrow)$ : Suppose any disjoint, closed subsets $A, B \subseteq X$ can be separated by a continuous function $f$ . Then note $f^{-1}([0, 1/2])$ and $f^{-1}((-1/2, 1])$ are disjoint open neighborhoods for $A$ and $B$ .

Note that the shrinking lemma applies to normal spaces, and so the proof for Urysohn's lemma cannot be lifted to regular spaces. That is, the statement that a point and a disjoint, closed set can be separated by a continuous function is false. Instead, we define a subclass of regular spaces for which this is true.

[4.4.5]Definition(Completely Regular)#

A space $X$ that satisfies the $T_1$ condition is called completely regular if any singleton subset $\{x_0\}$ and some disjoint, closed set $B$ can be separated by a continuous function.

Immediately, we note that a completely regular space is regular: if a singleton subset is separated from a disjoint, closed set by a continuous function $f$ , then $f^{-1}([0, 1/2))$ and $f^{-1}((1/2, 1])$ are the desired disjoint, open neighborhoods about the singleton and closed set. Furthermore, we note that a normal space is completely regular, as we simply take one closed set to be a singleton.

This class of spaces is particularly nice as well: the subspace and product operations leave its property intact.

[4.4.6]Theorem(Subspace and Product of Completely Regular Space is Completely Regular)#

Suppose $X$ is a completely regular space. Then any subspace $U \subseteq X$ is also completely regular. Moreover, suppose $\{X_\lambda\}_{\lambda \in \Lambda}$ is a collection of completely regular spaces. Then the product $X = \prod_{\lambda \in \Lambda} X_\lambda$ is also completely regular.

Proof.

Note that $U$ is regular—subspaces of regular spaces are regular—and thus $T_1$ . Let $x_0 \in U$ be any point that is disjoint from the closed set $B \subseteq U$ . Since $B$ is closed in $U$ , it contains all its limit points in $U$ and so $x_0$ is not a limit point for $B$ , meaning $x_0 \notin \Cl_X(B)$ . Thus $x_0$ is disjoint from the closed set $\Cl_X(B)$ , where both objects are treated as elements of $X$ . By complete regularity, there is a separation by a continuous map $f: X \to [0, 1]$ . Then $g = f|_U$ is a continuous map that separates $x_0$ and $\Cl_X(B) \cap U = \Cl_U(B) = B$ , making $U$ completely regular.

The proof for the product being completely regular, like the subspace, is extremely similar to the proof of products being regular, so we omit the proof.

[4.4.7]Remark(The Separation Axioms)#

The separation axioms have alternative names of the form $T_j$ . They are organized such that a $T_j$ space is automatically $T_i$ , and that there exists one $T_i$ space that isn't $T_j$ (for $j > i$ ). They are as follows.

$\mathbf{T}_0$ (Kolmogorov). Any two points are topologically distinguishable.
$\mathbf{T}_1$ (Accessible). Any two points have open neighborhoods that exclude the other point.
$\mathbf{T}_2$ (Hausdorff). Any two points can be separated.
$\mathbf{T}_3$ (Regular). A $T_1$ space for which any point and disjoint, closed set can be separated.
$\mathbf{T}_{3.5}$ (Completely Regular). A $T_1$ space for which any point and disjoint, closed set can be separated by a continuous function.
$\mathbf{T}_4$ (Normal). A $T_1$ space for which any pair of disjoint, closed sets can be separated.
$\mathbf{T}_5$ (Completely Normal). A $T_1$ space that is normal and has the property that all its subspaces are normal.