202303211003

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Urysohn Lemma

Info

This lemma roughly states that given two disjoint closed subsets of a normal space, we can separate them by real valued functions.

Statement of the lemma

Let $A,B$ be disjoint closed subsets of a normal space $X$. Then there exists a continuous map $f:X\to [0,1]$ such that $f(x)=0$ for all $x\in A$ and $f(x)=1$ for all $x\in B$.

Proof:

A dyadic number is a rational number of the form $\frac{a}{2^n}$, where $a,n$ are integers with $n\ge 0$. Then it is easy to check that these numbers are dense in $\mathbb{R}$.

We shall construct an open subset $U_r\subseteq X$ for each dyadic number $r\in [0,1]$, with $A\subset U_r\subset X\setminus B$. Such that for each pair $p<q$ of dyadics in $[0,1]$, we have that $Cl(U_p)\subset U_q$.

Let $U_1=X\setminus B$. Then since $X$ is normal, there is a nbhd around $A$, call this $U_0$, such that $Cl\left(U_0\right)\subset U_1$. Similarly, we get another open set $U_{\frac{1}{2}}$ such that $Cl(U_0)\subset U_{\frac{1}{2}}\subset Cl\left(U_{\frac{1}{2}}\right)\subset U_1$.

Now let $n\ge 2$. Assume we have constructed the required subsets for all $r$ in $[0,1]$ of the form $\frac{b}{2^{n-1}}=\frac{2b}{2^n}$. So we need to constuct $U_{\frac{2n+1}{2^n}}$, again use normality, and get $$ Cl\left(U_{\frac{2b}{2^{n}}}\right) \subset U_{\frac{2n+1}{2^{n+1}}} \subset Cl\left( U_{\frac{2n+1}{2^{n}}} \right) \subset U_{\frac{2n+2}{2^{n}}}

Hence, by induction, we construct $U_{r}$ for all dyadic $r$ in $[0,1]$. Now let $U_{r} = \phi$ for $r < 0$. and $U_{r} = X$ for $r > 1$. Now let $x \in X$, we define $$ D(x) := \{ r\ \mathrm{dyadic} : x \in U_{r} \}

Since $x\notin U_r$ for $r<0$, and $x\in U_r$ for $r>1$, $D(x)\ne \varphi$ and it is bounded below by 0. So $$ f(x) := \inf D(x)

exists and lies in $[0,1]$. Now see that $f(x) = 0$ for $x \in A$, and $f(x) = 1$ for $x \in B$. To show that $f$ is continuous, first observe that if $x \in Cl(U_{r})$ then $f(x)\le r$. And also, if $x \notin U_{r}$, then $f(x) \geq r$. Now let $x \in X$, pick a nbhd $(c,d)$ of $f(x)$. Take dyadic numbers $p,q$ such that $c < p <f(x)< q < d$. This means that $x \notin Cl(U_{p})$ but $x \in U_{q} \implies x \in U := U_{q} \setminus Cl(U_{p})$. But then for any $y \in U$, $y \notin U_{p}$, and so $f(y) \geq p$. Also since $y \in U_{q}$, $f(y)\le q$. This gives $f(U) \subset (c,d)$. Hence $f$ is continuous. --- # Related Problems [[Urysohn Metrization Theorem]] --- # References [[Normal Spaces]] [[Closure and Interior and Limit Points]] [[Continuous Functions]]