Urysohn's Lemma: Proof

Solution 1:

$\newcommand{\cl}{\operatorname{cl}}$I prefer a more topological proof.

Let $D$ be the set of dyadic rationals in $[0,1]$. We shall construct a family $\{U(d):d\in D\}$ of sets $A \subseteq U(d) \subseteq \Omega\setminus B$, all open with the possible exception of $U(0)$, satisfying the condition that $\cl U(d_0)\subseteq U(d_1)$ whenever $d_0,d_1\in D$ and $d_0<d_1$.

For $n\in\Bbb Z^+_0$ = set of non-negative integers let $D_n = \{\frac{k}{2^n} : k\in\{0,\ldots,2^n\} \}$. We have $D_{n} \subseteq D_{n+1}$ and $\bigcup_{n=0}^\infty D_n = D$. We shall construct the desired $U(d)$ inductively for $d \in D_n$.

For $n = 0$ let $U(0)=A$ and $U(1)=\Omega\setminus B$. Note that $U(0)$ is closed, but in general not open.

Now suppose the $U(d)$ have been constructed for $d \in D_n$. By normality, for each $k\in\{0,\ldots,2^n\}$ there is an open $U\left(\frac{2k+1}{2^{n+1}}\right)$ such that

$$\cl U\left(\frac{k}{2^n}\right)=\cl U\left(\frac{2k}{2^{n+1}}\right)\subseteq U\left(\frac{2k+1}{2^{n+1}}\right)\subseteq\cl U\left(\frac{2k+1}{2^{n+1}}\right)\subseteq U\left(\frac{k+1}{2^n}\right)\;.$$

Let

$$h:\Omega\to[0,1]:x\mapsto\begin{cases} 1,&\text{if }x\notin U(1)\\ \inf\{d\in D:x\in U(d)\},&\text{otherwise}\;. \end{cases}$$

Clearly $h[A]=\{0\}$ and $h[B]=\{1\}$, so it only remains to show that $h$ is continuous. First note that if $r\in D$ and $x\in U(r)$, then $h(x)\le r$, while if $x\in\Omega\setminus\cl U(r)$ then $h(x)\ge r$. Now let $x\in\Omega$ and $\epsilon>0$ be arbitrary.

• If $h(x)=0$, choose $r\in D$ so that $0<r<\epsilon$; then $V=U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq[0,\epsilon)$.

• If $h(x)=1$, choose $r\in D$ so that $1-\epsilon<r<1$; then $V=\Omega\setminus\cl U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq(1-\epsilon,1]$.

• If $0<h(x)<1$, choose $r,s\in D$ so that $h(x)-\epsilon<r<h(x)<s<h(x)+\epsilon$; then $V=U(s)\setminus\cl U(r)$ is an open nbhd of $x$ such that $h[V]\subseteq(h(x)-\epsilon,h(x)+\epsilon)$.

Thus, $h$ is continuous at every point of $\Omega$.

Solution 2:

Disclaimer: This is community wiki; feel free to correct, edit, expand etc.! ;)

(This is a slight modification taken from Rudin, Real and Complex Analysis.)

Regard for simplicity: $$C:=B\subseteq A^\complement=:U$$

Denote for construction the rationals: $$S:=\mathbb{Q}\cap[0,1]$$

By normality one can find inductively open $V_s$ for $s \in S$ such that $$C\subseteq V_1, \quad \overline{V_0}\subseteq U,\quad V_s\subseteq\overline{V_{s'}} \text{ for } s>s' .$$

Consider the upper and lower semicontinuous functions: $$\underline{h}_s:=s\chi_{V_s}:\quad\underline{h}_s^{-1}(a,\infty)=\varnothing,V_s,\Omega\in\mathcal{T}$$ $$\overline{h}_s:=(1-s)\chi_{\overline{V_s}}+s:\quad\overline{h}_s^{-1}(-\infty,b)=\varnothing,\overline{V_s}^\complement,\Omega\in\mathcal{T}$$

Pointwise suprema resp. infima preserve lower resp. upper semicontinuity: $$\underline{h}:=\sup_{s\in S}\underline{h}_s:\quad\underline{h}^{-1}(a,\infty)=\{\omega:\underline{h}(\omega)>a\}=\bigcup_{s\in S}\{\omega:\underline{h}_s(\omega)>a\}=\bigcup_{s\in S}\underline{h}_s^{-1}(a,\infty)\in\mathcal{T}$$ $$\overline{h}:=\inf_{s\in S}\overline{h}_s:\quad\overline{h}^{-1}(-\infty,b)=\{\omega:\overline{h}(\omega)<b\}=\bigcup_{s\in S}\{\omega:\overline{h}_s(\omega)<b\}=\bigcup_{s\in S}\overline{h}_s^{-1}(-\infty,b)\in\mathcal{T}$$

They approach each other as by contradiction: $$\underline{h}(\omega)<s<s'<\overline{h}(\omega)\implies\omega\in\overline{V_{s'}}\subseteq V_s\not\owns\omega$$ $$\underline{h}_s(\omega)>\overline{h}_{s'}(\omega)\implies\omega\in V_s\subseteq\overline{V_s}\subseteq V_{s'}\subseteq\overline{V_{s'}}\not\owns\omega\quad(s>s')$$

So they together become continuous: $$h:=\overline{h}=\underline{h}:\quad h^{-1}(c-\varepsilon,c+\varepsilon)=h^{-1}(-\infty,c+\varepsilon)\cup h^{-1}(c-\varepsilon,\infty)\in\mathcal{T}$$

Thus the limiting function is the desired bump: $$h(C)\subseteq h(V_1)\equiv1,\,h(U^\complement)\subseteq h(\overline{V_0}^\complement)\equiv0$$