# Abstract Nonsense

## Thoughts about separation (Urysohn’s lemma)

In this chapter we discuss one of the most important theorems in point-set topology. Please, forgive my in-eloquence is the description of the creation of the sets.

Theorem (Urysohn’s Lemma): Let $X$ be a normal space and $E,G\subseteq$ be disjoint and closed. Then, there exists some $\varphi:X\mapsto[0,1]$ which is continuous such that $\varphi(E)=0,\varphi(G)=1$.

Proof: We will do this in steps.

Step 1: Clearly $G'$ is a neighborhood of $E$ and thus by normality we may find some open $U_{\frac{1}{2}}$ such that $E\subseteq U_{\frac{1}{2}}\subseteq \overline{U_{\frac{1}{2}}}\subseteq G'$. Appealing to normality again we may find open $U_{\frac{1}{4}},U_{\frac{3}{4}}$ such that $E\subseteq U_{\frac{1}{4}}\subseteq\overline{U_{\frac{1}{4}}}\subseteq U_{\frac{1}{2}}\subseteq \overline{U_{\frac{1}{2}}}\subseteq U_{\frac{3}{4}}\subseteq\overline{U_{\frac{3}{4}}}\subseteq G'$.

So now we now we use the principle of recursive definition to form a class of sets $\Omega=\left\{U_d:d\in D\right\}$ such that $D=\left\{\frac{m}{2^n}:n\in\mathbb{N},\text{ }m=1,\cdots,2^n-1\right\}$ which is trivially dense in $[0,1]$.

So, define $U_0=E,U_1= G'$ and supposing that the first $n$ placement (the meaning of this term will be apparent in a second) define the $n+1$th placement as follows. Let $I$ be the set of all indexes of $U_k$‘s such that $U_k$ was created at or before the $n$th placement.

By normality we may find some open $U_j$ with $j=\frac{\min\text{ }I}{2}$ such that $E\subseteq U_j\subseteq\overline{U_j}\subseteq U_{\min\text{ }I}$. Similarly we may find some $U_\ell$ with $\ell=\frac{\max\text{ }I+1}{2}$ such that $\overline{U_{\max\text{ }I}}\subseteq U_\ell\subseteq\overline{U_\ell}\subseteq G'$. Also, for any $\overline{U_m},\text{ }m\in I-\{0,1\}$ we may find some $U_p$ with $p=\frac{\min{ }\left(I\cap (m,1)\right)+m}{2}$ such that $\overline{U_m}\subseteq U_p\subseteq \overline{U_p}\subseteq U_{\min\left(I\cap (m,1)\right)}$.

Doing this we clearly arrive at a set $\Omega=\left\{U_d:d\in D\right\}$ such that $z.

Step 2: We now define $\varphi:X\mapsto[0,1]$ by $\displaystyle \varphi(x)=\begin{cases}0 & \mbox{if} \quad x\in U_d,\text{ }\forall d\in D \\ \sup\left\{d:x\notin U_d\right\} & \mbox{if} \quad \text{otherwise} \end{cases}$.

It is clear that since $E\subseteq U_d$ for every $d\in D$ that $\varphi\left(E\right)=0$ and since $U_d\subseteq G'$ for every $d\in D$ we see that $\varphi\left(G\right)=1$. Thus, it remains to show that $\varphi$ is continuous.

Since the set of all $[0,\gamma),(\gamma,1]$ such that $0<\gamma<1$ form an open subbase for $[0,1]$ we must merely show that their inverse images are open.

Claim: $\displaystyle \varphi^{-1}\left([0,\gamma)\right)=\bigcup_{t<\gamma}U_t$.

Proof: Let $x\in \varphi^{-1}\left([0,\gamma)\right)$ then $0\leqslant \varphi(x)<\gamma$. By $D$‘s density in $[0,1]$ we know there exists some $d'\in D$ such that $\varphi(x). Now, suppose that $x\notin U_{d'}$, then $\sup\left\{d:x\notin U_d\right\}=\varphi(x)\geqslant d'$ which is clearly a contradiction. It follows that $x\in U_{d'}$ and consequently $\displaystyle x\in\bigcup_{t<\gamma}U_t$.

Conversely, let $\displaystyle x\in\bigcup_{t<\gamma}U_t$ then $x\in U_{\psi}$ for some $\psi<\gamma$. But, as stated earlier $\psi<\zeta\implies U_{\psi}\subseteq U_{\zeta}$ and so $x\in U_{\zeta}$ for every $\zeta\geqslant \psi$. It follows that $\varphi(x)<\psi<\gamma$ and so $x\in\varphi^{-1}\left([0,\gamma)\right)$. The conclusion follows. $\blacksquare$.

It follows that $\varphi^{-1}\left([0,\gamma)\right)$ is the union of open sets, and thus open.

Claim: $\displaystyle \varphi^{-1}\left((\gamma,1]\right)=\bigcup_{\gamma.

Proof: Let $x\in \varphi^{-1}\left((\gamma,1]\right)$ then $\gamma<\varphi(x)\leqslant 1$. By the density of $D$ we know that there exists some $d'\in D$ such that $\gamma. Now, assume that $x\notin\left(\overline{U_{d'}}\right)'\implies x\in \overline{U_{d'}}$ then clearly $x\in U_{d'}$ and so by earlier reasoning $x\in U_\psi$ for every $\psi>d'$ and so $\sup\left\{d:x\notin U_d\right\}=\varphi(x)\leqslant d'$ which is of course a contradiction. It follows that $x\in \left(\overline{U_{d'}}\right)'$ and so $\displaystyle x\in\bigcup_{t<\gamma}U_t$.

Conversely, let $\displaystyle x\in\bigcup_{\gamma then $x\notin \overline{U_{\beta}}$ and consequently $x\notin U_{\beta}$ for some $\beta<\gamma$. It follows that $\sup\left\{d:x\notin U_d\right\}=\varphi(x)\geqslant \beta>\gamma$. Thus, $x\in\varphi^{-1}\left((\gamma,1]\right)$. The conclusion follows. $\blacksquare$.

We thus see that $\varphi^{-1}\left((\gamma,1]\right)$ is the union of open sets and thus open. By previous comments this finishes the proof. $\blacksquare$

Corollary: If $X$ is normal and $E,G$ are closed disjoint subsets of $X$ there exists a continuous mapping $\varphi:X\mapsto[a,b]$ such that $\varphi(E)=a,\varphi(G)=b$.

Remark: Having, for example with $[a,b]=[0,1]$, $\varphi(E)=0$ and $\varphi(G)=1$ is preference. Taking $\psi=1-\varphi$ gives a continuous mapping such that $\psi(E)=1$ and $\psi(G)=0$

Corollary: If $X$ is normal then given any $x\in X$ and any neighborhood $U$ of $X$ there exists a function $\varphi:X\mapsto [0,1]$ such that $\varphi(x)>0$ and $\varphi(u)=0$ for every $u\in X-U$.

Proof: Let $x\in X$ be arbitrary and $U$ any neighborhood of $X$. By normality there exists some neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$. Thus, $\overline{V}$ and $X-U$ are disjoint closed sets and so there exists some $\varphi:X\mapsto [0,1]$ such that $\varphi(\overline{V})=1$ and $\varphi(X-U)=0$. The conclusion follows. $\blacksquare$

Corollary: Every normal space is completely regular.

There is in fact a nice function which works in place of the above for metric spaces.

Theorem: Let $X$ be a metric space and $E,G$ disjoint closed subspaces. Then, there exists a mapping $\rho:X\mapsto [0,1]$ such that $\rho(E)=0,\rho(G)=1$.

Proof: Define $\rho:X\mapsto [0,1]$ by $\displaystyle \rho(x)=\frac{d(E,x)}{d(E,x)+d(G,x)}$. It is clear that $\rho(X)\subseteq [0,1]$ (just noting that the mapping is well-defined).

We first remember from basic analysis that if $f,g$ are continuous real valued functions then $\displaystyle f+g,\frac{f}{g}$ are continuous provided that $g$ is never zero on $\text{Dom }g$. So, $\rho(x)$ is continuous (since as was proved the mapping $d(x,M)$ is continuous) as long as $d(x,E)+d(x,G)\ne 0$ for any $x\in X$. But, this is clear since $E=\overline{E}=\left\{y\in X:d(y,E)=0\right\}$ and similarly $G=\overline{G}=\left\{y\in X:d(y,g)\right\}$ and since $d(x,E),d(x,G)\geqslant 0$ to have $d(x,E)+d(x,G)=0$ we must have that $d(x,E)=d(x,G)=0$ and thus by previous comment $x\in E\cap G$ which of course contradicts the disjointness of these two sets.

It follows that $\rho:X\mapsto [0,1]$ is a continuous function, and since if $e\in E$ we have that $\displaystyle \rho(e)=\frac{d(e,E)}{d(e,E)+d(e,G)}=\frac{0}{0+d(e,G)}=0$; and if $g\in G$ we have that $\displaystyle \rho(g)=\frac{d(g,E)}{d(g,E)+d(g,G)}=\frac{d(g,E)}{d(g,E)+0}=1$ we may conclude that $\rho(E)=0,\rho(G)=1$. The conclusion follows. $\blacksquare$

While there are many more interesting problems/theorems consequent of Urysohn’s lemma we discuss only a few in the subsequent posts.

For example there is a “strong” Urysohn lemma where $f^{-1}(0)=E,f^{-1}(G)=1$ but this requires the introduction of the idea of $G_{\delta}$ sets which I am weary to do…considering the workload.