Abstract Nonsense

Crushing one theorem at a time

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+1th 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 nth 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<z'\implies U_z\subseteq U_{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)<d'<\gamma. 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<t}\left(\overline{U_t}\right)'.

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<d'<\varphi(x). 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<t}\left(\overline{U_t}\right) 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.

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March 10, 2010 - Posted by | General Topology, Topology | , ,

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