Regular and Normal Spaces

Regular and completely regular spaces

A Hausdorff space $X$ is said to be regular, if for every $x\in X$ and each closed subset $A\sbe X$ not containing $x$ we can find open subsets $U$ and $V$ such that $$ x\in U, A\sbe V\mbox{ and }U\cap V=\emptyset~. $$ A regular space $X$ is said to be completely regular, if for every $x\in X$ and each closed subset $A\sbe X$ not containing $x$ we can find a continuous function $f:X\rar[0,1]$, such that $f(x)=1$ and $f|A=0$.

1. Every subset $Y$ of a regular (completely regular) space $X$ is regular (completely regular). 2. The product of regular (completely regular) spaces is regular (completely regular).

Suppose $d_\a$, $\a\in I$ is a family of pseudo-metrics on a set $X$, which seperates point. Then the collection of subsets $$ {\cal S}\colon=\left\{B_\a(x,r):\, \a\in I,x\in X,r>0\right\} $$ forms a subbasis of a topology on $X$. The space $(X,d_\a,\a\in I)$ is called a uniform space.

Of course, this is not the purists way of introducing uniform spaces, but it's a convenient way to link them to topological spaces.

The following assertions are equivalent: 1. $X$ is completely regular. 2. $X$ carries the initial topology with respect to a family $f_\a:X\rar[0,1]$, which seperates points. 3. $X$ is a uniform space. 4. $X$ is homeomorphic to a subset of $[0,1]^I$.

A completely regular space with countable basis is metrizable.

Let $U_\a$, $\a\in I$, be an open cover of the topological space $X$. If for all $\a\in I$ the closure $\cl U_\a$ is Hausdorff or regular or completely regular, then so is $X$. If moreover $I$ is at most countable and all sets $\cl U_\a$ are separable or metrizable, then $X$ is separable or metrizable.

A Hausdorff space $X$ is regular if and only if for every open cover $U_\a$, $\a\in I$, we can find another open cover $V_\b$, $\b\in J$, such that for any $\b\in J$ there is some $\a\in I$ such that $\cl V_\b\sbe U_\a$.

Normal spaces

A Hausdorff space $X$ is said to be normal, if for each pair of closed and disjoint subsets $A$ and $B$ there exist disjoint open subsets $U$ and $V$, such that $A\sbe U$ und $B\sbe V$.

Every metric space $(X,d)$ is normal. Indeed suppose $A,B$ are disjoint, closed subsets, then $$ f(x)\colon=\frac{d_A(x)}{d_A(x)+d_B(x)} $$ is a continuous function satisfying $f|A=0$ und $f|B=1$. Hence the subsets $U\colon=[f < 1/2]$ and $V\colon=[f > 1/2]$ form disjoint open subsets such that $A\sbe U$ and $B\sbe V$.

Suppose $D$ is a dense subset of $\R_0^+$. For any $t\in D$ let $A_t$ be a subset of $X$ such that $s< t$ implies $A_s\sbe A_t$. Put $f(x)\colon=\inf\{t:x\in A_t\}$, then $f:X\rar[0,\infty]$ and for all $t\in\R_0^+$: $$ [f < t]=\bigcup\{A_s:\,s\in D,s < t\} \quad\mbox{und}\quad [f\leq t]=\bigcap\{A_s:\,s\in D,s > t\}~. $$

Suppose $X$ is normal and let $A,B$ closed, disjoint subsets of $X$. Then there exists a continuous function $f:X\rar[0,1]$, such that $f|A=0$ and $f|B=1$.

On a smooth manifold $M$ the function $f:M\rar[0,1]$ can be chosen to be smooth!

Let $X$ be normal, $A$ a closed subset of $X$ and $f:A\rar[0,1]$ continuous. Then there exists a continuous extension $F:X\rar[0,1]$, i.e. $F|A=f$.

Two continuous mappings $g:X\rar T$ and $h:Y\rar T$ can be attached to a single continuous mapping $G:X\cup_fY\rar T$ if and only if $h\circ f=g|A$. Let $Z$ be the disjoint union of $X$ and $Y$ and denote by $j_X$ and $j_Y$ the mappings $X\longrar Z\longrar X\cup_fY$ and $Y\longrar Z\longrar X\cup_fY$ respectively. Then $j_Y\circ f=j_X\circ i$, where $i:A\rar X$ denotes the canonical injection, and the mapping $G:X\cup_fY\rar T$ is uniquely determined by $G\circ j_X=g$ and $G\circ j_Y=h$. This actually characterizes attaching: Let $S$ be any topological space and $i_X:X\rar S$, $i_Y:Y\rar S$ continuous mappings, such that for every pair of functions $g:X\rar S$, $h:Y\rar S$ satisfying $i_Y\circ f=i_X\circ i$ there is exactly on continuous function $H:S\rar T$ such that $H\circ i_X=g$ and $H\circ i_Y=h$. Then $S$ is homeomorphic to $X\cup_fY$. This property is called the universal property of attaching: First of all there is exactly one continuous mapping $G:X\cup_fY\rar S$ such that $G\circ j_X=i_X$ and $G\circ j_Y=i_Y$. By assumption there is also a unique map $H:S\rar X\cup_fY$ such that $H\circ i_X=j_X$ and $H\circ i_Y=j_Y$. Hence $F\colon=G\circ H:S\rar S$ is continuous and we have: $F\circ i_X=G\circ j_X=i_X$ and $F\circ i_Y=G\circ j_Y=i_Y$; since $F$ is uniquely determined, we must have $F=id_S$ and analogously: $H\circ G=id_{X\cup_fY}$.

The attaching of two normal spaces $X$ and $Y$ is again normal.

Let $X$ be a normal space and let $f:X\rar Y$ be continuous, closed and onto. Then $Y$ is normal.

Suppose $R$ is an equivalence relation on a normal space $X$, such that the quotient map $\pi:X\rar X/R$ is closed. Then $X/R$ is normal.

Finally we are going to give another characterization of normal spaces in terms of open covers: for closed and disjoint subsets $A_1,A_2$ of a normal space $X$ we can find by definition two open, disjoint subsets $O_1$ and $O_2$ such that $O_i\spe A_i$. Since each of the pairs $A_i,O_i^c$ is again disjoint, we can find another pair of open sets $W_1,W_2$ such that $A_i\sbe W_i\sbe\cl{W_i}\sbe O_i$. Putting $U_i\colon=A_i^c$ and $V_i\colon=\cl{W_i}^c$, we infer that $U_1,U_2$ is an open cover of $X$ and $\cl{V_i}\sbe U_i$, moreover: $V_1\cup V_2=\cl{W_1}^c\cup\cl{W_2}^c=(\cl{W_1}\cap\cl{W_2})^c\spe(O_1\cap O_2)^c=X$. Thus we conclude, that for every open cover $U_1,U_2$ of a normal space there is another open cover $V_1,V_2$ satisfying $\cl{V_i}\sbe U_i$.

A Hausdorff space is normal if and only if for every open locally finite cover $U_\a$, $\a\in I$, there exists an open cover $V_\a$, $\a\in I$, such that for all $\a\in I$: $\cl V_\a\sbe U_\a$ ($V_\a$ may be empty).