# Abstract Nonsense

## Crushing one theorem at a time

We have discussed up until now Kolomogorov, $T_1$, Hausdorff, regular, and Tychonoff. We now discuss what (maybe bar Hausdoffness) the most fruitful separation axiom.

Normal ($T_4$): Let $X$ be a topological space. Then, $late X$ is called normal ($T_4$) if it is $T_1$ and given any disjoint closed $E,F\subseteq X$ there exists disjoint open $O_E,O_F$ which contain $E,F$ respectively.

We first prove a similar (and very functional) theorem that came up in regularity.

Theorem: Let $X$ be a topological space, then $X$ is normal if and only if given any closed $E\subseteq X$ and any neighborhood $N$ of $E$ there exists some neighborhood $U$ of $E$ such that $\overline{U}\subseteq N$.

Proof:

$\implies$: Since $X$ is normal and $E,N'$ are disjoint open sets there exists open sets $O_E,O_{N'}$ which respectively contain them. We claim that $\overline{O_E}\subseteq N$. But, this follows from an earlier problem since if $U,V$ are disjoint open sets then $v\in V\implies v\notin \overline{U}$. Thus, given any $n\in N'$ we have that $n\in O_{N'}\implies n\notin \overline{O_E}$. It follows that $N'\subseteq \overline{O_E}'\implies \overline{O_E}\subseteq N$. The conclusion follows.

$\Longleftarrow$: Conversely, let $E,F\subseteq X$ be disjoint and closed.  Clearly then $F'$ is a neighborhood of $E$ and by assumption there exists some neighborhood $U$ of $E$ such that $\overline{U}\subseteq F'$. Clearly then $E\subseteq U,F\subseteq\left(\overline{U}\right)'$ which are disjoint and open. The conclusion follows. $\blacksquare$.

The following theorem follows.

Theorem: Let $X$ be normal, then given any disjoint closed $E,F\subseteq$ there exists neighborhoods $U,V$ of them respectively such that $\overline{U}\cap\overline{V}=\varnothing$.

Proof: It follows from definition of normality that there exists disjoint open sets $O_E,O_F$ which contain each. But, by the previous theorem there exists neighborhoods $U,V$ of $E,F$ such that $\overline{U}\subseteq O_E,\overline{V}\subseteq O_F$. Clearly then $\overline{U}\cap\overline{F}\subseteq O_E\cap O_F=\varnothing$. The conclusion follows. $\blacksquare$.

Unfortunately normality doesn’t react as well with subspaces and products as the other separation axioms. Namely:

Theorem: Let $X$ be normal $E$ a closed subspace of $X$. Then $E$ is normal.

Proof: Let $G,F\subseteq E$ be disjoint and closed. Clearly then $G=E\cap G',F=E\cap F'$ for some closed $G',F'\subseteq X$. But, this clearly implies that $G,F$ are closed in $X$ and thus by its normality there exists disjoint open sets $O_G,O_F$ such that $G\subseteq O_G,F\subseteq O_F$. Clearly then $O_G\cap E,O_F\cap E$ are disjoint open sets in $E$ that contain $G,F$ respectively. The conclusion follows. $\blacksquare$

This in fact cannot be strengthened to a more general case. To come up with your own example try proving the general case and see where you run into issues. Things likes disjointness and openess in $X$ will be your main hinderances.

In fact, the product of just two normal spaces need not be normal. The spaces that is commonly used to illustrate this is complicated and so I suggest than an interested reader Google Sorgenfrey Plane.

The next theorem gives a nice result which enables many of the commonly encountered spaces to be automatically labeled normal.

Theorem: Let $X$ be a compact Hausdorff space, then $X$ is normal.

Proof: Let $G,F\subseteq X$ be disjoint and closed. By an earlier problem we know that since $X$ is Hausdorff and $F$ Hausdorff  for each $g\in G$ there exists disjoint open sets $O_g,O_{F_g}$ such that $g\in O_g,F\subseteq O_{F_g}$. Doing this for each $g\in G$ we obtain an open cover $\Omega=\left\{O_g\right\}_{g\in G}$ for $G$. But, by $G$‘s compactness there exists a finite subcover of $\Omega$, say $\left\{O_{g_1},\cdots,O_{g_n}\right\}$. Thus, if $\left\{O_{F_{g_1}}\right\},\cdots,O_{F_{g_n}}$ are the corresponding open sets containing $F$ but not $g_k$ we see that $\displaystyle U=\bigcup_{k=1}^{n}O_{g_k},V=\bigcap_{k=1}^{n}O_{F_{g_k}}$ are open sets such that $G\subseteq U,F\subseteq V$ and $U\cap V=\varnothing$. The conclusion follows. $\blacksquare$

Corollary: Since the product of Hausdorff spaces is Hausdorff and the product of compact spaces is compact we see that the previous theorem says that the product of compact Hausdorff spaces are normal.

The next theorem validates a strong suspicion one may have regarding normality and metric spaces.

Theorem: Any metric space (and thus any metrizable space) is normal.

Proof: Let $X$ be a metric space and $G,F\subseteq X$ be disjoint and closed. Since $G=\overline{G}=\left\{x\in X:d(x,G)=0\right\}$ we have that $f\in F\implies 2d(f,G)=\delta_f>0$. Similarly, define $\delta_g=2d(g,F)>0$. We leave it to the reader to verify that $\displaystyle G\subseteq\bigcup_{g\in G}B_{\delta_g}(g),F\subseteq \bigcup_{f\in F}B_{\delta_f}(f)$ and that these sets are disjoint. $\blacksquare$.

Our next theorem is fairly deep.

Theorem: Let $X$ be a second countable regular space, then $X$ is normal.

Proof: Let $E,G\subseteq X$ be disjoint and closed. For each $e\in E$ there exists a neighborhood $U_e$ of $e$ such that $U_e\cap G=\varnothing$. Using regularity we may find some neighborhood $V_e$ of $e$ such that $\overline{V_e}\subseteq U_e$. Clearly $\left\{V_e\right\}_{e\in E}$ covers $E$, but since every second countable space is Lindelof  this open cover has a countable subcover $\left\{V_{e_n}\right\}_{n\in\mathbb{N}}$. Form $\left\{K_{g_n}\right\}_{n\in\mathbb{N}}$ similary such that $\overline{K_{g_n}}\cap E=\varnothing$. The sets $\displaystyle V=\bigcup_{n=1}^{\infty}V_{e_n},K=\bigcup_{n=1}^{\infty}K_{g_n}$ are clearly open sets containing $E, G$ respectively. But, there is no need for them to be disjoint.

Thus, given $n$ define $\displaystyle \Omega_{e_n}=V_{e_n}-\bigcup_{j=1}^{n}\overline{K_{e_j}}$ and $\displaystyle \Sigma_{g_n}=K_{g_n}-\bigcup_{j=1}^{n}\overline{V_{e_j}}$. Clearly each $\Omega_{e_n},\Sigma_{g_n}$ is open. Also, $\left\{\Omega_{e_n}\right\}_{n\in\mathbb{N}}$ covers $E$ since for each $e\in E$, belongs to some $U_{e_n}$ and none of the sets $\overline{K_{g_n}}$. Similarly $\left\{\Sigma_{g_n}\right\}_{n\in\mathbb{N}}$ covers $G$.

But, if we define $\displaystyle \Omega=\bigcup_{n=1}^{\infty}\Omega_{e_n},\Sigma=\bigcup_{n=1}^{\infty}\Sigma_{g_n}$ then $\Omega\cap \Sigma=\varnothing$. To see this, suppose that $x\in \Omega\cap \Sigma$, then $\in \Omega_{e_n}\cap \Sigma_{g_m}$ for some $n,m\in\mathbb{N}$. Clearly either $n\leqslant m$ or $m\leqslant n$. But, if $n\leqslant m$ then it follows by definition that $x\in V_{e_n}$, and since $n\leqslant m$ it follows (once again by definition) that $x\notin \overline{V_{e_n}}$ which is of course a contradiction. A similar contradiction occurs if we assume that $m\leqslant n$. It follows that they are in fact disjoint. The conclusion follows. $\blacksquare$

We end this discussion on normality for now and begin next time with Urysohn’s lemma.