# Abstract Nonsense

## Thoughts about separation (Regularity and Complete Regularity)

So up until this point we’ve discussed spaces which are Kolomogorov, $T_1$, or Hausdorff. We now discuss the next property in the progression.

Regular ($T_3$): Let $X$ be a topological space. We call $X$ regular if it’s $T_1$ and given any closed subset $E$ of $X$ and any point $x\in X-E$ there exists disjoint open neighborhoods of both.

Remark: It is quite common to call a space regular if it only has the second property mentioned and $T_3$ if is also $T_1$. We, obviously, do not have that sentiment.

It clearly follows that every regular space is Hausdorff since its $T_1$ness guarantees that  points are closed sets.

We first discuss the most functional aspect of regular spaces.

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

Proof:

$\implies$: Suppose that $X$ is regular and let $x\in X$ be arbitrary. For any neighborhood $N$ of $x$ we have that $X-N$ is a closed set not containing $x$. By assumption then there exists disjoint open neighborhoods $U_x,U_{X-N}$ of each. We claim that $\overline{U_x}\subseteq N$.

To see this, let $x\notin N$ then $x\in U_{X-N}$ and since $U_x\cap U_{X-N}=\varnothing$ it follows by an earlier problem that $x\notin \overline{U_{x}}$. The conclusion follows.

$\Longleftarrow$: Let latex $E\subseteq X$ be closed and $x\in X-E$. Since they are disjoint we have that $E'$ is an open set containing $x$ and by assumption there exists some neighborhood $U$ of $x$ such that $\overline{U}\subseteq E'$. So, it follows quite readily that $U,\left(\overline{U}\right)'$ are the sets we seek.

$\blacksquare$

From this we get the following almost corollary.

Theorem: Let $X$ be regular and $x,y\in X$ be distinct. Then there exists neighborhoods $U,V$ of $x,y$ respectively such that $\overline{U}\cap\overline{V}=\varnothing$.

Proof: Since $X$ is Hausdorff there exists disjoint open neighborhoods $O_x,O_y$ of $x,y$ respectively. By the previous problem these neighborhoods contain neighborhoods $U,V$ such that $\overline{U}\subseteq O_x,\overline{V}\subseteq O_y$. These are clearly the neighborhoods we desired. $\blacksquare$

The following is the next step in the obvious progression which shows that in some sense compact sets act very similarly to points.

Theorem: Let $X$ be regular and $E,F\subseteq X$ be disjoint closed subsets such that $E$ is compact. There exists disjoint open sets $O_E,O_F$ such that $E\subseteq O_E,F\subseteq O_F$.

Proof: Clearly for each $e\in E$ we clearly have that $e\in X-F$ and by the regularity of $X$ there exists disjoint open sets $O_e,O_{F_e}$ such that $e\in O_e,F\subseteq O_{F_e}$. Clearly the class $\left\{O_{e}\right\}_{e\in E}$ is an open cover of $E$ and thus by $E$‘s compactness it has a finite subcover $\left\{O_{e_1},\cdots,O_{e_n}\right\}$. Clearly if $\left\{O_{F_{e_1}},\cdots,O_{F_{e_n}}\right\}$ are the corresponding open sets containing $F$ that $E\subseteq\displaystyle \bigcup_{j=1}^{n}O_{e_j}$ and $\displaystyle F\subseteq \bigcap_{j=1}^{n}O_{F_{e_j}}$ and both are disjoint. The conclusion follows. $\blacksquare$

We note that regularity is hereditary.

Theorem: Let $X$ be regular and $E$ a subspace of $X$ then $E$ is regular.

Proof: Let $G\subseteq E$ be closed and $x\in E-G$. Since $G$ is closed we have that $G=E\cap C$ for some closed set $C$ in $X$. Clearly $x\notin C$ otherwise $x\in C\text{ }x\in E\implies x\in C\cap E=G$. Thus, by $X$‘s regularity there exists open sets $O_x,O_C$ such that $x\in O_x,\text{ }C\subseteq O_C$ and $O_x\cap O_C$. Clearly then $E\cap O_x,E\cap O_C$ are disjoint sets which contain $x$ and $G$ respectively. The conclusion follows. $\blacksquare$

Next is the obvious.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty collection of regular spaces, then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is regular.

Proof: We actually need a little lemma first which we should have proved in the product topology section, namely:

Lemma: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a collection of topological spaces and $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$. Then, given any class $\left\{E_j\right\}_{j\in\mathcal{J}}$ such that $E_\ell\subseteq X_\ell,\text{ }\forall \ell\in \mathcal{J}$ then $\displaystyle \overline{\prod_{j\in\mathcal{J}}E_j}=\prod_{j\in\mathcal{J}}\overline{E_j}$.

Proof:

Let $\displaystyle \bold{x}\in \prod_{j\in\mathcal{J}}\overline{E_j}$ and let $N$ be any neighborhood of $x$. By assumption $\pi_j(\bold{x})\in\overline{E_j}$ for each $j\in\mathcal{J}$ and so there exists some point $e_j \in E_j$ in $\prod_{j}(N)$. It follows that $\displaystyle \prod_{j\in\mathcal{J}}\{e_j\}\in \overline{\prod_{j\in\mathcal{J}}E_j}\cap N$. Since $N$ was arbitrary it follows that $\bold{x}$ is an adherent point of $\displaystyle \prod_{j\in\mathcal{J}}E_j$ and so $\displaystyle \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}$.

Conversely, let $\bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}$. Let $\ell\in \mathcal{J}$ be arbitrary, and let $O_\ell$ be an open set containing $\pi_\ell(\bold{x})$. Since $\pi_\ell^{-1}(O_\ell)$ is open in $X$ it contains a point $\displaystyle \bold{y}\in\prod_{j\in\mathcal{J}}E_j$. Then $\pi_\ell(\bold{y})\in O_\ell \cap E_\ell$. The conclusion follows. $\blacksquare$

Let $\bold{x}\in X$ be arbitrary and let $N$ be any basic neighborhood of $\bold{x}$. Let $j_1,\cdots,j_n$ be the finite number of indices such that $\pi_{j_k}(\bold{x})\ne X_{j_k}$. It is clear that $\pi_{j_k}(N)$ is a neighborhood of $\pi_{j_k}(\bold{x})$ and by $X_{j_k}$‘s regularity there exists some $U_{j_k}\subseteq \pi_{j_k}(N)$ such that $\overline{U_{j_k}}\subseteq \pi_{j_k}(N)$. Clearly then $\displaystyle \prod_{j\in\mathcal{J}}G_j$ where $\displaystyle G_j=\begin{cases} U_{j_k} & \mbox{if} \quad j=j_k,\text{ }1\leqslant k\leqslant n \\ X_j & \mbox{if} \quad j\ne j_k\text{ }1\leqslant k\leqslant n\end{cases}$. This is clearly a neighborhood of $\bold{x}$ and $\displaystyle \overline{\prod_{j\in\mathcal{J}}G_j}=\prod_{j\in\mathcal{J}}\overline{G_j}\subseteq N$ since $\overline{U_{j_k}}\subseteq \pi_{j_k}(N)$. The conclusion follows. $\blacksquare$

We now prove a partial converse.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be a non-empty collection of topological spaces and $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ be under the product topology. If $X$ is regular then so is $X_\ell$ for any $\ell\in\mathcal{J}$

Proof: Let $F\subset X_\ell$ be closed and $x\in X_\ell-F$. Clearly then $\displaystyle \bold{F}=\prod_{j\in\mathcal{J}}E_j$ where $\displaystyle \begin{cases} F & \mbox{if} \quad j=\ell \\ X_j & \mbox{if} \quad j\ne \ell \end{cases}$ is a  closed subset of $X$  and $\displaystyle \bold{x}\in X-\bold{F}$ where $\displaystyle \bold{x}=\prod_{j\in\mathcal{J}}\{e_j\}$ with $\displaystyle e_j=\begin{cases} x & \mbox{if} \quad j=\ell \\ x_j & \mbox{if} \quad j\ne \ell \end{cases}$ where $\left\{x_j\right\}_{j\in\mathcal{J}-\{\ell\}}$ is a fixed but arbitrary set in $X-\bold{F}$. Thus, by $X$‘s regularity there exists disjoint open sets $O_{\bold{x}},O_{\bold{F}}$ such that $\bold{x}\in O_{\bold{x}}$ and $\bold{F}\subseteq O_{\bold{F}}$. Clearly then $x\in\pi_\ell(O_{\bold{x}})$ and $F\subseteq \pi_\ell(O_{\bold{F}})$. It thus remains to show that they are disjoint.

Thus, assume that $z\in \pi_\ell(O_{\bold{x}})\cap \pi_\ell(O_{\bold{F}})$ then $\displaystyle \bold{z}\in O_{\bold{x}}\cap O_{\bold{F}}$ where $\displaystyle \bold{z}=\prod_{j\in\mathcal{J}}\{y_j\}$ where $\displaystyle y_j=\begin{cases} z & \mbox{if} \quad j=\ell\\ x_j & \mbox{if} \quad j\ne\ell \end{cases}$ where $x_j$ is as before. This is of course a contradiction. The conclusion follows.

Theorem: Let $X$ be a regular space and $\varphi:X\mapsto Y$ be perfect. Then, $Y$ is regular.

Proof: Let $F\subseteq Y$ be closed and $\varphi(x)\in X-F$. It follows from $\varphi$‘s perfection that $\varphi^{-1}\left(\varphi(x)\right),\varphi^{-1}\left(F\right)$ are disjoint closed subsets of $X$ such that $\varphi^{-1}(\varphi(x))$ is compact. By an earlier theorem there exists open sets $O_x,O_F$ such that $\varphi^{-1}(\varphi(x))\subseteq O_x,\varphi^{-1}\left(F\right)\subseteq O_F$. Clearly then $X-O_x,X-O_F$ are disjoint closed sets containing $\varphi^{-1}\left(F\right),\varphi^{-1}(\varphi(x))$ respectively. Once again, by $\varphi$‘s perfection we see that $\varphi\left(X-O_x\right),\varphi\left(X-O_F\right)$ are closed sets containing $F,\varphi(x)$ respectively. Thus, $Y-\varphi\left(X-O_X\right),Y-\varphi\left(X-O_F\right)$ are open sets containing $\varphi(x),F$ respectively. All that remains is to show that they are disjoint.

So, let $\varphi(z)\in Y-\varphi\left(X-O_x\right)$ then $z\notin X-O_x\implies z\in O_x$ and so $z\notin O_F$ and so $z\in X-O_F$ and thus $\varphi(z)\in \varphi\left(X-O_x\right)$ and thus $\varphi(z)\notin Y-\varphi\left(X-O_x\right)$. A similar argument shows that $\varphi(z)\in Y-\varphi\left(X-O_F\right)\implies \varphi(z)\notin Y-\varphi\left(X-O_x\right)$. The conclusion follows. $\blacksquare$

Now, of course you might ask for an example a Hausdorff space which is not regular. Unfortunately, they are not as easy to come by. We outline such a topological space.

Example: Let $\mathbb{R}$ be such that $O\subseteq\mathbb{R}$ is open if  $O=U-C$ where $U$ is open in the usual topology and $C$ is countable.

Our next example is a much more analysis motivated concept.

Tychonoff (completely regular $T_\pi$): Let $X$ be a topological space, then it is called Tychonoff (completely regular/$T_\pi$) if it is $T_1$ and given any closed $F\subseteq X$ and any $x\in X-F$ there exists some $\varphi\in \mathcal{C}\left[X,[0,1]\right]$ such that $\varphi(x)=0$ and $\varphi(F)=1$

It is clear that every Tychonoff space is regular since given any closed $F\subseteq X$, $x\in X-F$, and the corresponding $\varphi\in\mathcal{C}\left[X,\mathbb{R}\right]$ the sets $\varphi^{-1}\left(\left([0,\frac{1}{2}\right)\right),\varphi^{1}\left(\left(\frac{1}{2},1\right]\right)$ are disjoint open sets containing $x,F$ respectively.

Theorem: Let $X$ be Tychonoff and $E$ a subspace of $X$. Then, $E$ is Tychonoff.

Proof: Let $F\subseteq E$ be closed and $e\in X-F$. Since $E$ is a subspace of $X$ we have that $F=E\cap G$ for some closed $G$ in $X$. Clearly $e\notin G$ otherwise $e\in G\cap E=F$. Thus, $e\in X-G$ and by $X$‘s complete regularity there exists some $\varphi:X\mapsto [0,1]$ which is continuous and $\varphi(x)=0,\varphi(G)=1$. Thus, $\varphi\mid E:E\mapsto [0,1]$ is clearly continuous and $\varphi(x)=0,\varphi(F)=1$. The conclusion follows. $\blacksquare$

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ then $\displaystyle X=\prod_{j\in\mathcal{J}}X_j$ is Tychonoff under the product topology if  $X_\ell$ is Tychonoff for each $\ell\in\mathcal{J}$.

Proof: Let $\bold{E}\subseteq X$ be closed and $\bold{x}\in X-\bold{E}$. We may clearly choose a basic open set $\mathcal{U}$ such that $\bold{x}\in\mathcal{U}\subseteq X-\bold{E}$. Let $j_1,cdots,j_n$ be the finitely many indices such that $\pi_{j_k}(\bold{U})\ne X_{j_k}$. Since by assumption each of $X_{j_1},\cdots,X_{j_n}$ are Tychonoff we may find some function $\varphi_{k}:X_{j_k}\mapsto [0,1]$ such that $\varphi\left(\pi_{j_k}(\bold{x})\right)=1,\text{ }\varphi\left(X_{j_k}-\pi_{j_k}(\bold{U})\right)=0$. Let $\psi_k=\varphi_k(\pi_k(\bold{y}))$ then $\psi_k:X\mapsto [0,1]$ continuously and vanishes outside of $\pi_{j_k}^{-1}\left(\pi_{j_k}\left(\bold{U}\right)\right)$. So, let $\varphi:X\mapsto [0,1]$ be given by $\varphi(\bold{y})=\psi_{1}(\bold{y})\cdots \psi_n(\bold{y})$. Then this is a continuous mapping of $X$ into $[0,1]$ such that $\varphi(\bold{x})=1\cdots 1=1$ and $\varphi\left(X-\bold{U}\right)=0$. The conclusion follows.

$\implies$: Let