# Prove that Normal $\implies$ Regular $\implies T_2 \implies T_1 \implies T_0$

Prove that Normal $\implies$ Regular $\implies T_2 \implies T_1 \implies T_0$

Proof: Let $\mathscr{X}$ be a topological space such that, for each $x \in \mathscr{X}$, the set $\{x\}$ is closed. We will show that if $\mathscr{X}$ is a normal space this implies $\mathscr{X}$ is a regular space. Suppose $\mathscr{X}$ is a normal space. By Definition 3.24 this means for every pair of disjoint closed sets $C, D \subseteq \mathscr{X}$, there are disjoint open sets $U$ and $V$ such that $C \subseteq U$ and $D \subseteq V$. Then let $C \subseteq \mathscr{X}$ and $D \subseteq \mathscr{X}$ be a pair of disjoint closed sets and let $U$ and $V$ be disjoint open sets such that $C \subseteq U$ and $D\subseteq V$. Since $C$ and $D$ are disjoint closed sets, we know that there exists some $x\in \mathscr{X}$ such that $x\in C$ but $x \notin D$. Let $y\in C$ such that the set $\{y\}$ is closed. Since $C$ and $D$ are disjoint closed sets we know that $y\in C$ implies that $y\notin D$. This implies that $y\in U$ but $y\notin V$. But by Definition 3.24 we have that $D\subseteq V$. Therefore, we have shown that $\mathscr{X}$ is a regular space because $y\in U$ and $D\subseteq V$, while for the closed set $D\subseteq \mathscr{X}, y\notin D$.

We will show that Regular$\,\Rightarrow T_{2}$. Suppose $\mathscr{X}$ is a regular space. Let $E\subseteq \mathscr{X}$ be a closed set such that for every point $z\notin E$, there are disjoint open sets $Q$ and $R$ such that $z\in Q$ and $E\subseteq R$. Let $b,c \in \mathscr{X}$ be a pair of disjoint points such that $b\in E$ and $c\notin E$. Let $Q$ and $R$ be disjoint open sets such that $c\in Q$ and $E\subseteq R$. Since $b\in E$, this implies $b\in R$. But since $Q$ and $R$ are disjoint open sets we know $b\notin Q$. But we have that $c\in Q$. Therefore, we have that $b\in R$ and $c\in Q$ by Definition 3.22 means that $\mathscr{X}$ is $T_{2}$, or a Hausdorff space.

We will show $T_{2}\,\Rightarrow T_{1}$. Suppose $\mathscr{X}$ is a $T_{2}$ space. Let $x,y \in \mathscr{X}$ be disjoint points such that, for disjoint open sets $U$ and $V$, $x\in U$ and $y\in V$. Since $U$ and $V$ are disjoint we know $x\in U$ implies $x\notin V$ and $y\in V$ implies $y\notin U$. By Definition 3.21, this means $\mathscr{X}$ is $T_{1}$.

We will show that $T_{1}\,\Rightarrow T_{0}$. Suppose $\mathscr{X}$ is $T_{1}$. Let $x,y\in \mathscr{X}$ be a pair of disjoint points such that $x\in U$ but $y\notin U$, and $y\in V$ but $x\notin V$. By Definition 3.20, we have $\mathscr{X}$ is $T_{0}$.

I believe the problem in this proof is where I show a regular space implies the space is $T_{2}$. I am looking for proof verification.

Definitions used for proof

• What is your question? Jul 16, 2017 at 18:00
• @Andres Mejia Is this proof correct? Jul 16, 2017 at 18:06
• You should mention what you think may be problematic with the write up, and mention somewhere in the body of the text that you are asking for a poof verification. Jul 16, 2017 at 18:08
• The definition for T1 is redundent. For all distinct x,y, some open U with x in U, y not in U is the usual definition and equivalent to the double statement given in the list of definitions. Exercise. Show a space is T1 iff for all x, {x} is closed. Jul 16, 2017 at 19:26

E.g.: Let $X$ be a normal space. To see it's regular note that in a normal space all sets $\{x\}$ are closed (and this is part of the requirements for regular as well). Let $x$ be a point not in the closed set $C$. Then $\{x\}$ and $C$ are disjoint closed sets in the normal space $X$, so they can be separated by disjoint open sets. So $X$ is regular.
It's really simple: normal implies regular because points are closed (in your definition), and so separating two disjoint closed sets implies separating a point disjoint from a closed sets, which implies separating two points ($T_2$) as well, it's all a matter of choosing $\{x\}$ as closed sets in the regular or normal definition. Your proofs of $T_2$ implies $T_1$ implies $T_0$ are OK.