# Why is the $T_1$ axiom necessary in this lemma?

I've problems understanding the necessity of the $T_1$ separation axiom in a certain lemma. First I'll the definitions we are using in class. Let $X$ be a topological space.

$X$ satisfies the $T_1$ separation axiom iff every singleton is closed.

also

$X$ is regular if for each point $x$ and closed $C$ such that $x\not\in C$ there are open disjoint neighborhoods $U,V$ of $x, C$ respectively.

Notice my definition of regularity doesn't ask $X$ to be $T_1$. I also understand that a regular space needs the $T_0$ axiom to be Hausdorff. My question is

Why is the $T_1$ axiom necessary in the proof of the following lemma?

Here's the lemma:

Lemma 31.1 Let $X$ be a $T_1$ topological space. Then $X$ is regular if and only if given a point $x\in X$ and a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline V \subset U$ (where $\overline V$ denotes the closure of $V$)

Proof: Suppose $X$ is regular. Take any $x\in X$ and $U$ any of its neighborhoods. Then $U^c$ is closed and by hypothesis there are open disjoint neighborhoods $V, W$ of $x,U^c$ respectively. If $y\in \overline V\cap U^c$, then $y\in W$ but $W\cap V=\varnothing$ Hence $y\not\in\overline V$ and thus, $\overline V\cap U^c=\varnothing \Rightarrow \overline V\subset U$ as we wanted.

For the converse, take any $x\in X$ and a disjoint closed $C$. Then $C^c$ is an open neighborhood of $x$ and by hypothesis there is a neighborhood $V$ of $x$ such that $\overline V\subset C^c$. Then $V$ and $\overline V ^c$ are disjoint open neighborhoods of $x, C$ respectively. Hence $X$ is regular. $\blacksquare$

I understand all the steps but I don't see where the $T_1$ axiom hypothesis plays a key role. I'm pretty sure the reason to ask for the $T_1$ axiom is to avoid pathological cases, but I couldn't think of any.

• A space in the indiscrete topology satisfies the condition of the lemma but is not regular if it has more than one point (provided regularity itself includes the $T_1$ assumption, which is not always the case). – Matt Samuel Nov 22 '15 at 22:25
• Isn't it $W\cap V$ instead of $W\cup V$ (and also after). – MoebiusCorzer Nov 22 '15 at 22:25
• @MoebiusCorzer You're right. Thanks for pointing it out. – EA304GT Nov 22 '15 at 22:28
• @MattSamuel But my definition of regularity does not include the $T_1$ assumption. Hence, the indiscrete topology is regular. – EA304GT Nov 22 '15 at 22:33
• You’ve misunderstood the definition of regularity that you’re using: since it requires singletons to be closed, it does imply $T_1$. (In other words, it’s what I consider the wrong definition.) – Brian M. Scott Nov 22 '15 at 22:38

Remember that a space is regular if $\textbf{all one-point set are closed}$ and each $\{x\}$ and closed subsets $A$ can be separated by disjoint open sets. The condition in the lemma is equivalent to the separation condition, but does not guarantee that the space is $T_1$. The fact that any pair $(\{x\},A)$ with $A$ closed, does not imply that the space is $T_1$.
• So, if we were working with the right definition (the one that doesn't assume $T_1$ axiom), then the $T_1$ hypothesis wouldn't be necessary for the lemma to be true? – EA304GT Nov 22 '15 at 22:52