On Wikipedia there is a claim: Sierpinski space is an example of a normal space that is not regular. https://en.wikipedia.org/wiki/Normal_space
I can see why it is not regular. I think the proof for by it is not normal requires "proof by vacuous truth" type of argument which I am not good at. Can someone check if this is legit.
Let $(S, \tau) = (\{0,1\}, \{\{\varnothing\}, \{1\}, \{0,1\}\})$ be the Sierpinski space. Since the only non-trivial closed set is $\{0\}$, and it is not disjoint from $\{\varnothing\}$ or $\{0,1\}$, therefore we fail to satisfy the condition required for normality, so $(S,\tau)$ is by default normal.
Is there some way to improve this argument? Thanks
By the way isn't $T_4$ = $T_1$ and $normal$ $\implies$ $T_3$ = $T_1$ and $regular$? So why doesn't being normal imply regular?
The argument is basically correct, but you’ve given the topology incorrectly: the first open set should be $\varnothing$, not $\{\varnothing\}$. Also, $\{0\}$ is disjoint from the closed set $\varnothing$; the reason that this doesn’t matter is that $X$ and $\varnothing$ are disjoint open nbhds of $A$ and $\varnothing$ for any $A\subseteq X$. Thus, normality is a non-trivial requirement only on disjoint non-empty closed sets.
$T_4$ does imply $T_3$. Suppose that $X$ is $T_4$, $x\in X$, and $F$ is a closed subset of $X$ not containing $x$. Since $X$ is $T_4$ and hence $T_1$, $\{x\}$ is closed. Thus, we can use normality to get disjoint open nbhds of $\{x\}$ and $F$, which are of course also disjoint open nbhds of $x$ and $F$.
Normality without the $T_1$ property, however, does not imply regularity, as this example shows.