Why Does Urysohn's lemma need the space to be Hausdorff?

Question

I've just finished a General Topology course. We used Dugundji's Topology book. There's one thing I don't understand regarding Urysohn's lemma: as stated in Dugundji's, it says: Let Y be a Hausdorff space. Then Y is normal if and only if for every $A,B\subseteq Y$ closed and disjoint there's a continuous function $f:Y\longrightarrow \mathbb{R}$ such that $0\leq f(x)\leq 1$, $f(A)=0$ and $f(B)=1$.

My question is: Why do we ask for $Y$ to be Hausdorff beforehand? Do I need the Hausdorff property for the $\implies)$ part of the proof or for $\impliedby)$? I really can't see at what point I use it.

Just in case it helps (I'm at a loss and don't know if it will), the author already asks $Y$ to be Haussdorf when defining "normal" space.

Thanks guys.

Answer 1

To prove the Urysohn lemma we do not need the extra separation axioms. This follows from studying the proof carefully, and is explicit in some topology texts (where a distinction is made between "normal" = separate disjoint closed sets by disjoint open sets and $T_4$ which is "normal plus $T_1$" (singletons are closed) which is the same as "normal plus Hausdorff" of course), where Urysohn's lemma is proved explicitly for "normal spaces". E.g. (if you read German) Boto von Querenburg, Mengentheoretische Topologie".

So Dugundji uses the word "normal" for what I called $T_4$ and so he has to assume Hausdorffness on $Y$ just to fulfill that extra separation axiom requirement. He probably knew full well that Hausdorffness was not needed for the proof, but wanted to exclude spaces like indiscrete spaces from the "nice class" of "normal" spaces. In practice, spaces we apply Urysohn's lemma in are indeed at least Hausdorff, and most topologists don't work in spaces which are not Hausdorff.

So TLDR; Urysohn's lemma and the closed-set-separation property are equivalent without Hausdorffness.

Answer 2

Based on your last paragraph, if $Y$ is normal, then it is also Hausdorff. So $\Rightarrow$ doesn't require this hypothesis.

On the other hand, there are non-Hausdorff spaces that satisfy the given property: for example take any normal space $Y$ and double its points: $X:=Y\times\{0,1\}$ with the topology inherited from $Y$ which glues $(y, 0)$ to $(y, 1)$, i.e. let $G\subseteq X$ be open iff $G=U\times\{0,1\} $ with $U$ open.