# Where is the Hausdorff condition used? (Folland's exercise 4.24)

I am doing Folland's exercise 4.24: Prove that if $$X$$ is a Hausdorff space then $$X$$ is normal (T$$_4$$) if-f $$X$$ satisfies the conclusion of Urysohn's lemma if-f $$X$$ satisfies the conclusion of Tietze's extension theorem.

Okay, so if $$X$$ is $$T_4$$ then (by the proofs of the theorems) both Urysohn's lemma and Tietze's extension theorem are satisfied.

So I would like to prove that $$T_2+$$Urysohn $$\implies T_4$$. I begin with $$A,B$$ being disjoint closed sets and take (by Urysohn's lemma) $$f\in C(X,[0,1])$$ such that $$f\vert_A=0, f\vert_B=1$$. I proceed by considering $$f^{-1}([0,1/2))$$ and $$f^{-1}((1/2,1])$$. Since $$f$$ is continuous ( $$[0,1]$$ has the subspace topology) these sets are disjoint and open and $$A,B$$ are both contained in one, hence $$X$$ is $$T_4$$. Observe that $$T_2$$ has not been used.

Now I would like to prove that $$T_2+$$ Tietze$$\implies T_4$$. Let $$A,B$$ be disjoint closed sets. Then $$A\cup B$$ is closed. Set $$f:A\cup B\to[0,1]$$ with $$f(x)=0$$ for $$x\in A$$ and $$f(x)=1$$ for $$x\in B$$. Now $$f$$ is continuous since the inverse image of any closed set is either $$\emptyset, A, B, A\cup B$$ (which are closed). By Tietze's theorem, $$f$$ extends to $$F\in C(X,[0,1])$$; that is $$F\vert_A=0, F\vert_B=1$$. By considering the same sets as above, we are done. Again, $$T_2$$ is nowhere to be seen.

Folland defines $$T_4$$ as $$T_1$$+ disjoint closed sets are separated by open sets. But even that is not enough of a reason, it would suffice for $$X$$ to be assumed $$T_1$$. Also, it is not a typo since $$X$$ is called Hausdorff and not $$T_2$$ (where maybe only the index was wrong). So where is $$T_2$$ secretly being used? Or is there something wrong in my proofs?

Indeed normal (closed disjoint sets can be separated by disjoint open sets) implies that Urysohn's lemma holds, which in turn implies that Tietze's extension theorem holds. And when Tietze holds your proof indeed shows that in $$X$$ disjoint closed sets can be separated. No argument there.
If we assume we're working in $$T_1$$ spaces we can, in the Tietze to normal implication conclude that we also have $$T_4$$. So $$T_1$$ is indeed enough to assume in context, and Hausdorff seems overkill. Maybe Folland is not striving for minimality here.
• I was thinking, maybe the exercise wants the implications in that order and $T_2$ is needed in the implication Tietze $\implies$ Urysohn. I can't work this through though, but I know (from the internet) that it can be done. – JustDroppedIn Apr 13 '19 at 21:14
• @JustDroppedIn Tietze $\impplies$ Urysohn is trivial even without Hausdorff. – Henno Brandsma Apr 13 '19 at 21:23