# Urysohn's extension theorem

Currently I am working my way through Ernest Michael's first article on continuous selections. Here, Urysohn's extension theorem is stated as follows:

For a $T_1$-space, the following properties are equivalent:

1. $X$ is normal.
2. The real line $\mathbb{R}$ is an extension space with respect to X. (i.e. every continuous function $f: A \rightarrow \mathbb{R}$, with $A \subseteq X$ closed, can be extended to a continuous function $f : X \rightarrow R$).

For the proof of this theorem, the author refers to Urysohn's original article, which is written in German (which is unfortunate since I don't speak German). If I am not mistaken, both Urysohn and Tietze proved similar properties. It seems to be that the so called Tietze extension theorem and its proof are widely known.

Since I am trying to prove Urysohn's extension theorem as stated above, I was wondering what the best approach would be. Is it in any way simpler (less technical) than Tietze's extension theorem? I'd appreciate any help, whether it'd be in the form of a hint or (an initiation to) a full proof.

• ??? The fact that 1 implies 2 is exactly the Tietze extension theorem proved at that place you link to. What's the problem? (It's trivial that 2 implies 1: Given disjoint closed sets $A_1$ and $A_2$ let $A=A_1\cup A_2$ and define $:A\to\Bbb R$ too equal $0$ on $A_1$ and $1$ on $A_2$...) – David C. Ullrich Jun 13 '16 at 16:37
• @DavidC.Ullrich My fault, I guess I should've read through the problem thoroughly before posting here. Now the stated Urysohn's extension theorem is "more specific" in the sense that it states the characterization of normality for $T_1$-spaces only. Since the implication of (1) by (2) doesn't use the fact that $X$ is a $T_1$-space (or am I missing something here?), it seems to be quite unnecessary to restrict the Tietze extension theorem to this specific case. This was why I was wondering if there was an alternate proof of the implication of (2) by (1), using the fact that $X$ is $T_1$. – Kasper Cools Jun 14 '16 at 18:48

As David C. Ullrich says in the comments, this is the Tietze extension theorem, which is also known as the Tietze-Urysohn theorem. I hadn’t seen Urysohn’s proof before, but I just tracked it down and took a look; he gave essentially the same proof that you’ve seen elsewhere. I never liked that proof, and years ago I came up with one that uses the same basic idea as the usual proof of Urysohn’s lemma. (Others may also have discovered this argument, though I’ve not seen it elsewhere.) You can find it in ‘A “More Topological” Proof of the Tietze-Urysohn Theorem’, The American Mathematical Monthly, Vol. $85$, No. $3$ (Mar., $1978$), pp. $192$-$193$. It’s still quite technical, but if you’ve worked through the proof of Urysohn’s lemma, the basic structure of the argument will be familiar.
• That's interesting, I will definitely take a look at your proof! As I pointed out in the comment section, I was wondering if the fact that Urysohn originally proved the theorem for $T_1$-spaces only, would have any utility? But as far as I can tell, it was just unnecessary and the theorem stated in my question can just be generalized to all topological spaces without making the proof any harder? – Kasper Cools Jun 14 '16 at 18:57
• @Kaspar: The Tietze extension theorem does not require that $X$ be $T_1$. My best guess is that Michael was using the alternate terminology in which normal means what I would call $T_4$, meaning that it implies $T_1$. Since $(2)$ only implies that $X$ is normal in my sense (i.e., not necessarily $T_1$), to get an equivalence with normality in the stronger sense he needs to assume that the space is $T_1$. – Brian M. Scott Jun 14 '16 at 19:16