Completely Hausdorff space is Urysohn

Question

I know that these two definitions are equivalent but I have no idea of how to prove this direction so I’m asking you for a hint or reference on the subject. Thanks

Answer 1

One definition of completely Hausdorff that I know is that for distinct $x \neq y$ in $X$ there is a continuous $f:X \to [0,1]$ so that $f(x)=0$ and $f(y)=1$. (in analogy to completely regular, which does a similar thing for a point and a closed set).

Such a space is indeed Urysohn in the sense that any two distinct $x,y$ have open neighbourhoods with disjoint closures:

Let $f$ be the promised function and let $U=f^{-1}[[0,\frac13)]$ and $V= f^{-1}[[(\frac23,1]]$, both are open by continuity and contain $x$ resp. $y$. Then $\overline{U}\subseteq f^{-1}[[0,\frac13]]$ and $\overline{V} \subseteq f^{-1}[[\frac23,1]$ so that $\overline{U} \cap \overline{V} = \emptyset$, as required.

There are $T_3$, Urysohn spaces $X$ such that all continuous real-valued functions on $X$ are constant. So the converse fails quite dramatically.

These notions as I know them are certainly not equivalent.