Counterexample to "Urysohn Lemma for Hausdorff Spaces"

Question

Take a Hausdorff topological space $X$, and two distinct points $p,q \in X$. Is there a continuous function $f: X \to \mathbb{R}$ such that $f(p) \neq f(q)$?

The answer is probably no, but I don't have a ready-made counterexample. As soon as we add the adjective "compact" the space $X$ becomes normal, so that the Urysohn lemma applies, and it gives us such a function.

So the place to look is Hausdorff spaces that are not regular/normal. However, the examples of these that I know are often constructed as finer topologies on the real line, and so come equipped with a ready made map to $\mathbb{R}$ that separates points.

Have you tried using the Deleted Tychonoff Plank? It is Hausdorff and not normal — M Turgeon, Jul 16 '14 at 17:53
@MTurgeon: The Deleted Tychonoff Plank is completely regular, and therefore completely Hausdorff, however. — user642796, Jul 16 '14 at 17:58

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

What you are looking for is a Hausdorff space which is not completely Hausdorff.

The examples provided as answers to this question would fit the bill. In particular, these Hausdorff spaces have the (stronger) property that there are distinct $x,y \in X$ such that $\overline{U} \cap \overline{V} \neq \varnothing$ for any open neighbourhoods $U,V$ of $x,y$, respectively.

Note that if $x \neq y$ can be separated by a continuous real-valued function $f : X \to [0,1]$ (i.e., $f(x) = 0$ and $f(y) = 1$) then the neighbourhoods $f^{-1} [ [0,\frac{1}{3}) ]$ and $f^{-1} [ ( \frac{2}{3} , 1 ] ]$ must have disjoint closures. (This is because for a continuous $f : X \to Y$ we must have that $\overline{f^{-1} [ B ]} \subseteq f^{-1} [ \overline{B} ]$ for all $B \subseteq Y$.)

Counterexample to "Urysohn Lemma for Hausdorff Spaces"

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