It is known that not every regular topological space is completely regular.

However, I do not see why the standard proof of the famous Urysohn's lemma for normal spaces becomes wrong if I assume only regularity and replace one of the closed sets in the proof with a point.

Can someone point me out which part of the proof is no longer true under these conditions?


1 Answer 1


For reference, I will use the proof outlined in Proof of Urysohn's lemma from Kelley's book.

So, assume $A$ is a point, and $X$ is merely regular, and let us see how the proof goes:

  • Put $F(1)=X-B$. Nothing can go wrong here.

  • Choose an open set $F(0)$ such that $A\subseteq F(0)$ and $\overline{F(0)}\cap B=\varnothing$. This works, as $X$ is regular, and $A$ is just a point.

  • Choose an open set $F(1/2)$ such that $\overline{F(0)}\subseteq F(1/2)\subseteq\overline{F(1/2)}\subseteq F(1)$. This works, as $X$ is regular, and $\overline{F(0)}$ is just a po..... Oops.

(If you decide to make $B$ the point rather than $A$, the proof survives one more step, but you’ll run into the same problem when trying to find $F(1/4)$.)


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