I have read that this statement is true. And in the wikipedia entry of regular spaces is also stated. But I fail trying to prove it, maybe because I don't understand fully the concept of local base of closed neighbourgoods.

For example, I think that $\mathbb{R}$, endowed with the cofinite topology, has a local base of closed neighbourgoods. In fact, as single points are closed, one could take, for each $x\in \mathbb{R}$, the closed set $\{x\}$ as the base, because any other open set $U$ that contains $x$ has this closed set contained in $U$. And it is clear that this topological space is not regular (is not even Hausdorff).

There must be something wrong in this reasoning, but I don't know what... Also if someone could provide a proof or a link to a proof, I will be very grateful. :)


What's wrong with that reasoning is that $\{x\}$ is not a neighbourhood of $x$ (with respect to that topology) and therefore $\bigl\{\{x\}\bigr\}$ is not a local base of closed neighbourhoods.

If a space $X$ is regular, let $x\in X$. For each open set $A$ such that $x\in A$, let $N_A$ and $O_A$ be an open sets such that $x\in N_A$, $O_A\supset X\setminus A$ and $N_A\cap O_A=\emptyset$. They exist, since $X$ is regular. But then $N_A\subset X\setminus O_A$ and therefore $X\setminus O_A$ is a closed neighbourhood of $x$. On the other hand, $X\setminus O_A\subset A$. Therefore set of all $X\setminus O_A$ is a local base of neighbourhoods of $x$.

Can you prove the reverse implication?

  • $\begingroup$ Ok, thanks a lot! I did not know that a neighbourhood of a point, although not necessarily open, needs to contain an open set that contains $x$. $\endgroup$ – Jaime Arboleda Castilla Jun 28 '18 at 12:14

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