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In Munkres we have the following lemma:

Lemma 31.1. Let $X$ be a topological space. Let one-point sets in $X$ be closed. (a) $X$ is regular ii and only if given a point $x$ of $X$ and a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\operatorname{cl}(V)\subseteq U$.

I do not see the use of $T_1$ assumption (Let one-point sets in $X$ be closed.) in the proof. I proved it myself before finding the result in Munkres without needing $T_1$.

(1) Could you confirm my statement?

(2) Can we say the same for the case of normal spaces (Lemma 31.1(b))? I've yet to start with the proof.

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Mohamed Ali
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  • If I recall correctly, Munkres' definition of a regular space requires $T_1$. Withtout assuming this, the "if" part will fail as you will not be able to conclude that $X$ is $T_1$, and so $X$ will not be regular (by this definition). – qualcuno Nov 23 '19 at 09:46
  • Different authors have used differing definitions of regular and completely regular. Some have assumed $T_1$; others have not. – DanielWainfleet Nov 23 '19 at 19:36

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Let $S = \{0,1\}$ be the Sierpinski space with $0$ being the open point. This satisfies the condition on neighbourhoods imposed by the lemma, and so we should have that $S$ is regular, which is not true if your definition of regular assumes $T_1$.

Otherwise, I think this should work, with the following proof:

Suppose that $X$ is regular and fix an open set $U \ni x$, so that there exist disjoint open sets $V \ni x, W \supset U^c$. This imples $V\subset W^c$, and so $cl(V) = cl(W^c) = W^c \subset U$.

Reciprocally, take $x \not \in F$ with $F$ closed. By hypothesis there exists $x \in V \subset F^c$ open such that $cl(V) \subset F^c$, and this implies that $cl(V)^c \supset F$. Now take $V$ and $cl(V)^c$ as your separating neighbourhoods.

qualcuno
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    I agree with that sentiment. Sometimes $T_3$ is used to mean regular (in the weak sense) + $T_2$. – qualcuno Nov 23 '19 at 10:02
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Munkres' definition of regularity requires that the space be $T_1$. Many authors do not include $T_1$ in the definition and call a space $T_3$ if it is regular (in the "weaker" sense) and $T_1$. See https://en.wikipedia.org/wiki/Separation_axiom

Anyway, to prove the core of the lemma, you do not need $T_1$. The same is true for normal spaces. It is again a matter of definitions.

Paul Frost
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    Quoting from the 2nd edition, page 195: he starts the definition paragraph with Suppose that one-point sets are closed in $X$. Later he remarks that he needs this assumption to get regular implies Hausdorff and normal implies regular. – Henno Brandsma Nov 23 '19 at 11:03