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If $A,B\subseteq X$ is closed in $X$ and $A\cap B=\emptyset$, then $\exists f:X\to [0,1]$ such that $f$ is continuous, $f(A)=\{0\}$ and $f(B)=\{1\}$ $\implies$ $X$ is normal.

Munkres’ defined normal space as $T_1$ and $T_4$. Showing $X$ is $T_4$ is trivial. Since $f$ is continuous, $f^{-1}([0,x)), f^{-1}((y,1])\in \mathcal{T}_X$, where $0\lt x\lt y\lt 1$. By elementary set theory, $A\subseteq f^{-1}(f(A))=f^{-1}(\{0\}) \subseteq f^{-1}([0,x))$ and $B\subseteq f^{-1}(f(B))=f^{-1}(\{1\})\subseteq f^{-1}((y,1])$. Hence $\exists f^{-1}([0,x)), f^{-1}((y,1])\in \mathcal{T}_X$ such that $A\subseteq f^{-1}([0,x))$, $B\subseteq f^{-1}((y,1])$ and $ f^{-1}([0,x))\cap f^{-1}((y,1]) =f^{-1}([0,x) \cap (y,1])=f^{-1}(\emptyset)=\emptyset$. Now how to show $X$ is $T_1$?

user264745
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    Are you sure you've got all the definitions/statements right? First, I assume you require $A$ and $B$ to be non-empty in the hypothesis, otherwise that implication can't be true (because $f(\emptyset) = \emptyset$). But then it would seem that the indiscrete topology on two points is a counterexample - that space is not $T_1$ but also does not have any disjoint non-empty closed sets, so the assumption is vacuously true. – Chris Eagle May 26 '22 at 19:42
  • I second @ChrisEagle's comment, and would add that what you've described is the opposite of the usual convention; to my knowledge $T_4$ is generally defined as "$T_1$ and normal". I can't speak to Munkres' definition but this is eg how wikipedia defines it – Atticus Stonestrom May 26 '22 at 19:46
  • @ChrisEagle I think I have got all definition/statements right. Non-emptyness of $A$ and $B$ are goes without saying. – user264745 May 26 '22 at 19:48
  • @AtticusStonestrom yeah. Wikipedia and Munkres’ definition of separation axiom don’t overlap much. I was feeling confused when I was reading Wikipedia definition. – user264745 May 26 '22 at 19:50
  • @ChrisEagle Is that means, we have to restrict to only $T_4$ condition (with respect to Munkres’ definition of normal)? – user264745 May 26 '22 at 20:00
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    The function statement you have is equivalent to being able to separate disjoint closed sets by open sets (what you seem to be calling "$T_4$", which some other people call "normal", but anyway without any assumption of $T_1$). You won't be able to get $T_1$ from the separation by functions. – Chris Eagle May 26 '22 at 20:06
  • @ChrisEagle yup. Different book use different definition. I asked this question because Munkres’ wrote normal instead of $T_4$. Thank you for clarifying. – user264745 May 26 '22 at 20:10
  • Does your converse of Urysohn lemma occur in Munkres? – Paul Frost Jun 03 '22 at 10:24

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Munkres does not claim that your converse is true. He only says that if all pairs of disjoint closed sets can be separated by a continuous function, then all pairs of disjoint closed sets can be separated by disjoint open sets (which is fairly trivial).

Unfortunately notation is not really standardized in the literature. $T_1$-spaces are certainly those whose one-point subsets are closed, but spaces which allow separation of disjoint closed sets by disjoint open sets are sometimes denoted as $T_4$ and sometimes as normal. Moreover, some authors (like Munkres) use "$T_4$" and "normal" as synonyms because they only define this concept in the realm of $T_1$-spaces.

Anyway, separation of disjoint closed sets by disjoint open sets (or by continuous functions) does not imply $T_1$. As an example take a space $X$ with the trivial topology and more than one point. Then $X$ does not have a non-trivial pair of disjoint closed sets, thus it trivially satisfies the "closed sets separation condition", but it not $T_1$.

Paul Frost
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  • Munkres sometimes assume normal as $T_4$ and sometimes as $T_1$ & $T_4$, for instance see. – user264745 Jun 03 '22 at 11:38
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    @user264745 I do not have access to Munkres first edition, perhaps he changed the definition in the second edition. But have a look at Wiki. In my opinion this is the most common understanding: Normal means "closed sets separation condition", $T_4$ means normal + $T_1$. But as I said, it is not standardized. See here for an alternative approach. Anyway, with my favorite interpretation your converse is true, and perhaps this was – Paul Frost Jun 03 '22 at 12:28
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    stated in Munkres first edition. Be cautious when you read "normal" or "$T_4$" and check which definition the author has in mind. – Paul Frost Jun 03 '22 at 12:30