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A subspace $A$ of a topological space $X$ is locally closed if for every $a \in A$ there exists a neighborhood $O$ of $a$ such that $A \cap O \subset O$ is closed. Prove that $A$ is locally closed if and only if $A$ is the intersection of open and closed set.

Assume first that $A$ is locally closed. Then for every $a \in A$ there exists $O_a$ such that $A \cap O_a$ is closed in $O_a$. Now the neighborhoods $O_a$ cover $A$ so we have that $\bigcup_{a \in A} O_a \supseteq A$. Can I use this to show that $A$ is the intersection of some open and closed set?

Ylvas
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  • What does it mean that a set is closed in $O_a$? – Henno Brandsma Jan 25 '22 at 10:36
  • If $A \cap O_a$ is closed in $O_a$, then $A \cap O_a = O_a \cap F$ for some $F$ closed in $O_a$? @HennoBrandsma – Ylvas Jan 25 '22 at 10:48
  • No, some $F$ closed in $X$ even. Otherwise you add no info. – Henno Brandsma Jan 25 '22 at 10:51
  • How does this help? Do I need to consider the closure of $A$ here? The proposed question that has answers relies on something called nets that I have not covered and I am looking for something that doesn't rely on nets/sequences. – Ylvas Jan 25 '22 at 10:54
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    In your text, is a neighbourhood always open? – Henno Brandsma Jan 25 '22 at 11:00
  • Maybe this. I have a result that if $E \subset A$, then $\operatorname{cl}A(E) = A \cap \overline{E}$, where $\overline{E}$ is the closure w.r.t to $X$. So in my case I have that $$A \subset \bigcup{a \in A} O_a \subset X.$$ Now applying this I get that $$\operatorname{cl}{\bigcup{a \in A} O_a} (A) = \bigcup_{a \in A} O_a \cap \overline{A}.$$ But $A$ is closed in $O_a$ for every $a$ so $\operatorname{cl}{\bigcup{a \in A} O_a} (A) = A$ implying that $$A = \bigcup_{a \in A} O_a \cap \overline{A}.$$ Would this work? Equality $\operatorname{cl}{\bigcup{a \in A} O_a} (A) = A$ concerns me. – Ylvas Jan 25 '22 at 11:05
  • doesn’t work in that form as unions do not commute with closures. You haven’t answered the neighbourhood question? – Henno Brandsma Jan 25 '22 at 11:30
  • Trying figure it out. By neighborhood I assume you meant some $O_a$? – Ylvas Jan 25 '22 at 11:35
  • You say there must be a neighbourhood (open?) $O$ such that $A\xap O$ is closed in it. What does neighbourhood mean there? – Henno Brandsma Jan 25 '22 at 11:36
  • For each $a \in A$ pick an open set $O_a$ containing $a$ so that $F_a \cap O_a = O_a \cap A$ for a closed set $F_a$ of $X$. Define $O=\bigcup_{a \in A} O_a$. I claim that $O \cap \overline{A} = A$. – Henno Brandsma Jan 25 '22 at 12:11

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