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here there are 2 definitions of locally closed sets:

$A$ is locally closed subset of $X$ if:

a) every element in $A$ has a neighborhood $V$ in $X$ such that $A\cap V$ is closed in $V$.

b) $A$ is open in its closure (in $X$)

why a) and b) are equivalent?

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3 Answers 3

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For me the definition is the following:

$c)$ $A$ is equal to $U\cap F$ where $U$ is open in $X$ and $F$ is closed in $X$.

Let us prove that this is equivalent to your definitions, that I write again here:

$a)$ For each $x\in A$ there is a neighbourhood $V$ of $x$ in $X$ such that $A\cap V$ is closed in $V$.

$b)$ $A$ is open in its closure $\overline{A}$ in $X$.

The proof is made in the following steps:

$c)\Rightarrow a)$ Just choose $V=U$, which is a neighbourhood of any point of $A$.

$a)\Rightarrow b)$ For each point $x\in A$ we use the given neigbourhood $V$ such that $A\cap V$ is closed in $V$. We can replace $V$ with an open neighbourhood $U\subset V$ (since $A\cap U=(A\cap V)\cap U$ is closed in $U$), and assume that $V$ is open.

The fact that $V$ is open implies that the closure of $A\cap V$ in $V$ is equal to $\overline{A}\cap V$. Indeed, $A=(A\cap V)\cup (A\cap (X\setminus V))$ so $\overline{A}\subset \overline{A\cap V}\cup \overline{A\cap (X\setminus V)}$. As $V$ is open we have $\overline{A\cap (X\setminus V)}\subset X\setminus V$, hence $V\cap \overline{A\cap (X\setminus V)}=\emptyset$, so we get $$\overline{A}\cap V\subset \overline{A\cap V} \cap V.$$ The other implication being clear, and using the hypothesis that $A\cap V$ is closed in $V$ we get$$\overline{A}\cap V=\overline{A\cap V} \cap V=A\cap V\subset A.$$

Hence, $\overline{A}\cap V$ is a neighbourhood of $x$ in $\overline{A}$ contained in $A$. Doing this for all points of $A$, this shows that $A$ is open in $\overline{A}$.

$b)\Rightarrow c)$ Since $A$ is open in $F=\overline{A}$, there exists an open subset $U$ of $X$ such that $\overline{A}\cap U=A=F\cap U$.

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  • $\begingroup$ "That since $A\cap V$ is closed in $V$ then $A \cap V = \overline{A} \cap V$" can you explain why? The closure of $A\cap V$ in $V$ is $\overline{A\cap V}\cap V$, isn't it? $\endgroup$
    – Daniel
    Apr 8, 2014 at 20:52
  • $\begingroup$ This works if $V$ is open, and we can assume this. I will edit the post. Thanks for the remark. $\endgroup$ Apr 9, 2014 at 9:26
  • $\begingroup$ You can directly see the closure of $A\cap V$ in 𝑉 is equal to $\bar{A}\cap V$, since the closure of union is the union of closure. Therefore $\bar{A}=\overline{A\cap V}\cup\overline{A\cap(X-V)}=\overline{A\cap V}$ and $\bar{A}\cap V=\overline{A\cap V}\cap V$. $\endgroup$
    – user832207
    Dec 3, 2020 at 16:53
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b) $\Rightarrow$ a)

Let $A$ be open in $\overline{A}$. Then $A=\overline{A}\cap U$ for some $U$ open in $X$ and $U$ serves as neighborhood for every element of $A$. Intersection $A\cap U$ equals $A=\overline{A}\cap U$ which is a subset of $U$ closed in $U$.

a) $\Rightarrow$ b)

Let $N$ be a neighborhood of $a\in A$ such that $N\cap A$ is closed in $N$. Its closure in $N$ is $N\cap\overline{A}$ so actually we have $N\cap A=N\cap\overline{A}$. That shows that $N\cap A$ is a neighborhood of $a$ in the subtopology on $\overline{A}$, and it is contained in $A\subset\overline{A}$. For any $a\in A$ such a neighborhood exists, so $A$ is open in $\overline{A}$.

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  • $\begingroup$ In b)$\Rightarrow$ a) you can just take $U=A$. $\endgroup$ Mar 31, 2021 at 7:46
  • $\begingroup$ @RobinBalean $A$ is a set that is open in its closure. That does not mean that $A$ is an open set (like $U$). Be aware that e.g. every closed set is open in its closure (because it coincides with its closure). So in general we cannot go for $A=U$. $\endgroup$
    – drhab
    Mar 31, 2021 at 8:18
  • $\begingroup$ Ah - thanks for pointing that out. I was a bit too hasty. $\endgroup$ Mar 31, 2021 at 18:14
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Just for the sake of completeness, I will state the complete list of equivalences:

Proposition. Let $X$ be a space and $A\subset X$ be a subspace. The following are equivalent:

  1. There is $U\subset X$ open such that $A\subset U$ and $A$ is closed in $U$.
  2. There is $F\subset X$ closed such that $A\subset F$ and $A$ is open in $F$.
  3. $A=U\cap F$, for some $U\subset X$ open and $F\subset X$ closed subsets.
  4. For all $x\in A$ there is an open neighborhood $U\subset X$ of $x$ such that $A\cap U$ is closed in $U$.
  5. For all $x\in A$ there is a neighborhood $U\subset X$ of $x$ such that $A\cap U$ is closed in $U$.
  6. $A$ is open in $\overline{A}$.

If any of these equivalent conditions happens, we say that $A$ is locally closed in $X$.

The proof of the equivalences 1$\Leftrightarrow$2$\Leftrightarrow$3 is immediate. The rest of the equivalences were already proven by Jérémy Blanc on his answer.

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