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?


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 '14 at 20:52
  • $\begingroup$ This works if $V$ is open, and we can assume this. I will edit the post. Thanks for the remark. $\endgroup$ – Jérémy Blanc Apr 9 '14 at 9:26

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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