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?


3 Answers 3


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 $\overline{A}\cap U=A=F\cap U$.

  • $\begingroup$ "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

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 $\overline{A}$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.