Came across this definition of a locally closed subspace in a book:

Any subspace $A$ of a space $X$ is called locally closed if each point of $A$ is contained in an open subset $G$ of $X$ such that $G ∩ A$ is closed in $G$.

If all points of $A$ are contained in $G$ then $G∩A$ is closed in $G$ should essentially just mean $A$ is closed in $G$ but is this really the case?

  • 1
    $\begingroup$ Right. Since $A \subseteq G$ we have $A\cap G=A$. $\endgroup$ Aug 19, 2020 at 11:49
  • $\begingroup$ So then how can we rectify this definition? $\endgroup$
    – Anisha
    Aug 21, 2020 at 6:37

2 Answers 2


Here is possibly a clearer equivalence of locally closed sets: Suppose $(X,\tau)$ is a topological space. A subset $A$ in $(X,\tau)$ is a locally closed subset of $X$ iff $A=U\cap \overline{A}$ for some open $U\in\tau$, i.e., $A$ is an open subset of its closure.

Here is a short proof of the equivalence:

Lemma: $A$ is locally closed iff for any $x\in A$, there is $V_x\in \tau$ such that $x\in V_x$ and $A\cap V_x=\overline{A\cap V_x}\cap V_x$.

Necessity is obvious. For sufficiency, let $\mathcal{V}$ the collection of open sets $V_x$ such that $x\in V_x$ and $A\cap V_x=\overline{A\cap V_x}\cap V_x$. $$ A=\big(A\cap V\big)\cup\big(A\cap(X\setminus V)\big) $$ and $\overline{A\setminus V}\subset X\setminus V$. Hence $$ A\cap V\subset \overline{A}\cap V\subset V\cap\overline{A\cap V}=A\cap V $$ Consequently $$ \overline{A}\cap\bigcup_{x\in V}V_x=\bigcup_{x\in X}\overline{A}\cap V_x=\bigcup_{x\in X}A\cap V_x=A\cap\bigcup_{x\in X}V_x=A $$

  • 3
    $\begingroup$ +1, let me add that there is some symmetry here. Given a closed set $C$, its open subsets are of the form $U \cap C$ where $U$ is open in the ambient space. Given an open set $U$, its closed subsets are of the form $U \cap C$ where $C$ is closed in the ambient space. To summarize, "open subset of a closed set" means the same thing as "closed subset of an open set" since both are the same thing as "intersection of an open set and a closed set". $\endgroup$
    – Mike F
    Aug 19, 2020 at 14:43

Yes you are correct. If $A \cap G$ is closed in $G$ and $A \subseteq G$, then this just means that $A \cap G = A$ is closed in $G$, as was already pointed out in the comments.

Let me give a little bit more information. Suppose that $A \subseteq X$ is a locally closed set. Each $x \in A$ has an open neighbourhood $G_x$ such that $G_x \cap A$ is closed in $G_x$. One can then put $G = \bigcup_{x \in A} G_x$, which is also an open set. By writing $G \setminus A = \bigcup_{x \in A} (G_x \setminus A)$, we see that $G \setminus A$ is open, i.e. $A$ is also closed in $G$.

Conclusion: A set $A \subseteq X$ is locally closed, in the sense you define, if and only if it is a closed subset of some open subset of $X$.

The definition that you give has a stronger flavour of "local", so I think it is probably the more natural definition, but it is sometimes convenient to dispense with quantifying over points of $A$ and just use that $A$ is closed in an open set.

Added: It might be nice to also see some simple examples. Any open set is locally closed because it is closed in itself. Any closed set is locally closed because it is closed in the whole space. Beyond these trivial examples, the kind of thing to have in mind is something like $X$ the $xy$-plane and $A$ the positive part of the $x$-axis. Then, $A$ is not closed or open in $X$, but it is closed in the open right half plane $G$.

  • $\begingroup$ Right. This makes a lot of sense, thank you. $\endgroup$
    – Anisha
    Aug 21, 2020 at 7:06

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