Proposition 5

For a subset $L$ of a topological space $X$, the following properties are equivalent:

a) $L$ is locally closed in $X$.

b) $L$ is the intersection of an open subset and a closed subset of $X$.

c) $L$ is open in its closure $\overline{L}$ in $X$.

If $L$ is locally closed, then, for each $x \in L$, there is an open neighbourhood $V_x$ of $x$ in $X$ such that $L \cap V_x$ is closed in $V_x$; $U = \bigcup_{x \in L} V_x$ is open in $X$, and Proposition 3 of no. 1 shows that $L$ is closed in $U$; therefore a) implies b). If $L = U \cap F$, where $U$ is open and $F$ closed in $X$, we have $L \subset F$; hence $L \subset U \cap L \subset U \cap F = L$, which shows that $L = U \cap L$ is open in $L$, so that b) implies c). Finally, if $L = U \cap L$, where $U$ is open in $X$, $L$ is closed in $U$, hence locally closed, and thus c) implies a).