Locally Closed Subset

In topology, a subset $E$ of a topological space $X$ is said to be locally closed if any of the following equivalent conditions are satisfied:

$$ \DeclareMathOperator{\id}{id} \DeclareMathOperator{\Hom}{Hom} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} $$

Proposition (局部闭集的等价定义)

For each $x \in E$, there is a neighborhood $U$ of $x$ such that $E \cap U$ is closed in $U$.
$E$ is an open subset of its closure $\overline{E}$.
$E$ is the intersection of an open set and a closed set in $X$.
$\overline{E} - E$ is closed in $X$.
$E$ is the difference of two closed sets in $X$.
$E$ is the difference of two open sets in $X$.

Note that (1) justifies the terminology locally closed and is precisely Bourbaki’s definition of locally closed.

Proof. (1) $\Rightarrow$ (2): $E\cap U$ is closed in $U$ 等价于 $E \cap U = \overline{E \cap U} \cap U$. 显然 $\overline{E \cap U} \cap U \subseteq \overline{E} \cap U$, 下面我们断言这里一定是等号. 设 $y \in \text{RHS}$, 因为 $y \in \overline{E}$, 所以对任意 $y$ 的开邻域 $O$ 都有 $O \cap E \ne \varnothing$. 因为 $U$ is open, 所以 $\{O \cap U\}$ 实际上是 $\{O\}$ 的一个子集 and thus $y \in \overline{E \cap U}$. 于是 $x \in U \cap E = U \cap \overline{E}$, 即 $x$ 是 $\overline{E}$ 的内点.

(2) $\Rightarrow$ (3): 因为 $E$ is open in $\overline{E}$, 所以存在开集 $O$ 使得 $E = O \cap \overline{E}$.

(3) $\Rightarrow$ (1): Write $E = A \cap B$ with $A$ open and $B$ closed. Take $U = A$, then $U \cap E = E = A \cap B$, which is of course closed in $A$.

(2) $\Rightarrow$ (4): $\overline{E} - E$ is closed in $\overline{E}$ and hence closed in $X$.

(3) $\Leftrightarrow$ (5), (6): 我们有等式 $A \cap B^c = A - B$.

(4) $\Rightarrow$ (5): 我们有 $E = \overline{E} - (\overline{E} - E)$. $\square$