closed set

closed under

Let $(X,\tau )$ be a topological space. Then a subset $C\subseteq X$ is closed if its complement $X\setminus C$ is open under the topology $\tau$.

Examples:

• In any topological space $(X,\tau )$, the sets $X$ and $\mathrm{\varnothing }$ are always closed.

• Consider $ℝ$ with the standard topology. Then $[0,1]$ is closed since its complement $(-\mathrm{\infty },0)\cup (1,\mathrm{\infty })$ is open (being the union of two open sets).

• Consider $ℝ$ with the lower limit topology. Then $[0,1)$ is closed since its complement $(-\mathrm{\infty },0)\cup [1,\mathrm{\infty })$ is open.

Closed subsets can also be characterized as follows:

A subset $C\subseteq X$ is closed if and only if $C$ contains all of its cluster points, that is, $C^{\prime }\subseteq C$.

So the set $\{1,1/2,1/3,1/4,\mathrm{\dots }\}$ is not closed under the standard topology on $ℝ$ since $0$ is a cluster point not contained in the set.