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 $\emptyset$ are always closed.
• Consider $\R$ with the standard topology. Then $[0,1]$ is closed since its complement $(-\infty,0) \cup (1,\infty)$ is open (being the union of two open sets).
• Consider $\R$ with the lower limit topology. Then $[0,1)$ is closed since its complement $(-\infty,0)\cup[1,\infty)$ 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, $C'\subseteq C$

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