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Why is every closed subset of a compact space compact?

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topology Because both a doughnut and a coffee cup have one hole (handle), they can be mathematically, or topologically, transformed into one another without cutting them in any way. For this reason, it has often been joked that topologists cannot tell the difference between a coffee cup and a doughnut.

In topology, a space is defined as compact if every open covering of the space contains a finite subcollection that also covers the space.1

To understand why every closed subset $F$ of a compact space $X$ is compact, consider an arbitrary open cover ${U_{α}}$ of $F$ . By the definition of the subspace topology, each $U_{α}$ is the intersection of $F$ with some open set $V_{α}$ in $X$ . Because $F$ is closed, its complement $X ⧵ F$ is an open set in $X$ .

By adding the open set $X ⧵ F$ to the collection ${V_{α}}$ , we obtain an open cover of the entire space $X$ . Since $X$ is compact, there exists a finite subcollection ${V_{1}, V_{2}, \dots, V_{n}}$ along with $X ⧵ F$ that covers $X$ . Consequently, the corresponding finite subcollection ${U_{1}, U_{2}, \dots, U_{n}}$ must cover $F$ . Because we have extracted a finite subcover from an arbitrary open cover, $F$ is compact.1