Compact Sets in Topological Space

# Compact Sets in Topological Space

Definition: Let $E$ be a topological space and let $A \subseteq E$. An Open Cover of $A$ is a collection of open sets whose union contains $A$. $A$ is said to be Compact if for every open cover of $A$ there is a finite subcollection that still covers $A$. |

Proposition 1: Let $E$ be a topological space with two topologies $\xi$ and $\eta$ and suppose that $\xi$ is finer than $\eta$. If $A \subseteq E$ is $\xi$-compact then it is $\eta$-compact. |

**Proof:**Suppose that $A$ is $\xi$-compact. Let $\{ U_{\alpha} : \alpha \}$ be an $\eta$-open cover of $A$. Since $\xi$ is finer than $\eta$, each of the $\eta$-open sets $U_{\alpha}$ are also $\xi$-open and thus $\{ U_{\alpha} : \alpha \}$ is a $\xi$-open cover of $A$. Since $A$ is $\xi$-compact, there exists $\alpha_1, \alpha_2, ..., \alpha_n$ such that $A$ is covered by $\{ U_{\alpha_1}, U_{\alpha_2}, ..., U_{\alpha_n} \}$. Thus $A$ is $\eta$-compact. $\blacksquare$

Proposition 2: Let $E$ be a topological space.(1) If $K_1, K_2, ..., K_n$ is a finite collection of compact sets then $\displaystyle{\bigcup_{i=1}^{n} K_i}$ is compact.(2) If $C$ is closed, $K$ is compact, and $C \subseteq K$ then $C$ is compact.(3) If $E$ is Hausdorff and $K$ is compact then $K$ is closed. |

**Proof of (1):**Suppose that $K_1, K_2, ..., K_n$ are compact and let $\{ U_{\alpha} : \alpha \}$ be an open cover of $\bigcup_{i=1}^{n} K_i$. Then it is an open cover for each $K_i$. So for each $K_i$, extract an finite subcollection of $\{ U_{\alpha} : \alpha \}$ that covers $K_i$ and union these subcollections together to get a finite subcollection of $\{ U_{\alpha} : \alpha \}$ which covers $\bigcup_{i=1}^{n} K_i$. Thus $\bigcup_{i=1}^{n} K_i$ is compact. $\blacksquare$

**Proof of (2):**Let $C$ be closed and let $K$ be compact. Let $\{ U_{\alpha} : \alpha \}$ be an open cover of $C$. Then $\{ U_{\alpha} : \alpha \} \cup \{ C^c \}$ is an open cover of $K$. So there exists $\alpha_1, \alpha_2, ..., \alpha_n$ such that $K$ is covered by $\{ U_{\alpha_1}, U_{\alpha_2}, ..., U_{\alpha_n}, C^c \}$ and $C$ is covered by $\{ U_{\alpha_1}, U_{\alpha_2}, ..., U_{\alpha_n} \}$ so that $C$ is compact. $\blacksquare$

**Proof of (3):**Let $K$ be compact and suppose that $x \not \in K$. For each $a \in K$, since $a \neq x$ and since $E$ is Hausdorff there exists open sets $U_a$ and $V_a$ such that $a \in U_a$ and $x \in V_a$. Then $\{ U_a : a \in K \}$ is an open cover of $K$ and so there exists $a_1, a_2, ..., a_n \in K$ such that $K \subseteq \bigcup_{i=1}^{n} U_{a_i}$.

- Then observe that $\bigcap_{i=1}^{n} V_{a_i}$ is a neighbourhood which contains $x$ but does not intersect $K$ (it is a finite intersection of open sets), so that $x \not \in \overline{K}$. Therefore $\overline{K} = K$ and so $K$ is closed. $\blacksquare$

Proposition 3: Let $E$ and $F$ be topological spaces and let $f : E \to F$. If $K \subseteq E$ is compact and $f$ is continuous then $f(K)$ is compact. |

**Proof:**If $\{ U_{\alpha} : \alpha \}$ is an open cover of $f(K)$ then $\{ f^{-1}(U_{\alpha}) : \alpha \}$ is an open cover of $K$ by the continuity of $f$ ensuring us that preimages of open sets in $F$ are open in $E$. By compactness of $K$, there exists $\alpha_1, \alpha_2, ..., \alpha_n$ for which $\{ f^{-1}(U_{\alpha_1}), f^{-1}(U_{\alpha_2}), ..., f^{-1}(U_{\alpha_n}) \}$ is an open cover of $K$, and so that $\{ U_{\alpha_1}, U_{\alpha_2}, ..., U_{\alpha_n} \}$ is an open cover of $f(K)$. Thus $f(K)$ is compact. $\blacksquare$.