[proofplan] We prove both assertions by passing to complements. The complement of an arbitrary intersection is an arbitrary union of complements, and the topology is closed under arbitrary unions. The complement of a finite union is a finite intersection of complements, and the topology is closed under finite intersections. [/proofplan] [step:Take complements to prove arbitrary intersections of closed sets are closed] Let $(F_i)_{i\in I}$ be a family of closed subsets of $X$. For each $i \in I$, define $U_i := X \setminus F_i$. Since $F_i$ is closed, $U_i \in \tau$. By De Morgan's law for complements in the ambient set $X$, \begin{align*} X \setminus \bigcap_{i\in I} F_i = \bigcup_{i\in I} (X \setminus F_i) = \bigcup_{i\in I} U_i. \end{align*} Because $(X,\tau)$ is a [topological space](/page/Topological%20Space), $\tau$ is closed under arbitrary unions. Hence $\bigcup_{i\in I} U_i \in \tau$, so $X \setminus \bigcap_{i\in I} F_i$ is open. Therefore $\bigcap_{i\in I} F_i$ is closed. [guided] We use the definition of a [closed set](/page/Closed%20Set): a subset of $X$ is closed exactly when its complement in $X$ is open. For each index $i \in I$, define the open complement \begin{align*} U_i := X \setminus F_i. \end{align*} Since each $F_i$ is closed by hypothesis, each $U_i$ belongs to the topology $\tau$. Now compute the complement of the intersection. For a point $x \in X$, the condition $x \notin \bigcap_{i\in I} F_i$ means that there exists at least one $i \in I$ such that $x \notin F_i$. This is exactly the condition that $x \in \bigcup_{i\in I} (X \setminus F_i)$. Thus De Morgan's law gives \begin{align*} X \setminus \bigcap_{i\in I} F_i = \bigcup_{i\in I} (X \setminus F_i) = \bigcup_{i\in I} U_i. \end{align*} The topology axiom we need is the arbitrary-union axiom: if every $U_i$ lies in $\tau$, then $\bigcup_{i\in I} U_i$ also lies in $\tau$. Therefore $X \setminus \bigcap_{i\in I} F_i \in \tau$. Since the complement of $\bigcap_{i\in I} F_i$ is open, the set $\bigcap_{i\in I} F_i$ is closed. [/guided] [/step] [step:Take complements to prove finite unions of closed sets are closed] Let $n \in \mathbb{N}$, and let $F_1,\ldots,F_n$ be closed subsets of $X$. For each $k \in \{1,\ldots,n\}$, define $U_k := X \setminus F_k$. Since $F_k$ is closed, $U_k \in \tau$. By De Morgan's law, \begin{align*} X \setminus \bigcup_{k=1}^{n} F_k = \bigcap_{k=1}^{n} (X \setminus F_k) = \bigcap_{k=1}^{n} U_k. \end{align*} Because $(X,\tau)$ is a [topological space](/page/Topological%20Space), $\tau$ is closed under finite intersections. Hence $\bigcap_{k=1}^{n} U_k \in \tau$, so $X \setminus \bigcup_{k=1}^{n} F_k$ is open. Therefore $\bigcup_{k=1}^{n} F_k$ is closed. [/step]