[proofplan] Each property follows by translating the open-set axioms of a topology into statements about closed sets via complementation. A set is closed if and only if its complement is open, and [De Morgan's laws](/theorems/622) convert unions to intersections and vice versa. Property (1) uses the fact that $X$ and $\varnothing$ are both open. Property (2) uses that arbitrary unions of open sets are open, which becomes: arbitrary intersections of closed sets are closed. Property (3) uses that finite intersections of open sets are open, which becomes: finite unions of closed sets are closed. [/proofplan] [step:Show $\varnothing$ and $X$ are closed by complementation] A set $F \subseteq X$ is [closed](/page/Closed%20Set) if and only if $X \setminus F \in \tau$. Since $X \in \tau$, the complement $X \setminus X = \varnothing$ is open, so $X$ is closed. Since $\varnothing \in \tau$, the complement $X \setminus \varnothing = X$ is open, so $\varnothing$ is closed. [/step] [step:Show arbitrary intersections of closed sets are closed via De Morgan's law] Let $\{F_i\}_{i \in I}$ be any collection of closed sets. For each $i$, $X \setminus F_i \in \tau$ (since $F_i$ is closed). By [De Morgan's law](/theorems/622): \begin{align*} X \setminus \bigcap_{i \in I} F_i = \bigcup_{i \in I} (X \setminus F_i). \end{align*} The right-hand side is an arbitrary union of open sets. By the topology axiom (arbitrary unions of open sets are open), $\bigcup_{i \in I}(X \setminus F_i) \in \tau$. Therefore $X \setminus \bigcap_{i \in I} F_i \in \tau$, which means $\bigcap_{i \in I} F_i$ is closed. [guided] The argument converts the intersection of closed sets into a union of open sets via De Morgan's law, applies the open-set axiom, and converts back. Each step uses a single fact. **Starting point:** Each $F_i$ is closed, which by definition means $X \setminus F_i \in \tau$ (the complement is open). **De Morgan conversion:** We want to show $\bigcap_{i \in I} F_i$ is closed, i.e., $X \setminus \bigcap_{i \in I} F_i \in \tau$. By [De Morgan's law](/theorems/622), $X \setminus \bigcap_{i \in I} F_i = \bigcup_{i \in I}(X \setminus F_i)$. **Open-set axiom:** Each $X \setminus F_i$ is open, and the axioms of a [topology](/page/Topology) state that an arbitrary union of open sets is open. Therefore $\bigcup_{i \in I}(X \setminus F_i) \in \tau$. **Conclusion:** Since $X \setminus \bigcap_{i \in I} F_i \in \tau$, the set $\bigcap_{i \in I} F_i$ is closed. Note the duality: the open-set axiom says arbitrary unions preserve openness, and the corresponding closed-set property is that arbitrary intersections preserve closedness. The "arbitrary" and "finite" qualifiers swap between unions and intersections under complementation. [/guided] [/step] [step:Show finite unions of closed sets are closed via De Morgan's law] Let $F_1, \ldots, F_n$ be closed. For each $k \in \{1, \ldots, n\}$, $X \setminus F_k \in \tau$. By [De Morgan's law](/theorems/622): \begin{align*} X \setminus (F_1 \cup \cdots \cup F_n) = (X \setminus F_1) \cap \cdots \cap (X \setminus F_n). \end{align*} The right-hand side is a finite intersection of open sets. By the topology axiom (finite intersections of open sets are open), $(X \setminus F_1) \cap \cdots \cap (X \setminus F_n) \in \tau$. Therefore $X \setminus (F_1 \cup \cdots \cup F_n) \in \tau$, which means $F_1 \cup \cdots \cup F_n$ is closed. [guided] The structure mirrors the previous step but with the roles of union/intersection swapped, and crucially, finiteness enters. **Goal:** Show $F_1 \cup \cdots \cup F_n$ is closed, i.e., $X \setminus (F_1 \cup \cdots \cup F_n)$ is open. **De Morgan conversion:** $X \setminus (F_1 \cup \cdots \cup F_n) = (X \setminus F_1) \cap \cdots \cap (X \setminus F_n)$. Each $X \setminus F_k$ is open since $F_k$ is closed. **Finite intersection axiom:** The topology axiom states that a finite intersection of open sets is open. This is the axiom that must be restricted to finitely many sets --- arbitrary intersections of open sets need not be open. Therefore $(X \setminus F_1) \cap \cdots \cap (X \setminus F_n) \in \tau$. **Conclusion:** Since $X \setminus (F_1 \cup \cdots \cup F_n) \in \tau$, the union $F_1 \cup \cdots \cup F_n$ is closed. The restriction to finitely many sets is essential: an infinite union of closed sets need not be closed. For example, $\bigcup_{n=1}^{\infty} [1/n, 1] = (0, 1]$ is not closed in $\mathbb{R}$, even though each $[1/n, 1]$ is closed. [/guided] [/step]