I have been thinking about this one for awhile, but I cannot crack it. I think the proof to this is similar to the proof that the topology containing all singleton sets is the discrete topology, in that that proof used the infinite union of singletons to build every subset. In this one I imagine that some how the use of finite intersections can create every finite subset of $X$, but that seems weird and I'm not sure how I would show that.
Is this correct? Can someone show me where to start?
Best Answer
I'm assuming that $X$ is infinite, otherwise the claim does not hold.
If suffices to prove that the singletons $\{x\}$ are open, for all $x \in X$.
$X \setminus \{x\}$ is also an infinite set so there exist infinte sets $A, B \subseteq X \setminus \{x\}$ such that $A \cap B = \emptyset$.
For example, let $\{a_n : n \in \mathbb{N}\}$ be a countable subset of $X \setminus \{x\}$ and then define $A = \{a_{2n} : n \in \mathbb{N}\}$ and $B = \{a_{2n - 1} : n \in \mathbb{N}\}$.
Now $A \cup \{x\}$ and $B \cup \{x\}$ are infinite subsets of $X$ so they are open.
Hence $$\{x\} = (A \cup \{ x \}) \cap (B \cup \{ x \})$$ is also open as an intersection of two open sets.
Now an arbitrary $S \subseteq X$ is just
$$S = \bigcup_{x \in S} \{x\}$$
which is a union of open sets.