Another topology problem help Let $X$ be infinite with the finite complement topology (a set is open if it is $\varnothing$ or its complement is finite). If $A$ is any infinite subset of $X$, show that the closure of $A$ is $X$.
I'm having trouble with this because I have no idea what the set is.
 A: If $A$ is an infinite set then so is $\operatorname{cl}(A)$. This as a consequence of $A\subseteq\operatorname{cl}(A)$. In the topology you mention all closed sets (with no more than one exception) are finite as complements of the open cofinite sets. The exception is the whole space $X$ as complement of the open empty set. So as closed and infinite set $\operatorname{cl}(A)$ must equalize $X$.
A: Let $x\in X$ be a point and let $U$ be a neighborhood of $x$. $U$ contains all but finitely many of the points in $X$, hence it must contain at least one point in any infinite subset. Thus given any infinite subset $A$, $x$ is in the closure of $A$, and since $x$ was arbitrary we must have that every point is in the closure.
A: You don’t need to know anything about $X$ beyond the fact that it is infinite. Suppose that $A\subseteq X$ is also infinite. You know that every point of $A$ is in the closure of $A$, so we need only worry about points of $X\setminus A$. Suppose that $x\in X\setminus A$, and $x$ is not in the closure of $A$. Then $x$ must have an open nbhd, say $U$, that is disjoint from $A$. But $U\cap A=\varnothing$ if and only if $A\subseteq X\setminus U$: $A$ is a subset of the complement of $U$. That’s impossible: $U$ is open, so its complement is finite, but $A$ is infinite. Thus, every point of $X\setminus A$ is in the closure of $A$, which is therefore the whole space $X$.
