Finite non-empty open sets as a topological property Show that having nonempty open sets containing only finitely many elements is a
topological property. That is, if $X$ is a topological space that has nonempty open
sets with finitely many elements and $f \colon X \to Y$ is a homeomorphism, then $Y$ has
nonempty open sets with finitely many elements.
I am really stuck on where to start here
If $f \colon X \to Y$ is a continuous map and Y has nonempty finite open sets, does $X$
have to have nonempty finite open sets?
My gut instincts tells me this isn't the case. Does this counter example work?
Let $f \colon \mathbb{R} \to \mathbb{Z}$ s.t. $f(x) = \lfloor x\rfloor$. Consider $\mathbb{R}$ with standard topology and $\mathbb{Z}$ with discrete topology. All open sets in standard topology are infinite since they are just open intervals. Obviously, the discrete topology has nonempty finite sets.
 A: Your proposed counterexample will not be continuous. To see this, note that $\mathbb{R}$ is connected and so if $f$ were continuous, its image $f(\mathbb{R})$ should be connected. However, the only connected subsets of $\mathbb{Z}$ with the discrete topology are the singleton subspaces, and $f$ is not constant.
You are right in thinking that continuous functions do not preserve this property. For example, the constant map $\mathbb{R} \to \ast$ is continuous, but $\mathbb{R}$ does not have finite open sets. In the same light, the constant map $c : \ast \to \mathbb{R}$ such that $c(\ast) = 0$ is also continuous.
You are missing a key hypothesis (which you have quoted!). You need to show that if $X$ has finite open sets and there exists homeomorphism $f : X \to Y$, then $Y$ has finite open sets. 
Try to think why a homeomorpshism should preserve both open sets and 'finiteness'. Spoilers below,

 Take $O$ finite (non-empty) and open in $X$. Since $f$ is a homeomorpshim, we know that it is an open map, and so $f(O)$ is an open set of $Y$. Moreover, this set is finite as $f$ is bijective. Note that we actually only need $f$ to be open for this to work, as we always have $|f(O)| \leq |O|$.

A: Hint: Use the fact that $f$ is a bijection for the finiteness part.
