Finite topological Hausdorff spaces Let $X$,$Y$ be two finite topological Hausdorff spaces, how can I prove that if the two spaces have the same number of points, they must be homeomorphic?
EDIT: so sorry I missed to add the hausdorff assumption in the body, I thought it was clear from the caption.
 A: As @David Mitra (in a comment) and @jon.sand (in an answer) have pointed out, this result is true for $T_2$ spaces. In fact, it's even true for $T_1$ spaces. For each positive integer $n$ there is exactly one $T_1$ topology (the discrete topology), up to homeomorphism, on an $n$-element set. This is because for a $T_1$ topological space each singleton set is a closed set, and since every subset of such a space is a finite union of singleton sets, it follows that every subset of such a space is a closed set, and hence every subset of such a space is an open set (i.e. the discrete topology). However, it's not true in a very strong way if we slightly weaken the separation axiom assumption to $T_0.$
A few weeks ago I happened to be looking over the paper cited below, which among other things shows how different the $T_0$ and $T_1$ separation axioms are for finite topological spaces. For example, on a $14$-element set, there are $98,484,324,257,128,207,032,183$ different $T_0$ topologies and $115,617,051,977,054,267,807,460$ different topologies. Or, to put it another way, on a $14$-element set there is exactly one $T_1$ topology and roughly $\frac{1}{6}$ Avogadro's number of $T_0$ topologies!
Marcel Erné and Kurt Stege, Counting finite posets and topologies, Order 8 #3 (1991), 247-265.
A: Hint : for a point $x$ in a Hausdorff space, what is the intersection of all open sets containing it ?
If the space is finite, what property does this intersection have ?
A: In general it is not true. Consider discrete topology and anti-discrete topology. But second one is not Hausdorff.
