TITLE: Are there uncountably many non homeomorphic ways to topologize a countably infinite set?
QUESTION [30 upvotes]: Today I was fooling around a bit trying to count the topologies on a finite set. I didn't make much progress, so I did some googling and quickly discovered it is an open problem to give a closed form expression for the number $T_n$ of topologies on a set with $n$ elements. At this point I asked myself "if I can't compute $T_n$, can I perhaps compute $T_\omega$?". 
First I wondered whether it is even obvious that there are uncountably many ways to topologize a countably infinite set $X$. It is. For each $S \subset X$, $\{\varnothing, S, X\}$ is a topology so there are at least $c$ ways to topologize $X$. This is a bit of a cheat though since, up to homeomorphism, this topology depends only on the cardinalities of $S$ and $X-S$ for which there are only countably many possibilities. Thus we are lead naturally to the question in the title:

Are there uncountably many non homeomorphic topologies on a countably infinite set?

The most obvious approach that I could think of was to look at order topologies. Distinct ordinal numbers are not order-isomorphic and there are uncountably many countable ordinals so surely by considering the countable ordinals in their order topologies we exhibit uncountably many nonhomeomorphic spaces... right? There is a gap though because distinct ordinals can be homeomorphic. For instance, if $\alpha \geq \omega$ then $\alpha + 1, \alpha +2, \alpha +3, \ldots$ are all homeomorphic (we can sneak the $n$ isolated points at the end of $\alpha + (n+1)$ into the copy of the $\omega$ at the beginning of $\alpha + 1$). To close the gap, we need to prove that for any countable ordinal $\alpha$ there exists a countable ordinal $\beta > \alpha$ homeomorphic to no ordinal in $[0,\alpha]$. I would be flabbergasted if this did not hold, but to prove it would probably require more set theory than I have at my disposal.
If my title question is easily answered (in the affirmative) then, as a follow up question, I would also be interested to know if we can actually determine the number of topologies on a countable set (either topologies proper, or topologies up to homeomorphism). At the moment it would seem I can do little more than supply the obvious upper bound of $2^c$ for both quantities...
Added: There have been several nice answers. Nate (whose answer I accepted) provided probably the slickest way to resolve the title question. 
Martin answers the follow up question by showing that the existence of $2^c$ ultrafilters on $\omega$ (this is proved at his first link) implies the existence of $2^c$ topologies on $\omega + 1$ (hence $2^c$ topologies up to homeomorphism since there are only $c$ permutations of a countable set). Actually it seems the proof that there are $2^c$ ultrafilters can easily be adapted to prove the existence of $2^c$ topologies directly. I'll sketch the argument below while I convince myself I understand what's going on.
Let $X$ be the (countable) set of all subsets of $\mathbf{R}$ which are finite unions of intervals with rational endpoints. For any of the $2^c$ sets $S \subset \mathbf{R}$ we get a basis $\mathscr{B}_S$ for a toplogy on $X$ consisting of all sets $U_S(F) := \{ x \in X : x \cap F = S \cap F \}$ as $F$ ranges over finite subsets of $\mathbf{R}$. This is a basis since if $x \in B(F_1) \cap B(F_2)$ then $x \in B(F_1 \cup F_2) \subset B(F_1) \cap B(F_2)$. It remains to check that different subsets of $\mathbf{R}$ give rise to different topologies. In fact, if $\mathscr{B}_T$ is a refinement of $\mathscr{B}_S$ then $S \subset T$. Suppose that $s \in S$. If $\mathscr{B}_T$ refines $\mathscr{B}_S$ then there should exist a finite subset $F \subset \mathbf{R}$ so that $U_T(F) \subset U_S(\{s\})$. Wlog $s \in F$ since $U_T(F \cup \{s\}) \subset U_T(F)$. If $s \notin T$ then we have $s \notin x$ for all $x \in U_T(F)$ while $s \in x$ for all $x \in U_S(\{s\})$ so these basis elements are disjoint. Thus $s \in T$ must obtain and we are done.

REPLY [22 votes]: Here's a simple, though non-optimal, answer.
In any topological space $(X,\tau)$, let us say an open set $U$ is minimal if for every open $V \subset U$, either $V=U$ or $V=\emptyset$.  (That is, it has no nonempty proper open subsets; it is in some sense an "atom".)  For a natural number $n$, let $F_\tau(n)$ be the number of minimal open sets which contain $n$ elements.  It's easy to see that the function $F_\tau$ is preserved by homeomorphisms, so topologies with different $F_\tau$s are non-homeomorphic.
If $X = \{x_1, x_2, \dots\}$ is a countable set, for any function $F : \mathbb{N} \to \mathbb{N}$, we can produce a topology $\tau$ on $X$ with $F_\tau = F$.  Just choose any partition $P$ of $X$ which contains $F(n)$ sets of size $n$, and take the topology $\tau$ generated by $P$ (and note that in $\tau$, the sets of $P$ are minimal open).  For instance, if $F(n) = n$, we could take a partition like
$$\{ \{x_1\}, \{x_2,x_3\},\{x_4,x_5\}, \{x_6,x_7,x_8\}, \{x_9, x_{10}, x_{11}\}, \{x_{12},x_{13},x_{14} \}, \cdots \}.$$
Since $\mathbb{N}^\mathbb{N}$ is uncountable, this produces uncountably many non-homeomorphic topologies on $X$.  
Of course there are only $\mathfrak{c}$ of these, not $2^\mathfrak{c}$, but at least they are simple.