Are there uncountably many non homeomorphic ways to topologize a countably infinite set?

Solution 1:

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.

Solution 2:

For any (ultra)filter $\mathcal F$ on $\omega$ you can define a topology on $\omega\cup\{\omega\}$ given by the base $\{\{a\}; a\in\omega\} \cup \{\{\omega\}\cup A; A\in\mathcal F\}$. (I.e. the neighborhoods of the point $\omega$ are given by the sets from the filter and other points are isolated.)

It should not be very difficult to show that such two spaces are homeomorphic if and only iff the corresponding (ultra)filters are isomorphic. (In the sense that there is a bijection transforming one filter into another. In fact, it suffices to notice that we can recover the filter from the topology by taking all neighborhoods of the point $\omega$ and removing this point.)

Since it is known that there are $2^{\mathfrak c}$ non-isomorphic ultrafilters on $\omega$, we get that there are $2^{\mathfrak c}$ non-homeomorphic countable topological spaces. (If I remember correctly, this result is due to Prikry Pospíšil; I wasn't able to find a reference, but quick google search gives e.g. this discussion.)

However, here I have used an advanced fact about ultrafilters, I hope that someone will come up with a more elementary solution. (And I also hope that I didn't make a mistake there.)

EDIT: Now I found in Comfort, Negrepontis: The Theory of Ultrafilters, Corollary 7.4, which claims that there are $2^{2^{|X|}}$ free ultrafilters on a set $X$. It is attributed to Pospíšil.

EDIT: Now I had time to have a closer look on the link I provided and I can see that the same reasoning as for ultrafilter works for topologies too.

If we know that there exists $2^{\mathfrak c}$ topologies, and we notice the fact that in each homeomorphism class can be only $\mathfrak c$-many of them (since $\mathfrak c$ is the number of bijections from $\omega$ to $\omega$); there are precisely $2^{\mathfrak c}$ classes.

The fact that on a set $X$ there exists $2^{2^{|X|}}$ topologies is mentioned as Theorem 1.4 in paper Roland E. Larson and Susan J. Andima, The lattice of topologies: A survey equations, Rocky Mountain J. Math. Volume 5, Number 2 (1975), 177-198, doi: 10.1216/RMJ-1975-5-2-177. The reference for this result given in this paper is Otto Fröhlich, Das Halbordnungssystem der topologischen Räume auf einer Menge, Math. Ann. 156 (1964), 79-95. (Perhaps for the case $|X|=\aleph_0$ there is a simple proof.)

Solution 3:

Any partial order, $(P,<)$ determines a topology on $P\,$ by defining a subset $D$ open if:

$$\forall x\in D,\text{ if } y<x, \text{ then } y \in D$$

This topology has the advantage that if the topology on two partially ordered sets are homeomorphic, then the partially ordered sets are isomorphic.

So are there uncountably many partial orders on a countable set that are non-isomorphic?

Let $S\subset \mathbb{N}$. Define elements of the countable order:

$$\{s,a_1,a_2,a_3,...,b_{1,1},b_{2,1},b_{2,2},b_{3,1}...\}$$

with $b_{n,i}$ having the condition $i\leq n$ and the rules:

$a_n <_S\, s$ if $n\in S$.

$b_{n,i}<_S \, a_n$

The $b_{n,i}$ ensure that the elements $a_n$ each have a different number of smaller elements, so there is no non-trivial automorphism, and the $s$ element ensures that this order uniquely encodes $S$.