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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.

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"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 don't have the time to sit through the detail, but I believe it has something to do with the order type of limit ordinals below $\alpha$. – Asaf Karagila Apr 24 '11 at 11:19
@Asaf I think you're right. This paper claims to classify ordinal topologies using the Cantor normal form and the Cantor-Bendixson rank. I didn't read it because I don't understand the machinery, but, to quote the second paragraph, "The intuition behind this classification is that ordinal topologies are mainly determined by their limit points". – Mike F Apr 24 '11 at 22:55
consider a partition of $\mathbb{N}$ (this determines the minimal open sets of a topology on $\mathbb{N}$). for every non-decreasing sequence of $\mathbb{N}\cup\{\infty\}$ we get a distinct topology. – yoyo Apr 25 '11 at 21:07
up vote 21 down vote accepted

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.

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Such a nice example! – Mike F Apr 24 '11 at 15:34
@Nate Eldredge : For $F(n) = n$, don't you want a partition more like this one? $$ \{ \{x_1\}, \{x_2,x_3\},\{x_4,x_5\}, \{x_6,x_7,x_8\}, \cdots, \{x_{12},x_{13},x_{14} \}, \cdots \} $$ i.e. one subset of size $1$, two subsets of size $2$, three of size $3$, etc... your example seems to match with $F(n) = 1$. – Patrick Da Silva Sep 1 '14 at 20:58
@PatrickDaSilva: Thanks for the correction, I incorporated it. – Nate Eldredge Sep 17 '14 at 21:28

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 atributed 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.)

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Another way to find the cardinality of the set of all ultrafilters on $\omega$ (or on any given set) is to use topological arguments for the space $\beta\omega$, see here. – Martin Sleziak Apr 21 '12 at 7:34

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:


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 $S$.

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Thanks! Alternatively we could use that there are uncountable many countable ordinals – Mike F Apr 24 '11 at 15:50
I wonder if there is a way to see there are $2^c$ nonisomorphic partial orderings of a countable set. Actually I guess you only need to show there are $2^c$ partial orderings proper since there are only $c$ bijections of a countable set! – Mike F Apr 24 '11 at 15:55
@user14111: ah, fair call. – Mike F May 17 '13 at 3:56

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