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Let $X =$ {0, 1, 2, ...} and $T$ = { $\emptyset$, $X$, {0}, {1}, {0,2}, {0,1,3}, {0,1,2,4}, {0,1,2,3,5}, ... } $\cup$ {{0,1,2}, {0,1,2,3}, {0,1,2,3,4}, ... }. It is easily verified that $T$ forms a topology on $X$. Burdick has shown (Amer. Math. Monthly January 2006 p. 83) that the singleton {0} generates infinitely many distinct sets in this space under the operations of closure, complement and union. Q: Is this the only countably infinite topological space (up to homeomorphism) in which a finite seed set generates an infinite family under those three operations?

The answer seems to be yes. Burdick's and my solutions to the above Monthly problem both used this space. Even before seeing his solution I was led to think it's the only one that works.

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(edited April 9 to correct the topology description)

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  • $\begingroup$ how is the space $X$ above Hausdorff? $\endgroup$
    – Toink
    Commented Apr 7, 2013 at 16:48
  • $\begingroup$ Toink, I should have said $T_0$ instead of Hausdorff, but I've moved my comment to an answer. $\endgroup$
    – Joel David Hamkins
    Commented Apr 7, 2013 at 18:27
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    $\begingroup$ Mathematrucker, your family is not a topology, since it is closed neither under unions (not even finite unions) nor finite intersections. Do you intend the topology generated by those sets? $\endgroup$
    – Joel David Hamkins
    Commented Apr 7, 2013 at 18:55
  • $\begingroup$ @Joel I corrected the topology description above, but you and Toink have shown that my original question didn't capture any uniqueness I thought I observed in my computer experiments. The only change I would suggest is to restrict to a finite space and singleton seed set and require that the power set of the space be generated under the three operations. If this still doesn't imply a topologically unique space (for each finite cardinality), then what I thought might be a bonus discovery in my computer experiments, was evidently a bogus one. $\endgroup$
    – mathematrucker
    Commented Apr 9, 2013 at 18:35

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If you replace each point in your space with 2 points (or more), and use the induced topology, then you still have the same generating property, but the resulting space has no isolated points, and hence is not homeomorphic to your space.

Perhaps you want to insist upon a weak separation axiom.

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Joel showed you how to get infinitely many non-$T_0$ non-homeomorphic spaces as you want. But you can even get infinitely many $T_0$ non-homeomorphic spaces as you want just by isolating points $0$ through $k$ in your topology ($k \geq 1$). The resulting space will have exactly $k$ many isolated points and every $\{j\}$, for $j \geq k$ can be generated from the finite set $\{0,1, \dots, k-1\}$ by closure, complement and union.

You cannot wish to get a Hausdorff, not even a $T_1$ space as you want because a space is $T_1$ if and only if singletons are closed. So, a finite set in a $T_1$ space will only generate finitely many sets under the operations of closure, complement and union.

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