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Is there an example of a sigma algebra that is not a topology? If this is not the case, is it possible to prove that all sigma algebras are topologies?

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The Borel sets in $[0,1]$, obviously. It contains all points but not all subsets. –  t.b. Jul 13 '11 at 14:37
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@Claudia: Didn't you pose exactly the same question here yesterday? Maybe I'm going mad... –  Stefan Walter Jul 13 '11 at 14:45
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@Stefan: I had the same déjà-vu, don't worry about your mental health :) I see only now that Joel made more or less the same point as I did in a comment –  t.b. Jul 13 '11 at 14:50
    
I do not recall topologies being $\sigma$-algebras at all... They are not usually closed under complements nor countable intersections. –  Asaf Karagila Jul 13 '11 at 14:58
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See also mathoverflow.net/questions/70137/… same question and basically the same answer. –  plusepsilon.de Jul 13 '11 at 15:17

2 Answers 2

If a $\sigma$-algebra contains all one-point sets then the topology it generates is the discrete topology, so the only thing we need to ensure is that we have a $\sigma$-algebra in which not every set is measurable.

The very first example that comes to mind is already an example. As there are non-Borel sets in $[0,1]$ the Borel sets are not the discrete topology, but the smallest topology that contains the Borel sets is the discrete topology by the above paragraph.

(I'm ignoring choice issues here, as I lack the expertise).


Added:

A similar argument works for the Lebesgue $\sigma$-algebra.

In a positive direction: Every $\sigma$-algebra on a countable set is a topology.

To sum up:

  • On finite or countable sets every $\sigma$-algebra is a topology.
  • In view of Pete's answer below, on every uncountable set there is a $\sigma$-algebra that isn't a topology, namely the countable-cocountable $\sigma$-algebra. This example is probably the optimal one in terms of simplicity.

For the sake of completeness:

  • Pete L. Clark's answer in this thread uses choice in the very weak form "the countable union of countably many countable sets is countable", as noted by Nate Eldredge.

  • Andrés Caicedo argues beautifully in this MO-thread why some choice is necessary to ensure that there exist non-Borel sets.

  • If I interpret François G. Dorais's answer here correctly, there are models of ZF with the property that the $\sigma$-algebra generated by the singleton subsets of a set is equal to the power set. (Please correct me if this is too naïve a rendering)

  • Asaf Karagila adds in a comment below: "[...] an even freakier statement is given in Jech's The Axiom of Choice, Ch. 5, Exercise 14. There is an extension of every transitive model, with the same $\aleph$-cardinals, and for every $\alpha$ there is a set $X$ which is a countable union of countable sets, and $P(X)$ can be partitioned into $\aleph_{\alpha}$ nonempty sets."

  • Jay adds: As John B. S. Haldane might have said, "Set theory without the axiom of choice is not only queerer than we suppose, but queerer than we can suppose."

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@Asaf: No I mean that I don't know how to construct a non-Borel set without assuming at least some choice. And I mean that I ignore choice issues in general. –  t.b. Jul 13 '11 at 16:32
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I have no choice but to agree with Andres... :-) –  Asaf Karagila Jul 13 '11 at 16:45
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@Asaf: I guess so :) You might also be interested in François G. Dorais's answer here. By the way, I have nothing against asking about and investigating choice, but I'm a lazy person and stick to good ol' ZFC - don't try to tell an old man to change his ways... Foundational issues are interesting but more a hobby of mine than anything else. –  t.b. Jul 13 '11 at 17:17
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I saw this question before, an even freakier statement is given in Jech's The Axiom of Choice, Ch. 5, Exercise 14. There is an extension of every transitive model, with the same $\aleph$-cardinals, and for every $\alpha$ there is a set $X$ which is a countable union of countable sets, and $P(X)$ can be partitioned into $\aleph_\alpha$ nonempty sets. Creepy, right? :-) –  Asaf Karagila Jul 13 '11 at 17:37
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As John B. S. Haldane might have said, "Set theory without the axiom of choice is not only queerer than we suppose, but queerer than we can suppose." –  Jay Jul 14 '11 at 0:31

Let $S$ be any uncountable set, and let $\mathcal{A}$ be the collection of all subsets of $S$ which are either countable or have countable complement.

This collection is evidently closed under complementation. If I have a countable union of elements of $\mathcal{A}$, all of which are countable, then the union is countable. Otherwise, at least one element is cocountable, hence so is the union. A similar argument works for intersections. So $\mathcal{A}$ is a $\sigma$-algebra.

It is not a topology because it contains all the singleton sets but not all subsets of $S$ -- in particular, it contains no set which is uncountable with uncountable complement.

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Ah very nice! That's much simpler than the Borel sets. I guess I let myself get carried away... –  t.b. Jul 14 '11 at 1:23
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Of course, this argument is also using the axiom of choice (a countable union of countable sets is countable). –  Nate Eldredge Jul 14 '11 at 3:52
    
@Theo: If the space is $T_1$ then every countable set is $\Sigma^0_2$ and every cocountable set is $\Pi^0_2$, therefore the $\sigma$-algebra denoted by Pete as $\mathcal A$ is a sub-algebra of $\Delta^0_3$. It is simpler than Borel sets, but the Borel sets tend to include them. (I know both you as well Pete are aware of that, this is a side remark for the less familiar reader :-)) –  Asaf Karagila Jul 14 '11 at 22:08
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@Nate: Yes, I am assuming the axiom of choice. I am (like most mathematicians?) in fact always assuming the Axiom of Choice, although (unlike many mathematicians?) I agree that it is nice to know what happens in the absence of AC. But really, how far does one get in measure theory without the "fact" that countable unions of countable sets are countable? (This is not a rhetorical question.) –  Pete L. Clark Jul 15 '11 at 1:37

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