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Very basic question: I can't get my head around the Cantor space. It has a basis of clopen sets. Finite unions of closed sets are closed, and unions of open sets are open, so a finite union of basis elements of a Cantor space is clopen. The only open sets in a Cantor space that are not closed, if there are any, are infinite unions of basis elements. An example of an open set that's not closed is ... what?

Not a homework question; I managed to stump myself. Thanks for your help.

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Hint: Look at the complement of a point—the Cantor space is not discrete.

Later: Since the Cantor set is homeomorphic to a countable product $\prod_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ of the cyclic group of order two (use the identification of the cantor set with the points in $[0,1]$ whose infinite ternary expansion contains no $1$), it is homogeneous. This means in particular that the Cantor set has no isolated points and hence it has no open points.

Now note that a basic open set is of the form $\prod_{n=1}^{\infty} X_n$ with all but finitely many $X_n = \mathbb{Z}/2\mathbb{Z}$. But this means that a basic open set contains a space homeomorphic to the entire Cantor space, hence non-empty open sets are uncountable (in fact of cardinality $\mathfrak{c}$). In particular we see that a convergent sequence (which is of course closed) can't be open. Passing to complements we get an open set that's not closed, as required.

[Meta: Thanks to ccc for pointing this out and to Brian for making me think again.]

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    Great. I type slowly on my iPhone while you answer!2011-08-01
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    Hold it horizontally (landscape view, I mean), use both your thumbs :)2011-08-01
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    I don't like doing that and I type better without that. :-)2011-08-01
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    How about convergent sequences? These are compact subsets of a Hausdorff space, and therefore closed, right?2011-08-01
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    @gary: Yes, they are closed. However, why are they not open? Do you have a simple argument for that? I haven't thought deeply but I don't see an immediate answer avoiding Baire.2011-08-01
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    @Theo: I did not think it thru too much either; it is 3 a.m here; maybe tomorrow.2011-08-01
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    (Non-empty) open sets are uncountable.2011-08-01
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    @ccc: yes, and why? I must be missing something obvious.2011-08-01
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    It's enough to show that basic open sets are uncountable. In this case you can do this directly by saying you have infinitely many coordinates of wiggle room in your binary sequences and invoke the standard Cantor diagonalization argument.2011-08-01
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    @ccc: Okay, thanks! Still not as simple as I'd like it to be.2011-08-01
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    @Theo: Well you of course have the stronger fact that each basic open set is homeomorphic to Cantor space itself (and in particular is uncountable), but I don't know if you'd consider that simpler.2011-08-01
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    @ccc: Thanks again! You're probably right that your first argument *is* simpler and I *do* like the second one. I was naïvely hoping for an argument I'd expect an average student with only very basic knowledge in topology to come up with.2011-08-01
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    @Theo: The observation that a basic open set in the product space $2^\omega$ is homeomorphic to $2^\omega$ is pretty elementary; the problem is that too often students see only the middle-thirds Cantor set in elementary courses. (And of course that means that every non-empty open set is not just uncountable, but of cardinality $\mathfrak{c}$.)2011-08-02
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    @Brian: Thanks, you're right. I had what we call a "plank in front of my head" in German.2011-08-02
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    @ccc: I now implemented your suggestion. Thanks again.2011-08-02
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    @Theo: A good student familiar only with the middle-thirds Cantor set might still come up with an example like $[0.1/9]\cup\bigcup_{n>0}\left[\sum_{k=1}^n 2/3^k,\sum_{k=1}^n 2/3^k+1/3^{n+2} \right]$, which is clearly open, being a union of clopen pieces of the construction. but which has $1$ as a fairly obvious limit point in its (non-trivial) complement.2011-08-02
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    @Brian: Perfect! That's exactly the sort of thing I was looking for.2011-08-02
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    @Theo: Das Brett vorm Kopf ist ein Berufsrisiko! :-)2011-08-02
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    @Brian: Wow... ich bin beeindruckt! So einen Kommentar hätte ich jetzt nicht erwartet :) This seems to require serious command of the ["awful German language"](http://www.crossmyt.com/hc/linghebr/awfgrmlg.html) (Twain). May I ask: are you fluent in German?2011-08-02
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    @Theo: Not really. I read it tolerably well (e.g., DER SPIEGEL most weeks), but my listening comprehension is very rusty, and my writing and speaking are painfully slow. I post very occasionally to de.sci.mathematik, but only when I’ve time and energy to test wordings against what I can find on the web -- for some reason I'd almost rather get the mathematics wrong than the language!2011-08-02
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    @Brian: Thanks for the clarification and "middle-thirds" example (I somehow forgot that people even identify Cantor space with the middle-thirds set, since the latter is so messy to work with). .@Theo: You're welcome! .@Both: Now I will leave before the comment section is overrun by long, intimidating German words. :-)2011-08-02
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Just make a sort of fishbone, and remove the spine: $$\bigcup_{i=1}^{\infty} [10^i].$$