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I am trying to prove that an infinite product of discrete spaces may not be discrete. I tried taking the simplest nontrivial discrete space, $X:=\{0,1\}$ with the discrete topology, and tried to find a sequence of $0$s and $1$s in $\displaystyle\prod_{i=1}^\infty X_i$ that can't be produced by taking intersections of preimages of open sets in the factor spaces under the projection maps, but couldn't come up with anything.

Is this a good approach, or is this actually too simple of an example?

How would you go about solving this problem? I'd be interested in hearing the thought process behind the proof as well. Thanks.

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2 Answers 2

up vote 9 down vote accepted

The product you describe is compact by Tychonoff's theorem, but the only compact discrete spaces are the finite ones.

The thought process behind this proof is strongly informed by the material in this blog post. A product of finite discrete spaces is an example of a profinite set or Stone space, and these behave in very particular ways.

Alternatively, the product you describe is second-countable, which an uncountable discrete space isn't.

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To supplement, $\{0,1\}^{\mathbb N}$ is homeomorphic to the Cantor set, en.wikipedia.org/wiki/Cantor_set. The homeomorphism is the obvious one: For example, the sequence $001101\dots$ of zeros and ones is identified with the unique point in the Cantor set that is on $[0,1/3]$, $[0,1/9]$, $[2/27,1/9]$, $[8/81,1/9]$, $[8/81,25/243]$, $[74/729,25/243],\dots$ –  Andres Caicedo Dec 31 '12 at 2:14
    
@Qiaochu: This is much nicer than just finding a subset that's not open. That's the sort of thing I need to work on coming up with. Thanks. –  Alex Petzke Jan 1 '13 at 17:13
    
@Alex: I should also mention that working through Munkres helps. IIRC there is a lot of emphasis on proving that spaces aren't homeomorphic by proving that one satisfies a topological property the other doesn't. –  Qiaochu Yuan Jan 1 '13 at 21:02
    
@Qiaochu: Thanks, I will keep that in mind. –  Alex Petzke Jan 2 '13 at 0:27

More elementarily: The open sets in $\prod_{\mathbb N}\{0,1\}$ are precisely the (possibly infinite) unions of cylinder sets. Because even a single cylinder set is infinite, so is every nonempty open set. In particular, a singleton cannot be open, so the space is not discrete.

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Is the term "cylinder" a standard? I did graduate work in topology, and I've never seen it before in this sort of context. –  Thomas Andrews Dec 31 '12 at 12:41
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@ThomasAndrews: Standard enough for someone to have written a Wikipedia article using it. –  Henning Makholm Dec 31 '12 at 12:43
    
Yeah, I just mean, the person asking is probably working from a textbook that made certain definitions, and adding new terms which requires looking up on Wikipedia doesn't seem like a pedagogically sound plan –  Thomas Andrews Dec 31 '12 at 12:51
    
@ThomasAndrews: Well, since I learn these things from Wikipedia, I don't know what are in people's textbooks. The wikipedia article gave me the impression that this was a standard term people use for talking about bases for infinite product topologies, so I used it. –  Henning Makholm Dec 31 '12 at 13:17
    
@Thomas: I think that the term comes from measure theory. I don’t remember encountering it in the topological sense until fairly recently. –  Brian M. Scott Dec 31 '12 at 18:10

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