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Let $X$ be a connected topologoical space. Is it true that the countable product $X^\omega$ of $X$ with itself (under the product topology) need not be connected? I have heard that setting $X = \mathbb R$ gives an example of this phenomenon. If so, how can I prove that $\mathbb R^\omega$ is not connected? Do we get different results if $X^\omega$ instead has the box topology?

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$\mathbb R^\omega$ with the product topology is not only connected but path connected. The "natural" linear path works. This is not the case for the box topology, where, say $t\mapsto(t,t,t,\ldots)$ is not continuous. (Wikipedia claims that $\mathbb R^\omega$ as a box product is not connected). –  Henning Makholm Nov 28 '11 at 15:10
    
More precisely, the set of all sequences that converge towards 0 is clopen in the box topology. –  Henning Makholm Nov 28 '11 at 15:24

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Maybe this should have been a comment, but since I don't have enough reputation points, here it is.

http://planetmath.org/encyclopedia/ProofOfProductsOfConnectedSpaces.html

On this webpage, you will find a proof that the product of connected spaces is connected (using the product topology).

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The first part of your question - about connectedness of $\mathbb R^\omega$ with the usual product topology - has already been answered.

To show that box product $\mathbb R^\omega$ is not connected we only need find a clopen subset $U$ of this topological space (different from $\emptyset$ and the whole space). Here are two examples of such sets:

  • $U=$ set of all sequences, that converge to $0$; as suggested by Henning's comment, see also here here. Indeed, if $x_n\to 0$ then $V=\prod(x_n-1/2^n,x_n+1/2^n)$ is a neighborhood of $x$ such that $V\subseteq U$, therefore $U$ is open. Similar argument shows that the complement of $U$ is open in the box topology.

  • $U=$ set of all sequences that are bounded; see e.g. Example 10.16 here. The argument is similar, here we can even use open intervals of the same length on each coordinate.

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