Infinite product of connected spaces may not be connected?

Question

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?

Answer 1

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

On this webpage, you will find a proof that the product of connected spaces is connected (using the product topology). In case of another broken link in the future, the following summary (copied from here) could be useful:

The key fact that we use in the proof is that for fixed values of all the other coordinates, the inclusion of any one factor in the product is a continuous map. Hence, every slice is a connected subset.

Now any partition of the whole space into disjoint open subsets must partition each slice into disjoint open subsets; but since each slice is connected, each slice must lie in one of the parts. This allows us to show that if two points differ in only finitely many coordinates, then they must lie in the same open subset of the partition.

Finally, we use the fact that any open set must contain a basis open set; the basis open set allows us to alter the remaining cofinitely many coordinates. Note that it is this part that crucially uses the definition of product topology, and it is the analogous step to this that would fail for the box topology.

Answer 2

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.