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?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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).
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: