My guess is that $[0,1]^\omega$ is not path-connected in box topology. I was able to show that $[0,1]^\omega$ is path-connected in product topology by using the general result that product of path-connected spaces is path-connected. I was trying to show that there is no path from $ x = (0, 0, \ldots)$ to $y = (1, 1, \ldots)$ in box $[0,1]^\omega$ by finding a contradiction, but I couldn't.
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1Possibly helpful: https://math.stackexchange.com/q/2798417/42969, https://math.stackexchange.com/q/86395/42969. – Martin R Jun 15 '25 at 15:52
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1I also calculate all the quasi-components/components/path-components of $\mathbb{R}^\omega$ (and so also $[0, 1]^\omega$) here – Jakobian Jun 15 '25 at 23:45
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The space isn't connected (with the box topology). For instance, if $U$ is the set of convergent sequences and $V$ is the set of sequences that don't converge, then these are disjoint, open, and $[0,1]^{\omega}=U\cup V$. This is a variation of an answer from a comment or, e.g., the argument at https://topospaces.subwiki.org/wiki/Connectedness_is_not_box_product-closed
yoyo
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It isn’t connected with the box topology, but it is connected with the product topology? – J. W. Tanner Jun 15 '25 at 16:45
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