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Let $C([0,1]^\omega)$ denote the set of continuous functions $f:[0,1]^\omega \to \mathbb{R}$. We endow $C([0,1]^\omega)$ with two topologies. Let $\tau$ be the pointwise topology on $C([0,1]^\omega)$ and let $\sigma$ be the weak topology.

Are $(C([0,1]^\omega), \tau)$ and $(C([0,1]^\omega), \sigma)$ homeomorphic?

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    $\begingroup$ By weak topology on $C(X)$ you mean the initial topology w.r.t. all functions from the dual $C^*(X)$ (i.e., all linear continuous functionals on $C(X)$)? $\endgroup$
    – Martin Sleziak
    Commented Jun 18, 2015 at 10:30
  • $\begingroup$ @MartinSleziak That's correct - thanks for asking! $\endgroup$
    – Dominic van der Zypen
    Commented Jun 18, 2015 at 10:40
  • $\begingroup$ This answer provides an example showing that pointwise convergence and weak convergence are not equivalent in $C([0,1])$. I suppose that the same example should work for product of countably many copies of $[0,1]$. (The answer was given by Theo Buehler.) $\endgroup$
    – Martin Sleziak
    Commented Jun 18, 2015 at 10:57
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    $\begingroup$ Do you mean isomorphic as topological vector spaces? $\endgroup$
    – Eric Wofsey
    Commented Jun 18, 2015 at 11:15
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    $\begingroup$ Isomorphic as vector spaces, yes. But not as topological vector spaces. Martin is correct: the method used for $C[0,1]$ also works for this. More generally, try proving it for $C(K)$, where $K$ is an infinite compact metric space. $\endgroup$
    – Gerald Edgar
    Commented Jun 18, 2015 at 11:38

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