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I was trying to disprove (or also prove) whether $\mathbb{R}^{[0,1]}$ is separable. My intuition tells me it's a disprove. I thought perhaps proving that $\mathbb{R}^{[0,1]}$ is sequentially compact will assist?

Or maybe that won't help?

And my second question is in regard to $\mathbb{R}^{[0,1]}$ being regular? In here I have no intuition at all, unfortunately.

Zhan I.s.
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  • What is $\mathbb{R}^{[0,1]}$? Is it the interval $[0,1]$ in $\mathbb{R}$? If it is then the set $[0,1]\cap \mathbb{Q}$ is dense and countable. – Enigma Oct 19 '14 at 21:59
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    @S.S: I think he means all functions from $ [0,1] $ to $\mathbb R$ – user136592 Oct 19 '14 at 22:02
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    For your second question: A product of spaces is $T_1$, Hausdorff, regular, completely regular if and only if every factor has that respective property. – Stefan Hamcke Oct 19 '14 at 22:03
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    That $\Bbb R^{[0,1]}$ is separable is a special case of the Hewitt-Marczewski-Pondiczery theorem; see this answer for a statement of the theorem and references. – Brian M. Scott Oct 20 '14 at 02:35

1 Answers1

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Assuming that you intend $\mathbb R^{[0,1]}$ to have the product topology, it is separable. More generally, the product of continuum many (or fewer) separable spaces is always separable. This (or a more general result) is the Engelking-Karlowicz theorem; see http://blog.assafrinot.com/?p=2054 and specialize the theorem as stated there by taking $\kappa=\lambda=\aleph_0$ and $\mu=2^{\aleph_0}$. (Note that, though you're interested in the product of separable spaces, it suffices to consider the product of countable spaces; i.e., it suffices to check that $\mathbb Q^{[0,1]}$ is separable because it's dense in $\mathbb R^{[0,1]}$.)

Andreas Blass
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