Countable Dense Subset – Proving the Space Has One

general-topologyseparable-spaces

I'm doing this exercise in Munkres book and got no clue about the solution. Hope some one can help me solve this.

Show that the product space $R^{I}$, where $I = [0,1]$, has a countable dense subset. If $J$ has cardinality greater than $\mathscr{P}(Z_{+})$, then the product space $R^{J}$ does not have a countable dense subset

With this problem, I even can't imagine the space $R^{I}$. So if some one can talk something about this space, I really appreciate it (as I know, this space is rather important, right?)

Thanks so much

Best Answer

Note that the structure of $I$ or $J$ is irrelevant for topology of $\mathbb{R}^I$ or $\mathbb{R}^J$. Only the cardinality matters. So we want to show that product of continuum copies of $\mathbb{R}$ is separable but product of more than continuum copies of $\mathbb{R}$ is not separable. Actually the following holds: Product of $≤ 2^κ$ spaces of density $≤ κ$ has density $≤ κ$ (for $κ$ infinite cardinal). This is called Hewitt–Marczewski—Pondiczery theorem (e. g. Engelking 2.3.15). On the other hand if all spaces forming a product contain two disjoint open subsets (e. g. at least 2 point Hausdorff spaces) then there is injection from index set to powerset of a dense set so product of more than $2^κ$ such spaces cannot have density $≤ κ$. For proofs and more details see http://dantopology.wordpress.com/2009/11/06/product-of-separable-spaces/.

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