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This question is a follow up on my comment-question in this thread.

It appears that there was some resolution in the end, but I would still like to know more about this. An internet search turned up surprisingly little.

What, morally, is the difference between $\mathbb{R}^{\mathbb{R}}$ and $\mathbb{R}^{\omega_1}$?

Similarly, what is the difference between $\mathbb{R}^{\mathbb{N}}$ and $\mathbb{R}^{\omega}$?

Do the natural topologies on these pairs differ, and if so how?

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There is no difference at all between $\Bbb R^{\Bbb N}$ and $\Bbb R^\omega$; they are two names for the same space. (I use $\Bbb N$ to mean the set of non-negative integers. If you use it to mean $\Bbb Z^+$ instead, then $\Bbb R^{\Bbb N}$ and $\Bbb R^\omega$ are trivially homeomorphic, because the index sets $\Bbb Z^+$ and $\omega$ have the same cardinality, but they aren’t literally the same index set.)

Whether there is a topological difference between $\Bbb R^{\Bbb R}$ and $\Bbb R^{\omega_1}$ depends on your set-theoretic hypotheses. If you assume the continuum hypothesis, i.e., that $2^\omega=\omega_1$, then $\Bbb R^{\Bbb R}$ and $\Bbb R^{\omega_1}$ are homeomorphic spaces: they are both products of $\omega_1$ copies of $\Bbb R$. If you assume the negation of the continuum hypothesis, then $|\Bbb R|=2^\omega>\omega_1$, and $\Bbb R^{\Bbb R}$ and $\Bbb R^{\omega_1}$ are not homeomorphic.

Added: If $2^\omega>\omega_1$, then $\Bbb R^{\omega_1}$ has a base of cardinality $\omega_1$, so every point is the intersection of at most $\omega_1$ open sets. In $\Bbb R^{\Bbb R}$, however, no point is the intersection of $\omega_1$ open sets. The base for $\Bbb R^{\omega_1}$ is the natural product base built from a countable base in each factor. If a point of $\Bbb R^{\Bbb R}$ were the intersection of $\omega_1$ open sets, it would be the intersection of $\omega_1$ basic product open sets. Those open sets could restrict only $\omega_1$ coordinates, leaving $2^\omega$ coordinates free to vary, so they could not pin down a single point.

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    I looked through some of Willard's book in the preview and I like the way it's written. I think I'll go ahead and purchase it. I'd have a hard time turning down any nice mathematics book at that price! Thanks again.2012-09-07