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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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Alex Petzke
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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.

Brian M. Scott
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  • Even if the continuum hypothesis holds, one might give $\mathbb{R}$ and $\omega_1$ different topologies, e.g., the usual topology on $\mathbb{R}$ and the order topology on $\omega_1$. I'm not sure what topologies are usually put on the function spaces though. – Trevor Wilson Sep 06 '12 at 01:53
  • @Trevor: These are just products, so there is no topology on the set in the exponent. – Brian M. Scott Sep 06 '12 at 01:55
  • I see. I guess the more interesting topologies I was trying to remember are topologies on the set of continuous functions, not on the set of all functions. – Trevor Wilson Sep 06 '12 at 02:01
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    It is possible for the sets $\mathbb{R}^\mathbb{R}$ and $\mathbb{R}^{\omega_1}$ to have the same cardinality even if CH fails, so perhaps you could say why they are not homeomorphic in that case? (If this is something that is obvious to all topologists then I apologize.) – Trevor Wilson Sep 06 '12 at 02:11
  • Thank you for the answer. For me to understand this further will take some additional reading on my own. Where would one first encounter a construction like $\mathbb{R}^{\omega_1}$? It's not in my introductory topology book, and various internet searches turned up surprisingly little. Do you know of any good (free) internet resources covering uncountable products like this? – Alex Petzke Sep 06 '12 at 20:55
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    @Alex: That’s a tough one. This sort of material is generally graduate level and is more likely to be found in papers than anywhere else. I shouldn’t be surprised if you could find some relevant material in the archives of Ask A Topologist, though you have to be willing to read mathematics written in plain ASCII. If you’re willing to spend a little, I’d consider getting Willard’s General Topology, which is a very good first-year graduate level text and reference. (Cont’d.) – Brian M. Scott Sep 06 '12 at 21:41
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    @Alex: (Cont’d.) It actually deals covers normality of $\Bbb N^{\omega_1}$ in Exercise 21C, with extensive hints. The book is available in an inexpensive edition and is well worth owning if you’re interested in general topology. (There’s even more in Ryszard Engelking’s General Topology, but it’s out of print, expensive, and hard to find.) – Brian M. Scott Sep 06 '12 at 21:42
  • 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. – Alex Petzke Sep 07 '12 at 01:28