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Question: Why is the topological space $\mathbb{R}^\infty$ defined to be the subset of $\prod_{i=1}^\infty \mathbb{R}_i$ consisting of sequences $(a_i)_{i=1} ^{\infty}$ such at most finitely many $a_i\neq 0$? Why does one insist on the condition that $a_i\neq0$ for at most finitely many $i$?

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    $\begingroup$ Because that is a different thing with a different name. You might as well ask why we don't define "France" to mean Germany. $\endgroup$ Feb 27, 2013 at 14:37
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    $\begingroup$ mathoverflow.net/questions/73246/… This might be helpful $\endgroup$
    – Stahl
    Feb 27, 2013 at 14:38
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    $\begingroup$ @Chris: If you asked me what $\Bbb R^\infty$ was as a topological space, I’d understand it to be the product of $\omega$ copies of $\Bbb R$, not the $\sigma$-product. It’s in a more functional analytic context that it becomes something other than the topological product. $\endgroup$ Feb 27, 2013 at 14:39
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    $\begingroup$ @chriseagle I think I understand why France and Germany are different. The problem is one of notation $\prod_{i=1} ^n \mathbb{R}=\mathbb{R}^n$ but not for $\infty$. Perhaps I am too hung up on notation. $\endgroup$ Feb 27, 2013 at 14:41
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    $\begingroup$ @Holdsworth88 Note that your notation $\prod_{i=1}^n \mathbb{R} = \mathbb{R}^n$ implies that $n < \infty$ so there is only a finite number of $a_i \neq 0$. In this sense the standard definition of $\mathbb{R}^\infty$ is a direct extension... $\endgroup$
    – gt6989b
    Feb 27, 2013 at 14:44

2 Answers 2

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This condition makes $\mathbb{R}^\infty$ a CW-complex. This basically means it is a "good" topological space.

It also makes $\mathbb{R}^\infty$ the coproduct in the category of topological spaces (i.e. direct sum) as compared to the product (Cartesian product) $\prod_{n\in\mathbb{N}} \mathbb{R}^n$. Compare with the difference between the coproduct (direct sum) of infinitely many abelian groups, for example, and the product (direct product).

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Another more elementary reason is the following theorem:

Let $f: A \rightarrow \prod_{\alpha \in J} X_{\alpha}$ be given coordinate-wise, i.e. $f(a) = (f_{\alpha}(a))_{\alpha \in J}$ where $f_{\alpha}:A \rightarrow X_{\alpha}$ with the product topology (i.e. the finite support condition you described) we have that $f$ is continuous if and only if $f_{\alpha}$ is.

This fails if we do not insist the finite support condition and the simplest counterexample $f: \mathbb{R} \rightarrow \prod_i \mathbb{R}_i$ given by $f(t) = (t, t, ..., )$ works

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  • $\begingroup$ This follows directly from the definition of the coproduct as well. $\endgroup$
    – user314
    Feb 27, 2013 at 15:36
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    $\begingroup$ coproduct in the category of spaces is the internal disjoint union actually :) But you right in that this is exactly the universal property for the product in Top. $\endgroup$ Feb 27, 2013 at 17:54
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    $\begingroup$ Could you explain why $\displaystyle f : \mathbb{R} \to \prod_{i=1}^{\infty} \mathbb{R}$ given by $f(t) = (t, t, \ldots)$ is not continuous? $\endgroup$
    – Adayah
    Dec 9, 2017 at 20:41

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