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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 $i$?

Holdsworth88
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  • I don't understand this question at all. That's what $\Bbb{R}^\infty$ is. If we didn't insist on that condition, we'd have a different space. – Chris Eagle Feb 27 '13 at 14:33
  • My confusion is why one doesn't define $\mathbb{R}^\infty$ simply as the product of infinitely many copies of $R$. – Holdsworth88 Feb 27 '13 at 14:35
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    Because that is a different thing with a different name. You might as well ask why we don't define "France" to mean Germany. – Chris Eagle Feb 27 '13 at 14:37
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    http://mathoverflow.net/questions/73246/why-is-mathbbr-infty-defined-the-way-it-is This might be helpful – Stahl Feb 27 '13 at 14:38
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    @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. – Brian M. Scott Feb 27 '13 at 14:39
  • I certainly have seen the notation $\mathbb{R}^\infty$ being applied to the countably infinite product of $\mathbb{R}$ with itself under the product topology. There seems to be no unanymity here. – Michael Greinecker Feb 27 '13 at 14:40
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    @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. – Holdsworth88 Feb 27 '13 at 14:41
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    @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... – gt6989b Feb 27 '13 at 14:44
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    One obvious reason I can think of would be that $\mathbb{R}^{\infty}$ the way you defined it has a nice basis when regarded as a vector space, unlike the infinite product $\prod_{i=1}^{\infty}\mathbb{R}_i$. – Andy Brandi Feb 27 '13 at 14:45

2 Answers2

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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

Elden Elmanto
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  • This follows directly from the definition of the coproduct as well. –  Feb 27 '13 at 15:36
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    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.

     – Elden Elmanto Feb 27 '13 at 17:54
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    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? – Adayah Dec 09 '17 at 20:41