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 $i$?

  • 7
    $\begingroup$ 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
  • 3
    $\begingroup$ mathoverflow.net/questions/73246/… This might be helpful $\endgroup$
    – Stahl
    Feb 27, 2013 at 14:38
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @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


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).


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

  • $\begingroup$ This follows directly from the definition of the coproduct as well. $\endgroup$
    – user314
    Feb 27, 2013 at 15:36
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.