Let $\mathbb{R}^{\infty}=\bigcup\limits_{n=1}^{\infty}\mathbb{R}^{n}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all sequences which are nonzero for only finitely many terms. Give $\mathbb{R}^{\infty}$ the weak topology, that is, $A\subset \mathbb{R}^{\infty}$ is open iff $A\cap \mathbb{R}^{n}$ is open in $\mathbb{R}^{n}$ for each $n$. Is $\mathbb{R}^{\infty}$ a subspace of $\mathbb{R}^{\omega}$ with the box topology?

The reason why I think this is true is because it looks like the box topology is the weak topology with respect to the natural inclusions $\mathbb{R}^n\to\mathbb{R}^{\omega}$ [Edit: I have doubts about this particular claim.]

  • $\begingroup$ What have you tried? Have you attempted anything using the definitions, for example? This sounds a lot like you're asking for an answer to a homework or exam problem without showing any effort. $\endgroup$
    – Steve Kass
    Dec 24, 2020 at 23:32
  • $\begingroup$ @Steve Kass Homework...on Christmas Eve... $\endgroup$ Dec 24, 2020 at 23:33
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    $\begingroup$ @SihOASHoihd: Well, it’s not absolutely impossible: my linear algebra final my first semester in college back in 1965 was a take-home over Christmas break. :-) But I’ve no problem with the question: you did in fact offer some evidence of having thought about it. $\endgroup$ Dec 24, 2020 at 23:39

1 Answer 1


Yes, the weak topology and the box topology on $\mathbb{R}^\infty$ are the same. One direction is easy: it is clear that every basic open subset in the box topology has open intersection with each $\mathbb{R}^n$ and so is open in the weak topology.

The converse requires more work. Let me first remark that both topologies are translation-invariant: this is obvious for the box topology, and for the weak topology, it follows from the fact that any translate of $\mathbb{R}^n$ by an element of $\mathbb{R}^\infty$ is contained in $\mathbb{R}^N$ for some $N$. Now suppose $U\subseteq\mathbb{R}^\infty$ is an open neighborhood of a point $p$ in the weak topology; we may translate to assume $p$ is the origin. Then $U\cap\mathbb{R}$ contains some interval $[-\epsilon_1,\epsilon_1]$. Since $[-\epsilon_1,\epsilon_1]$ is compact, openness of $U\cap\mathbb{R}^2$ implies it actually contains a box $[-\epsilon_1,\epsilon_1]\times[-\epsilon_2,\epsilon_2]$ by the tube lemma. Continuing this process, we get a sequence $(\epsilon_n)$ of positive numbers such that $U\cap\mathbb{R}^n$ contains $\prod_{i=1}^n[-\epsilon_i,\epsilon_i]$ for each $n$. This implies that $U$ contains $\prod_{i=1}^\infty(-\epsilon_i,\epsilon_i)\cap\mathbb{R}^\infty$ and so is a neighborhood of the origin in the box topology as well.

(Alternatively, instead of using a translation to assume $p$ is the origin, we could have started by picking $n$ such that $p\in\mathbb{R}^n$ and taking a compact neighborhood $K$ of $p$ in $\mathbb{R}^n$ contained in $U\cap\mathbb{R}^n$, and then found a set of the form $K\times\prod[-\epsilon_i,\epsilon_i]$ contained in $U$.)

More generally, a similar argument shows that given a sequence of locally compact pointed spaces $(X_n)$, the box topology on the "direct sum" $\bigoplus X_n$ (i.e. the subset of the product $\prod X_n$ consisting of points which are the basepoint on all but finitely many coordinates) is the same as the colimit topology considering $\bigoplus X_n$ as the colimit of the finite products.

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    $\begingroup$ To apply the tube lemma the way you do, don't you need to know that $U\cap ([-\epsilon_{1},\epsilon_{1}] \times \mathbb{R})$ contains a slice of the form $\{x_{0}\}\times [-N,N]$? $\endgroup$ Dec 25, 2020 at 1:22
  • $\begingroup$ By our choice of $\epsilon_1$, we know that $U\cap\mathbb{R}^2$ contains $[-\epsilon_1,\epsilon_1]\times\{0\}$. So, it must contain $[-\epsilon_1,\epsilon_1]\times[\epsilon_2,-\epsilon_2]$ for some $\epsilon_2>0$. $\endgroup$ Dec 25, 2020 at 1:41
  • $\begingroup$ @EricWofsey I do not quite understand how $[-\varepsilon_1, \varepsilon_1]\subseteq U\cap \mathbb{R}$ implies that $[-\varepsilon_1, \varepsilon_1]\times \{0\} \subseteq U\cap \mathbb{R}^2.$ Wouldn't you need to make $\varepsilon_1$ potentially smaller? $\endgroup$ Dec 25, 2020 at 1:53
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    $\begingroup$ @SeverinSchraven: That is true by definition (though we are abusing notation here so your misunderstanding is understandable). Here $\mathbb{R}^n$ really refers to the subset $\mathbb{R}^n\times\{(0,0,0,\dots)\}$ of $\mathbb{R}^\infty$. So $\mathbb{R}$ is identified with the subset $\mathbb{R}\times\{0\}$ of $\mathbb{R}^2$. $\endgroup$ Dec 25, 2020 at 1:56
  • $\begingroup$ Ahh, sorry, I was thinking about projections the whole time. $\endgroup$ Dec 25, 2020 at 2:00

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