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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.]

Henno Brandsma
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SihOASHoihd
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1 Answers1

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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.

Eric Wofsey
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    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 ${x_{0}}\times [-N,N]$? – SihOASHoihd Dec 25 '20 at 01:22
  • 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 $\epsilon_2>0$. – Eric Wofsey Dec 25 '20 at 01:41
  • @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? – Severin Schraven Dec 25 '20 at 01:53
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    @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 $\mathbb{R}^2$. – Eric Wofsey Dec 25 '20 at 01:56
  • Ahh, sorry, I was thinking about projections the whole time. – Severin Schraven Dec 25 '20 at 02:00
  • @EricWofsey I follow you now – SihOASHoihd Dec 25 '20 at 02:09