# Is the weak topology on $\mathbb{R}^{\infty}$ the same as the box topology?

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.]

• 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. Dec 24, 2020 at 23:32
• @Steve Kass Homework...on Christmas Eve... Dec 24, 2020 at 23:33
• @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. Dec 24, 2020 at 23:39

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.

• 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 $\{x_{0}\}\times [-N,N]$? Dec 25, 2020 at 1:22
• 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 $\epsilon_2>0$. Dec 25, 2020 at 1:41
• @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? Dec 25, 2020 at 1:53
• @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 $\mathbb{R}^2$. Dec 25, 2020 at 1:56
• Ahh, sorry, I was thinking about projections the whole time. Dec 25, 2020 at 2:00