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.]
Best Answer
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.