Is there a simple characterization of compact subspaces of the "unit sphere" in $\mathbb{R}^\omega$? Let $\mathbb{R}^\omega$ be the countably infinite product of $\mathbb{R}$ with itself in the product topology. $\mathbb{R}^\omega$ is metrizable but the metric doesn't arise from a norm. A natural analogue to a unit sphere is the quotient of $\mathbb{R}^\omega - \{0\}$ under the action of $(0, \infty)$ (that is, we say two elements of $\mathbb{R}^\omega$ are equivalent if and only if they are positive multiples of one another).
Denote the quotient space by $S$. In this answer we see that $S$ is not metrizable, but I'm trying to figure out if compactness can still be characterized nicely in $S$. I've started by characterizing compactness in $\mathbb{R}^\omega$: a subspace $K$ is compact if and only if it is closed and contained in some set of form $\Pi_{i=1}^\infty [a_i, b_i]$ where for all $i$ $a_i$ and $b_i$ are real numbers such that $a_i \leq b_i$. This in turn gives us a lot of sets in $S$ that are compact, but I'm looking for necessary and sufficient conditions about as simple as those found for $\mathbb{R}^\omega$. For example, "$K \subseteq S$ is compact if and only if it is the image under the quotient map of a compact set in $\mathbb{R}^\omega$" would be very nice.
Any answers, advice, or comments are appreciated!
 A: Let $U_n\subseteq S$ be the image of the set of sequences such that at least one of the first $n+1$ coordinates is nonzero.  The $U_n$ form a nested sequence of open sets that cover $S$, so any compact subset of $S$ is contained in some $U_n$.  Thus, it suffices to understand compact subsets of $U_n$.
Now I claim that in fact $U_n\cong S^n\times\mathbb{R}^\omega$.  Indeed, considering $S^n\times\mathbb{R}^\omega$ as the subset of $\mathbb{R}^\omega$ consisting of sequences whose first $n+1$ coordinates form a point of $S^n$, the quotient map $q:\mathbb{R}^\omega\setminus\{0\}$ restricts to a bijection $f:S^n\times\mathbb{R}^\omega\to U_n$ (a sequence in $q^{-1}(U_n)$ can be uniquely scaled so that its first $n+1$ coordinates form a point of $S^n$).  To see that $f^{-1}$ is continuous, note that $q$ restricts to a quotient map $q^{-1}(U_n)\to U_n$ and the composition of this quotient map with $f^{-1}$ is the map $q^{-1}(U_n)\to S^n\times\mathbb{R}^\omega$ that takes a sequence $(x_i)\in q^{-1}(U_n)$ and divides each coordinate by $\sqrt{x_0^2+\dots+x_n^2}$, which is clearly continuous.
Since $S^n$ is compact, we see that a subset of $U_n\cong S^n\times\mathbb{R}^\omega$ is compact iff it is closed and it is bounded on each coordinate of the $\mathbb{R}^\omega$ factor.  Or, in terms of $S$ as a quotient of $\mathbb{R}^\omega\setminus\{0\}$, this means a closed set $K\subseteq S$ is compact iff there exists $n$ such that every point of $K$ has a nonzero coordinate within the first $n$ coordinates, and if you normalize the sequences representing elements of $K$ such that the first $n$ coordinates form a point of the unit sphere, each coordinate after the first $n$ coordinates is bounded.  In particular, this shows that your hope is correct: every compact subset $K$ of $S$ is the image of a compact subset of $\mathbb{R}^\omega$ under the quotient map (namely, the compact set you get by picking representatives of elements of $K$ that are normalized so that the first $n$ coordinates form a point of the unit sphere).
