TITLE: A compact subset of $\mathbb{R}^\infty$ must be contained in some $\mathbb{R}^n$
QUESTION [5 upvotes]: I'd like to prove that a compact subset of $\mathbb{R}^\infty$ must be contained in some finite dimensional space $\mathbb{R}^n$. Here, $\mathbb{R}^\infty$ is the set of all eventually zero sequences of reals, and the topology on $\mathbb{R}^\infty$ is defined as: $U\subset\mathbb{R}^\infty$ is open iff $U \cap \mathbb{R}^n$ is open in each $\mathbb{R}^n$. I see that here we can show every convergent sequence in $\mathbb{R}^\infty$ (and its limit) must lie in some $\mathbb{R}^n$.
So I guess to prove the statement, we can suppose for a contradiction that a compact set $K$ is not contained in any $\mathbb{R}^n$. Then there is a sequence in $K$ that does not lie in any $\mathbb{R}^n$. So by the above, this sequence does not converge. Does this reach a contradiction? (Does compactness imply sequential compactness in $\mathbb{R}^\infty$?)

REPLY [3 votes]: Given $x \in \mathbb{R}^{\infty}$, let $m(x)$ be the smallest $m$ for which $x \in \mathbb{R}^m$, i.e., $m(x)=\min\{i \mid \pi_i(x)=0\}$. Let's call such $m(x)$ the dimension where $x$ lives. We then have the following intermediate lemma:

Lemma: If $A\subset \mathbb{R}^{\infty}$ has no two points living in the same dimension, then $A$ is closed and discrete.

Proof: The hypothesis tells us that $A \cap \mathbb{R}^n$ is finite for every $n$, hence closed. Since this holds for every $n$, we have that $A$ is closed. 
Note now that if $A$ has no two points living in the same dimension, then the same is true for $A \backslash \{x\}$ for any $x \in A$, and therefore $A \backslash \{x\}$ is also closed. Since $A$ is already closed, it follows that $A \backslash \{x\}$ is closed with respect to $A$, and thus $\{x\}$ is clopen in $A$. Thus, $A$ is discrete. $\blacksquare$
Now, take a compact $K$. If $K$ is not contained in some $\mathbb{R}^n$, then there are infinitely many $x_i \in K$ such that all $x_i$ live in different dimensions. The set $A:=\{x_i\}$ then satisfies the hypothesis of the previous lemma, which tells us that it must be closed and discrete. Since it is a closed set inside a compact one, it is itself compact. Being also discrete, it is finite, a contradiction.

The above is an adaptation of the proof that if $X$ is a CW-complex then every compact set is contained in a finite union of open cells.