# Open sets and noncompactness definition

Let $$\mathbb{R}^{\infty}$$ be the set of all "infinite-tuples" $$x=(x_1,x_2,\ldots)$$ of real numbers that end in an infinite string of $$0$$'s. Define an inner product on $$\mathbb{R}^{\infty}$$ by the rule $$(x,y)=\sum x_iy_i$$. Let $$\|x-y\|$$ be the corresponding metric on $$\mathbb{R}^{\infty}$$. Define $$e_i=(0,\ldots,0,1,0,\ldots,0,\ldots)$$, where $$1$$ appears in the $$i$$th place. Let $$X$$ be the set of all the points $$e_i$$. Show that $$X$$ is non-compact.

I do not understand the following answer:

To see that $$X$$ is not compact, define $$A_i$$ to be the open $$1$$-ball around $$e_i$$ in $$U$$ for all $$i$$. This gives, after intersecting with $$X$$, an open cover of $$X$$ (clearly each $$A_i$$ is open in $$U$$, so its intersection with $$X$$ is open in $$X$$, and clearly it's a cover, as each $$e_i$$ lies in $$A_i$$. But $$\{A_i\}$$ has no finite subcover - indeed, by what we did to show that $$X$$ is closed, we see that the $$A_i$$ are disjoint, and so $$\{A_i\}$$ has no proper subcovers at all, so certainly has no finite ones. Thus, $$X$$ is not compact.

I have two confusions:

1) what is the open cover? is it {Ai}?

2) why does the subcover not exist and have to do with being disjoint? does 1 ball mean a singleton? if so, why does it have to be proper, it can just be a subcover according to the definition.

The open cover of $$X$$ in the subspace topology is formed by the sets $$A_i\cap X$$.
The fact that the $$A_i$$ are pairwise disjoint in particular implies that $$e_i\notin A_j$$ for $$j\ne i$$ (otherwise $$e_i\in A_i\cap A_j$$, in contradiction to those two sets being disjoint). In other words, $$A_i\cap X = \{e_i\}$$ (which, in turn is an open set in $$X$$)
Thus written more explicitly, the open cover is $$\{\{e_i\}\mid i=1,2,3,\ldots\}$$
In particular, this is an infinite set, therefore any finite subcover is a proper subset. But since each element of that cover covers only a single point of $$X$$, a proper subset cannot cover $$X$$ as at least one point would be missing.
The open cover is formed by the open balls $$B(e_i,1)$$. Check that $$\|e_i-e_j\|=\sqrt 2$$ for $$i \neq j$$. Hence the only point of $$X$$ inside the ball $$B(e_i,1)$$ is $$e_i$$ itself. This is open in $$X$$ and there is no finite subcover because any finite subcover can include $$e_i$$'s for only a finite number of $$i$$'s.