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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.

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2 Answers 2

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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.

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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.

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