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I am reading this proof about compact closed unit ball and finite dimentional space

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I am confused about that last paragraph because I am not sure what would change in the proof if $\dim X<\infty$. Would we still have $||x_m-x_n||\geq \frac{1}{2}$?

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If the dimension of $X$ is finite, you cannot construct an infinite number of $x_n$, at a point $Vect(x_1,...,x_n)=X$.

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The infinite dimension assumption says that whenever we have finitely many $v_1,\ldots,v_n$, the subspace they generate, $X_n$, cannot contain all of $X$. There must be points outside of it, which allows us to apply Riesz, as the subspace is proper. So the recursive construction of the sequence can never halt, we keep on finding new points on $M$ with the norm conditions.

But if it never halts, we have an infinite sequence without a convergent subsequence, allowing the final contradiction based on this assumption.

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  • $\begingroup$ This suggests the Q: In finite dimensions: How many points, each two them at a distance of at least $d\in (0,2]$ can you put on the unit sphere? In $\mathbb R^2$ with the Euclidean metric, and $d=1/2,$ we can put $12 .$ With the $\sup$ norm and $d=1/2$ we can put $16.$ What is the upper bound, for all norms,for a given dimenion, and a given $ d$? $\endgroup$ – DanielWainfleet Dec 28 '16 at 1:35

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