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Let $V$ be any infinite dimensional normed space. Is it always possible to find $\{x_1, x_2, \dots\}$ be a countbly infinite subset of the closed unit ball, such that $||x_i-x_j||>1$ whenever $i\neq j$?

So far I have tried a few sequence spaces and $C[0,1]$ under various norms and I can find such a subset. I have not proved it or found an counterexample.

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If you only demand that $\Vert x_i-x_j\Vert>\epsilon$, for a fixed $0<\epsilon<1$, then this is a consequence of Riesz's lemma. Cliff Kottman has shown that the answer to your question is "yes". This can be found in the notes to chapter one of Diestel's Sequences and Series in Banach Spaces. See, also, the comments to this post. –  David Mitra Feb 6 '13 at 15:11
    
This was also shared in David Mitra's comments at another question (but that one was not a duplicate of this one because it only asked for $\geq 1$). @David: Would you mind posting an answer to one or both of these questions? –  Jonas Meyer Feb 6 '13 at 15:13
    
@DavidMitra I am reading Kottman's paper but I think there seems to be a bug in there. –  Montez Feb 6 '13 at 15:17
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up vote 3 down vote accepted

If you only demand that $\Vert x_i-x_j\Vert>\epsilon$, for a fixed $0<\epsilon<1$, then this is a consequence of Riesz's lemma.

Cliff Kottman has shown that the answer to your question is "yes". This result can be found in the notes to Chapter one of Diestel's Sequences and Series in Banach Spaces.

For completeness, here is the argument given in Diestel (which is attributed to Bob Huff and Tom Starbird):


Theorem:. If $X$ is an infinite dimensional normed linear space, then there is a sequence $(x_n)$ of norm-one elements of $X$ for which $\Vert x_n-x_m\Vert>1$ whenever $n\ne m$.


Proof: We define the required sequence $(x_n)$ by induction.

To start, choose $x_1\in X$ of norm one and $x^*\in X^*$ such that $\Vert x_1^*\Vert= x_1^* x_1=1$.

Now suppose appropriate linearly independent norm-one vectors $x_1^*,\ldots x_k^*$ from $X^*$ and norm-one elements $x_1,\ldots,x_k$ of $X$ have been chosen.

Choose $y\in X$ so that $$\tag{1}x_i^* y<0\text{ for each }i=1,\ldots k,$$ and let $x$ be a non-zero vector such that $$\tag{2}x\in\bigcap\limits_{i=1}^k \text{ker}\ x_i^*.$$ Choose $K$ so that $$\tag{3}\Vert y\Vert <\Vert y+Kx\Vert.$$

Using $(2)$, one easily verifies that for a non-trivial linear combination $\sum\limits_{i=1}^k\alpha_i x_i^*$ of the $x_i^*$, we have: $$ \biggl|\sum_{i=1}^k \alpha_i x_i^*(y+Kx)\biggr|\le \biggl\Vert\sum_{i=1}^k \alpha_i x_i^* \biggr\Vert \Vert y\Vert; $$ whence by $(3)$ $$\tag{4} \biggl|\sum_{i=1}^k \alpha_i x_i^*(y+Kx)\biggr|<\Vert y+Kx\Vert. $$

Now define $$ x_{k+1} ={y+Kx\over\Vert y+Kx\Vert}. $$ Using Hahn-Banach, choose $x_{k+1}^*$ to be a norm one element of $X^*$ satisfying $$ x^*_{k+1} x_{k+1}=1.$$

Then by $(4)$ it follows that $x_{k+1}^*$ is not a linear combination of $x_1^*,\ldots, x_k^*$ (if $x_{k+1}^*=\sum_{i=i}^k\alpha_i x_i^*$, evaluate both sides at $x_{k+1}$).

Finally, we have, for $1\le i\le k$ $$ \Vert x_{k+1}-x_i\Vert\ge |x_i^*( x_{k+1}-x_i)|=|x_i^* x_{k+1} - x_i^* x_i|. $$ Now, $x_i^* x_i=1$. Also, $x_i^* x_{k+1}<0$ by the definition of $x_{k+1}$, $(1)$ and $(2)$. Thus $\Vert x_{k+1}-x_i\Vert >1$, as desired.



See, also, the comments to this post. A result due to John Elton and Edward Odell shows that one can actually find an $\epsilon>0$ (which depends on the space) and a sequence of norm one elements $(x_n)$ such that $\Vert x_n-x_m\Vert\ge1+\epsilon$ whenever $n\ne m$!

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I must confess, I do not see how to find the $y$ in $(1)$. I'm probably overlooking something simple here... –  David Mitra Feb 6 '13 at 15:26
    
Since the $x_{i}^\ast$ are linearly independent, the map $X \to \mathbb{R}^k, x \mapsto (x_1^\ast(x),\dots, x_k^\ast(x))$ is onto, otherwise you have a non-trivial linear relation between them. –  Martin Feb 6 '13 at 15:43
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