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Show that in the infinite dimensional normed space $X$, for every $\alpha > 0$ exist $x_1, x_2,\ldots$ such that $||x_n|| \leq 1$ and $||x_n-x_m|| > 1 - \alpha$.

I have no idea how to solve this one.

Edit: I'm not sure. Let $D = \{x \in X : ||x|| \leq 1 \}$ and let $x_1 \in D$. Obviously, $span\{x_1\} \neq X$, since $span\{x_1\}$ is a finite dimensional space and $X$ is not. Also, it is closed. So, we can use the Riesz's lemma, i.e., there exists $x_2 \in D$ such that $||x_2-x_1|| \geq 1 - \alpha$. Next, $span\{x_1,x_2\} \neq X$ and we repeat the same strategy again and again. Am I right?

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    $\begingroup$ Use Riesz's Lemma. $\endgroup$ – David Mitra Jan 4 '15 at 13:19
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    $\begingroup$ en.wikipedia.org/wiki/Riesz%27s_lemma and use the fact that every finite dimensional subspace is closed $\endgroup$ – Prahlad Vaidyanathan Jan 4 '15 at 13:19
  • $\begingroup$ For fun, prove you can do it with $\alpha=0$. $\endgroup$ – David Mitra Jan 4 '15 at 13:21
  • $\begingroup$ Thanks. Still I'm not sure what to do. See the edited post. $\endgroup$ – stolikp Jan 4 '15 at 15:24
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    $\begingroup$ It looks fine. But you should say exactly what "the strategy" is: your new $x_n\in D$ satisfies $\Vert x_n-y\Vert$ for all $y$ in $\text{span}\,\{x_1,\ldots ,x_{n-1}\}$. $\endgroup$ – David Mitra Jan 4 '15 at 22:43

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