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Let $(X,\lVert \cdot \rVert)$ be a normed space and $X$ is not finite dimensional. Prove that $\overline{B_1}(X)=\{x\in X: \lVert x \rVert \leq 1\}$ is not compact.

Now I want to use Riesz's lemma to prove it. By selecting $x_n$ in the unit ball such that $dist(x_n,Y_{n-1})>\frac{1}{2}$ where $Y_{n-1}=span\{x_1,\dots,x_{n-1}\}$. But how to prove that each $Y_{n}$ is closed and such $x_n$ is always in the unit ball?

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    What are your thoughts on the question? What have you tried so far? Do you know about Riesz's lemma? – 2017-02-09
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    Yes, I want to use Riesz's lemma to prove this one. – 2017-02-09

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Hint: The classic proof of this fact uses Riesz's lemma.

Construct a sequence $(x_n)$ as follows: for $n = 1,2,3,\dots$, let $Y^{(n)} = \text{span}(x_1,\dots,x_n)$ (so $Y^{(0)} = \{0\}$). For each $n$, select $x_n \in X$ such that $\|x_n\| = 1$ but $\inf_{y \in Y^{(n-1)}} \|x - y\| \geq 1/2$. The existence of such a sequence proves that the unit ball fails to be compact (why?).

Each $Y^{(n)}$ is closed because every finite dimensional normed vector space is complete.

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    why choose $x_n \in X$? I know we want to construct a sequence that doesn't have a convergent subsequence. But do we need to choose $x_n$ in the unit ball? – 2017-02-09
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    "**such that** $\|x_n\| = 1$" – 2017-02-09