How will we prove that the closed unit ball in $\ell^2$ is closed, bounded, convex but not compact
Closed Unit ball in $\ell^2$ is not compact?
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real-analysis
functional-analysis
compactness
lp-spaces
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2[Proving that the unit ball in $\ell^2(\mathbb{N})$ is non-compact](http://math.stackexchange.com/q/115344) – 2012-11-26
2 Answers
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In addition to the above, convexity is straightforward. Pick $x,y$ in the closed unit ball and $0 < t <1$. We need to show that $\|tx + (1-t)y\| \le 1$. This follows from the observation:
$\|tx + (1-t)y\| \le \|tx\| + \|(1-t)y\| = t\|x\| + (1-t)\|y\| \le \max(\|x\|,\|y\|) \le 1 $
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Denote by $e_n \in \ell^2$ the sequence $(0, \ldots, 0, 1, 0, \ldots)$ with the $1$ at position $n$. Then $\|e_n - e_m\|_2 = \sqrt 2, \qquad n \ne m $ Hence $(e_n)$ cannot have any subsequence which is Cauchy. (The closed unit ball is moreover bounded by 1 and closed).