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$\ell^2$ is not locally compact.

How to prove this, I just know the definition of locally compact. I am finding it hard to find any trick. $\ell^2$ is the sequence of all square summable sequences. Any help would be appreciated. Thanks in advance.

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"Locally compact" means every point has an open neighborhood whose closure is compact. What does an open neighborhood of $\mathbf 0 =(0,0,0,\ldots)\in\ell^2$ look like? This is a metric space, so every open neighborhood $G$ of $\mathbf 0$ must include some open ball centered at $\mathbf 0.$ Let $r$ be the radius of such an open ball. Then the points $ x_k = \Big( \quad \ldots0,0,0,\underbrace{\quad\dfrac r 2, \quad}_{\large k\text{th component}} 0,0,0, \ldots \quad \Big)$ for $k=1,2,3,\ldots$ are all within the open neighborhood $G$. If the closure of $G$ is compact, then so is every closed subset of the closure of $G$, including $\{x_1,x_2,x_3,\ldots\}.$

Cover the set $\{x_1,x_2,x_3,\ldots\}$ with open balls of radius $r/2,$ one centered at each of $x_1,x_2,x_3,\ldots.$ There can be no finite subcover, because each of the open balls of radius $r/2$ about the points $x_k$ can cover only one of $x_1,x_2,x_3,\ldots$, since the distance between any two of $x_1,x_2,x_3,\ldots$ is $r/\sqrt2 > r/2.$

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you need to show that closed balls are not compact. For simplicity you can show Unit closed ball is not compact. Take the sequence $\{e_n\}$ then $ \|e_n - e_m\|^2 = 2$ for all distinct natural numbers $m$ and $n$. Therefore $\{ e_n \}$ does not contain any convergent subsequence.

Note that your claim is even true for any normed space.

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  • $\begingroup$ Well, not just any normed space. $\Bbb R^2$ happens to be locally compact $\endgroup$ – Hagen von Eitzen May 20 '17 at 20:34
  • $\begingroup$ You never gave a definition of $e_n$. $\endgroup$ – Michael Hardy May 20 '17 at 20:36
  • $\begingroup$ Oh Yes I was dummy!! actually any infinite dimensional normed space! I have tendency to think about infinite Dimensions when talking about Normed spaces .... $\endgroup$ – Red shoes May 20 '17 at 20:38
  • $\begingroup$ @MichaelHardy Since the vector $e_n$ is a standard notation in Hilbert spaces, readers will understand what I mean by that . $\endgroup$ – Red shoes May 20 '17 at 20:45
  • $\begingroup$ @nonlinearthought : Perhaps, but how did you conclude that $\|e_n-e_m\|=1\text{?}$ If $\|e_n\| = \|e_m\| = 1$ and $\langle e_n,e_m\rangle =0$ then you have $\|e_n-e_m\| = \sqrt 2. \qquad$ $\endgroup$ – Michael Hardy May 20 '17 at 20:48

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