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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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2 Answers 2

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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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  • $\begingroup$ What argument could I use to say $\{x_1,x_2,x_3,\ldots\}$ is closed? $\endgroup$
    – edamondo
    Mar 4, 2021 at 17:26
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    $\begingroup$ @edamondo : It is closed if it contains all of its limit points, and it does, since it has no limit points. Or you can say it is closed because its complement is open. Just let $y$ be some point not in that set and think about finding an open ball centered at $y$ that does not contain any of the points in this set in question. $\endgroup$ May 8, 2021 at 17:15
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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 $\endgroup$ May 20, 2017 at 20:34
  • $\begingroup$ You never gave a definition of $e_n$. $\endgroup$ May 20, 2017 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, 2017 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, 2017 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 $\|e_n-e_m\| = \sqrt 2. \qquad$ $\endgroup$ May 20, 2017 at 20:48

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