$\ell^2$ is not locally compact

Question

$\ell^2$ 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.

Answer 1

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.

Answer 2

"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 $r/\sqrt2 > r/2.$