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Im taking a course in functional analysis and im trying to prove that in infinite dimensions there is no compact unit ball. I've read some results following the Riesz Lemma but i seem to not quiet understand. Can someone show another approach to the problem or try explain me the Riesz Lemma approach? i would really appreciate it! Thank you for your atention.

riemannfanboy
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    Take a vector $x_1\in X$ of norm $1$. Denote with $X_1$ the span of $x$. Let $x_2'\in X$ be any vector that is not in the span of $X_1$, denote with $X_2$ the span of $x_2'$ and $X_1$. Now find a norm $1$ vector $x_2\in X_2$ so that $\sup_{y\in X_1}|x_2-y|≥\frac12$. The existence of such a vector is given by the Riesz Lemma, but it is a bit overkill since this is a finite dimensional problem. Continue in this way to get a sequence $x_n$ of norm $1$ with mutual distances $≥\frac12$, so this sequence cannot admit a convergent subsequence. – s.harp Sep 11 '19 at 14:47

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Here's another approach.

How far apart are $(1,0,0,\dots)$ and $(0,1,0,\dots)$, two points in (on the boundary of) the closed unit ball? If we put an open ball of that radius on each of $(\pm 1, 0, 0, \dots)$, $(0, \pm 1, 0, \dots)$, $\dots$, do we cover the closed unit ball with open sets? Can I find a finite subcover among these balls? (Hint: how many of these open balls cover each center of an open ball?)

A related method: Consider the sequence $(1,0,0, \dots)$, $(0,1,0,\dots)$, $(0,0,1,0,\dots)$, $\dots$ and then treat it as a set of points. Does this set contain an infinite number of points of the closed unit ball? Does this set have a limit point? (An infinite subset of a compact set must contain a limit point.)

Eric Towers
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I will give you one example, where the unit ball (closed) need not be compact, when we consider an infinite dimensional space.

Consider $\ell^{2}$, the space of real sequences $\left( x_i \right)$ such that $\sum\limits_{i = 1}^{\infty} \left| x_i \right|^2$ is convergent. The norm of any vector in $\ell^2$ is given as

$$\left| \left| \left( x_i \right) \right| \right| = \left( \sum\limits_{i = 1}^{n} \left| x_i \right|^2 \right)^{\frac{1}{2}}$$

Now, consider the closed unit ball

$$B \left( 0, 1 \right) = \left\lbrace \left( x_i \right) \in \ell^2 | \left| \left| \left( x_i \right) \right| \right| \leq 1 \right\rbrace$$

If this set was to be compact, then every sequence would have a convergent subsequence. We will try to construct one sequence which will not have any convergent subsequence so that our purpose will be satisfied.

Therefore, we consider the usual (standard) Schauder basis for $\ell^2$ as our sequence $\left\lbrace \left( x_i^{\left( n \right)} \right) \right\rbrace$, where $x_i^{\left( n \right)} = \begin{cases} 1, & i = n \\ 0, & otherwise \end{cases}$. Notice that we use superscript to denote the "running index" of the sequence.

Clearly, $\left| \left| \left( x_i^{\left( n \right)} \right) \right| \right| = 1$ for each $n \in \mathbb{N}$ so that $\left( x_i^{\left( n \right)} \right) \in B \left( 0, 1 \right)$.

However, if we take two distinct elements of this sequence, say $\left( x_i^{\left( m \right)} \right)$ and $\left( x_i^{\left( n \right)} \right)$, where $m \neq n$, then

$$\left| \left| \left( x_i^{\left( m \right)} \right) - \left( x_i^{\left( n \right)} \right) \right| \right| = \sqrt{2}$$

so that this sequence has no convergent subsequence. This does our job!

Aniruddha Deshmukh
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