Examples of compact sets that are infinite dimensional and not bounded

1. A compact set in a metric space must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ and a fixed point $x_0$ such that $d(x_n,x_0) \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence.

2. A compact set in a metric space (also in a Hausdorff space) must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will have no convergent subsequence (within the set, at least).

3. A closed and bounded set in a metric space need not be compact. In an infinite dimensional Banach space, closed balls are not compact. For example, in $\ell^p$, $\{ e_1,e_2,\dots,e_n,\dots \}$ is a sequence in the closed unit ball which has no convergent subsequence. In an infinite dimensional Hilbert space we can essentially copy this example by finding an orthonormal sequence. The result holds in any infinite dimensional Banach space, but the details of the construction are not quite so trivial.

4. A compact subset of an infinite dimensional Banach space can be infinite dimensional, in the sense that it is not contained in any finite dimensional subspace. One way to generate infinite dimensional compact sets is to ensure that any sequence of linearly independent vectors converges to zero. So for example, if $X=\ell^p$, consider the set $\{ 0,e_1,e_2/2,\dots, e_n/n,\dots \}$. This is clearly infinite dimensional, but it is also compact, since any sequence in this set either converges to $0$ or has a constant subsequence.

5. In general, a subset $A$ of a metric space $X$ is compact if and only if it is complete and totally bounded. A subset of a metric space is totally bounded if for any $\varepsilon > 0$ there exists $N \in \mathbb{N}$ and points $x_1,\dots,x_N$ such that $A \subset \bigcup_{n=1}^N B(x_n,\varepsilon)$. In a Banach space $X$, this is equivalent to saying that it is bounded and, for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ and a subspace $V$ of $X$ whose dimension is $N$ such that $(\forall x \in A) \, d(x,V) < \varepsilon$.

The following result is available:

Let $\left(X,\lVert\cdot\rVert\right)$ be a Banach space and $S$ a subset of $X$. Then $S$ is compact if and only if the following two conditions hold:

• $S$ is closed and bounded;
• for each positive $\varepsilon$, there exists a finite dimensional vector space $F=F(\varepsilon)$ such that for all $x\in S$, we have $d(x,F)=\inf\left\{\left\lVert x-y\right\rVert, y\in F\right\}\lt\varepsilon$.

These conditions are necessary because if $\varepsilon$ is fixed, there is an integer $n$ and $x_1,\dots,x_n \in S$ such that $S\subset\bigcup_{j=1}^nB(x_j,\varepsilon)$ and we take $F$ as the vector subspace generated by the $x_j$'s. Taking $\varepsilon=1$ we get that $S$ is bounded.

The converse is a little bit more tricky. Let $M$ be such that $\lVert x\rVert\lt M$ for each $x\in S$. Fix $\varepsilon\gt 0$. Then $B(0,M+\varepsilon)\cap F(\varepsilon)$ has a compact closure, hence there is an integer $N$ and $y_1,\dots, y_N$ such that $B(0,M+\varepsilon)\cap F(\varepsilon)\subset\bigcup_{j=1}^NB(y_j,\varepsilon)$ and we conclude that $S\subset \bigcup_{j=1}^NB(y_j,\varepsilon)$.

For Banach spaces there is the following theorem by Grothendieck:

A closed subset $K$ of a Banach space $X$ is compact if and only if $K$ is contained in the closed absolutely convex hull of a sequence $(x_n)$ such that $||x_n||\to 0$.