In an infinite dimensional Banach space, does a compact subset have to be finite dimensional? I know it cannot contain any infinite dimensional balls, if this mean it has to be finite dimensional, then why do I sometimes see phrases like "for any compact subspaces of some vector space"?

Also, do compact sets in infinite dimensional Banach spaces have to be bounded?

Is there a general result that tells us what compact subsets of infinite dimensional vector spaces look like?

Thanks so much!


5 Answers 5

  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 $S\subset \bigcup_{j=1}^NB(y_j,\varepsilon)$.

  • $\begingroup$ Ok. To explain why $S \subset F$ is not allowed, could you give a definition of finite dimensional space? Why is $(0,1)$ not a "space". $\endgroup$ Apr 15, 2019 at 19:45
  • $\begingroup$ @ViktorGlombik I meant vector space. It indeed needed to be added. Thanks! $\endgroup$ Apr 15, 2019 at 20:31
  • $\begingroup$ Can $S$ also be finite dimensional? $\endgroup$ Apr 15, 2019 at 20:41

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 $||x_n||\to 0$.


To answer the first question, Does a compact subset have to be finite dimensional?:

Tychonoff's Theorem states that an arbitrary cartesian product of compact spaces is compact.

So for instance in infinite-dimensional separable Hilbert space, the cartesian product of the closed intervals [-1/n, 1/n], n = 1, 2, 3, ... is a compact subset.


The question is also interesting when the infinite-dimensional vector space is not a Banach space but a topological vector space like the spaces of distributions $\mathcal{S}'(\mathbb{R}^d)$ and $\mathcal{D}'(\Omega)$. There a subset is compact for the subspace topology iff it is closed and bounded. That's called the Montel property.


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