Is there a way to prove Compact $\Leftarrow$ Closed and Bounded in a finite dimensional normed space WITHOUT this famous Lemma

compactnessnormed-spacesreference-request

Context. It is well known that in a finite dimensional normed space $X = (X,\|\cdot\|)$, a subset $M \subset X$ is said to be compact iff $M$ is closed and bounded.

The $(\Rightarrow)$ is pretty much straightforward, since we know that a compact subset of a metric space is closed and bounded (and a (finite dimensional) normed space is a metric space with the metric induced by the norm).

On the $(\Leftarrow)$ implication: with some easy search, one can find several proofs of this (even on YouTube, or here, really). But all of these have one thing in common: They use the following auxiliary Lemma.

Lemma. Let $\{x_1,\dots,x_n\}$ be a linearly independent set of vectors in a normed space $X$ (of any dimension). Then there is a number $c>0$ such that for every choice of scalar $\alpha_1,\dots,\alpha_n$ we have
$$ \| \alpha_1x_1 + \dots + \alpha_nx_n\| \leqslant c(|\alpha_1| + \dots + |\alpha_n|).$$
And I was wondering: Certainly there must be a way to prove the $(\Leftarrow)$ implication without using the Lemma above, but I can't find no proof like this online.

Basically I am looking for some references (or the proof itself) of this implication without using the Lemma mentioned above.

Thanks for any help in advance.

Best Answer

TlDr; just use Heine-Borel for $\Bbb R^n$, after appealing to the linear homeomorphism $X\cong\Bbb R^n$.

If you want a more birds-eye approach, there is a general Heine-Borel-Bolzano-Weierstrass theorem:

Let $(X,\rho)$ be a metric space. All three conditions are equivalent:

  • $X$ is topologically compact
  • $X$ is sequentially compact
  • $X$ is complete and totally bounded

Our finite dimensional vector space $X$ is, if you take the ground field to be $\Bbb R$ (say), complete (that's because $\Bbb R^n$ is, if you like). Any closed subset thereof is also complete.

All you need to do is determine that a bounded subset $C$ of $X$ is also totally bounded, meaning:

For all $\epsilon>0$ there exists finitely many points $(x_n)_{n=1}^m$ of $C$ and balls $B(x_n,\epsilon)$ which cover $C$ in their union.

This is like a special case of compactness: such covers are called $\epsilon$-nets.

Let's use the linear-and-topological isomorphism $X\cong\Bbb R^n$. The bounded subset $C$ is mapped to an isomorph $C'\subseteq\Bbb R^n$ which is also bounded. We must verify that $C'$ is totally bounded; this more-or-less follows from the Pythagorean identity:

Fix $\epsilon>0$ and $N\in\Bbb N$ such that $\frac{\sqrt{n}}{N}<\epsilon$. $C'$ is contained within some box $[-M,M]^n$, where $M>0$ is a constant integer. This box allows a partition into lattice points of the form $x=(x_1,\cdots,x_n)$ where each coordinate has $Nx_i\in\Bbb Z\cap[-M,M]$. We can place, at every such lattice point $x$, a ball $B(x,\epsilon)$. Note that the cube: $$\{(y_1,\cdots,y_n)\in[-M,M]^n:|y-x_i|\le1/N\}$$Is completely covered by $B(x,\epsilon)$ by choice of $N$. Taking the union of the $B(x,\epsilon)$ over all lattice points $x$, we have a cover of $[-M,M]$ and thus a cover of $C'$ via a finite $\epsilon$-net, as desired.

So every bounded subset of $\Bbb R^n$ is totally bounded, every bounded subset of $X$ is totally bounded (it is an isomorphism invariant). We could have done the same proof within $X$ after first fixing an orthonormal basis too.

Every closed, bounded subset of $X$ is then complete and totally bounded hence compact by the main theorem. Really, by the isomorphism, it suffices to check the Heine-Borel theorem for subsets of $\Bbb R^n$ - maybe you know a proof for that already.

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