If $\dim X=n$ then for any norm in $X$, $X$ is complete.

Question

I know there are standard proofs for this theorem, but I need to prove it by contradiction or proving that $\dim X=\infty$. I thought maybe using Hahn-Banach?

Thanks.

Answer 1

Proving by contraposition doesn't seems very friendly... You can do it like that :

Let $x_n$ be a Cauchy sequence that doesn't converge.

First, notice that $x_n$ is bounded :

Indeed, there exist $n$ such that

$$\forall m>n, \ \|x_n - x_m \| < 1$$

So you have

$$\forall m>n, \|x_m\| = \| x_n + (x_m - x_n) \| \leq \|x_n\| + \|x_m -x_n \| \leq \|x_n\|+1 $$

Now, define

$$M = \max_{k\leq n} \|x_k\|$$

Then

$$ \forall k, \|x_k\| \leq M +1$$

Second step :

A Cauchy sequence that has a subsequence that converge converge.

Indeed, let $x_{n_k}$ be such a converging subsequence (to $x$).

Then, $\forall \epsilon > 0$,

$$\exists k_0, \forall k > k_0, \| x_{n_k} - x \| \leq \epsilon$$

And

$$\exists n_0 \geq n_{k_0}, \forall n,m \geq n_0, \| x_n-x_m \| \leq \epsilon$$

By combining the two, you get

$$\forall n > n_0, \| x_n - x \| \leq \| x_n - x_{n_n} +x_{n_n} - x \| \leq \| x_n - x_{n_n}\| + \|x_{n_n} - x \| \leq \epsilon + \epsilon $$

And $x_n$ converge.

Third step :

$(x_n)$ has no converging subsequence, and $(x_n)$ is contained in the closed ball of radius $M+1$ so this ball is not sequentially compact.

This imply that the unit ball is not compact, hence the space is infinite dimensional (by Riesz's Lemma)