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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.

Modus Operandi
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    Won't you also need that the underlying scalar field is complete? – parsiad Aug 07 '15 at 23:03
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    You can't prove the statement "$\operatorname{dim} X = n \Rightarrow X \text{ is complete}$" by starting with the assumption $\operatorname{dim} X = \infty$. This would be trying to prove $P\Rightarrow Q$ by assuming $\neg P$. The only thing you could prove with that assumption would be the converse, $Q\Rightarrow P$, which is equivalent to $\neg P \Rightarrow\neg Q$. – Jack Lee Aug 07 '15 at 23:09
  • @NormalHuman : he is specificaly asking for a proof by contradiction, a strange idea indeed, but that is not covered in your link – Tryss Aug 07 '15 at 23:33
  • @Tryss I leave rewriting the proof as an exercise for the OP. –  Aug 07 '15 at 23:36
  • @JackLee, you are right, I'll change it. – Modus Operandi Aug 08 '15 at 18:44

1 Answers1

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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)

Tryss
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