# Proving a normed space is complete with directly using a Cauchy sequence.

Consider the following question:

Let $$X$$ be a normed space and $$T: X\longrightarrow X''$$ is the canonical mapping. Prove the range of T, denoted $$\mathscr{R}(T)$$, is closed iff $$X$$ is complete.

Where I want to stick with the "$$\implies$$" part of the of the proof. Here I'm concerned with—as the title implies—whether I can get a away with the following proof.

## Proof

Let $$x \in X$$ be arbitrary, then $$Tx \in \mathscr{R}(T)$$. Since $$\mathscr{R}(T)$$ is closed, $$Tx$$ must be one of its limit points. Since $$Tx$$ is a limit point of $$\mathscr{R}(T)$$ then there is a sequence $$(x_n'')$$ in $$\mathscr{R}(T)$$ such that $$x_n'' \longrightarrow Tx$$. Since each element $$x_i''$$ of $$(x_n'')$$ is in $$\mathscr{R}(T)$$, it can be written $$x_i'' = Tx_i$$. In other words, $$Tx_n \longrightarrow Tx$$ in $$\mathscr{R}(T)$$. Formally, for any given $$\epsilon > 0$$ there exists an integer $$N(\epsilon)$$ such that for any $$n > N$$ we have

$$\|Tx_n - Tx\| < \epsilon$$

and since canonical mappings are norm preserving (and linear) we have

$$\|x_n - x\| = \|T(x_n - x)\| < \epsilon$$

So $$x_n \longrightarrow x$$ in $$X$$.

Since $$x'' \in \mathscr{R}(T)$$ was arbitrary, every point in $$X$$ has a convergent sequence in $$X$$. Convergent sequences in metric spaces are Cauchy, and therefore $$X$$ is complete.

## Discussion

Here I didn't directly prove that a Cauchy sequence in $$X$$ implies that it is converges in $$X$$. But I think I proved that every vector in $$X$$ has a convergent sequence to it, which I believe is equivalent. Is this true and valid?

• No, every point in $\mathbb{Q}$ has a sequence converging to it, but $\mathbb{Q}$ is not complete. – Floris Claassens Apr 3 at 16:04
• In general, I agree. But what about in the application above? – Zduff Apr 3 at 16:18
• I can't come up with a counter-example of the top of my head. But I would still say no. I would suggest just proving this using Cauchy sequences. – Floris Claassens Apr 3 at 16:26

Your proof is wrong, and it misses the point. You start with $$x''=Tx$$, and then choose a sequence $$\{x''_n\}$$ with $$x''\to Tx$$. What prevents you from taking $$x_n''=x''$$ for all $$n$$? Then you are proving nothing.
Besides the above, the argument is unnecessarily complicated, if you already know that $$T$$ is an isometry.
You want to show that $$X$$ is complete, so you should start with a Cauchy sequence $$\{x_n\}\subset X$$, not with a point in the range of $$T$$. As $$\|Tx_n-Tx_m\|=\|x_n-x_m\|$$ for all $$n,m$$, the sequence $$\{Tx_n\}$$ is Cauchy, so there exists $$x$$ with $$Tx=\lim Tx_n$$. Then $$\|x-x_n\|=\|Tx-Tx_n\|\to0.$$ And that's it. And, by the way, this proves that the range of any isometry with complete domain is complete, this has nothing to do with double duals or anything.
• Yeah I see your point now. But previously I started the way you suggest, but I'm hung on some assumption in your proof. You say $(Tx_n)$ is Cauchy , so there exists a $Tx$ that it converges to. Is this because the range of $T$ is closed? Because I'm under the impression that this is only true if the range of $T$ is finite dimensional. – Zduff Apr 3 at 17:35
• Yes, it's because the range of $T$ is closed, and it that has nothing to do with dimension: it's topology of metric spaces. – Martin Argerami Apr 3 at 18:09