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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?

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  • $\begingroup$ No, every point in $\mathbb{Q}$ has a sequence converging to it, but $\mathbb{Q}$ is not complete. $\endgroup$ – Floris Claassens Apr 3 at 16:04
  • $\begingroup$ In general, I agree. But what about in the application above? $\endgroup$ – Zduff Apr 3 at 16:18
  • $\begingroup$ 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. $\endgroup$ – Floris Claassens Apr 3 at 16:26
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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.

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  • $\begingroup$ 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. $\endgroup$ – Zduff Apr 3 at 17:35
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    $\begingroup$ 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. $\endgroup$ – Martin Argerami Apr 3 at 18:09
  • $\begingroup$ So every closed metric space is complete? $\endgroup$ – Zduff Apr 3 at 18:36
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    $\begingroup$ Not in abstract. But here your metric space is a subset of a complete space (the double dual) so closed is the same as complete. $\endgroup$ – Martin Argerami Apr 3 at 18:39

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