Closed subspace of the second dual of a Banach space

Question

I'm reading the following passage in Rudin's Functional Analysis, and I'm trying to understand why the last sentence holds: why is $\phi(X)$ closed in $X^{**}$?

Is there some obvious reason for this? The only way I can see it holding is by using what Rudin calls the "dilation principle":

Suppose $(X, d_1), (X, d_2)$ are metric spaces and $(X, d_1)$ is complete. If $E$ is closed in $X,$ if $f:E \to Y$ is continuous, and if $d_2(f(x'), f(x'')) \ge d_1(x', x'') \;\; \forall x', x'' \in E,$ then $f(E)$ is closed in $Y$.

I believe this applies to the above setting, with $X =X$ and $Y=X^{**}$, since $\phi$ was just shown to be an isometry. Is this correct? Is there a better way to see why it holds?

Answer 1

Yes. The image of a linear isometry is always closed: if $\{\phi(x_j)\}$ is Cauchy, then since $$ \|x_k-x_j\|=\|\phi(x_k-x_j)\|=\|\phi(x_k)-\phi(x_j)\|, $$ the net $\{x_j\}$ is also Cauchy. So $x_j\to x$ for a certain $x$, and then $\phi(x_j)\to\phi(x)$, so the image of $\phi$ is closed.