Every normed space has a completion?

Question

So I know that a completion of $X$ is a Banach space $Y$ such that $X$ is isometrically isomorphic to a dense subset of $Y$, say $A$.

So we need to prove that we can always find a $T \in L(X,A)$ such that T is bounded and $\|T\| = 1$. How do we construct this?

And if we have more completions, are these completions isomorphic as well?

Kees

Answer 1

You can construct the completion of a normed vector space $X$ by using a construction of the completion of a general metric space and then defining a normed vector space structure on that completion.

But there's a better way, or at least a way that a lot of people might regard as simpler and more elegant: If $X$ is a normed vector space then the dual $X^*$ is a Banach space, and $X\subset X^{**}$ just as if $X$ were a Banach space; the completion of $X$ is just its closure in $X^{**}$.

(More precisely, the completion of $X$ is the closure of $j(X)$, where $j:X\to X^{**}$ is the canonical injection; so $j$ is exactly the $T$ you ask for.)