In our lecture on functional analysis, we had the following statement: "Each normed space $X$ is isometric isomorph to a dense subset of a Banach space."

We were also give a proof, which goes as follows:

"Define $Y$ as the closure of $i(X) \subseteq X''$ (where $i: X \rightarrow X''$ is the linear isometry between a normed space $X$ and its bidual space $X''$). Since $Y$ is closed and $X''$ is a Banach space, we get that $Y$ is a Banach space."

I do understand the statements in the proof as such, but I don't understand how my statement actually follows from this proof. I think there might be just one final line of argument missing.

Thank you for your explanations.

  • 2
    $\begingroup$ $X$ is isometrically isomorphic to its image under $i$, and $i(X)$ is dense in $Y$, which is a Banach space. $\endgroup$ – mathworker21 Mar 22 '17 at 7:47
  • $\begingroup$ @mathworker21 Thank you for your answer! That perfectly answered my question. $\endgroup$ – SallyOwens Mar 25 '17 at 9:21

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