generalization of a normed space I study analysis and  have a problem:
I have a normed space for example $(X,M)$ that is not complete,  how can I complete the space $X$ with respect to norm $M$?
please help me
Thanks
 A: The procedure is the same as completing any metric space, and should be in any functional analyis textbook. Formally, one defines an equivalence relation on Cauchy sequences by decreeing that $(x_{n}) \sim (y_{n})$ if and only if only if $\lim_{n \to \infty}(x_{n}-y_{n}) = 0.$ The set of equivalence classes under $\sim$ becomes a linear space with operations $[(x_{n})] + [(y_{n})] = [(x_{n}+y_{n})]$ and $\lambda [(x_{n})] = [(\lambda x_{n})],$ which are well-defined. The new space becomes a normed linear space by setting
$M([x_{n})] = \lim_{n \to \infty} M(x_{n})$ (recall that Cauchy sequences are bounded). The new space is complete and $X$ is isometrically embedded in it by identifying $x$ with the (class of the) constant sequence whose terms are all $x$.
A: It is known that $\mathcal{B}(X,Y)$ is complete whenever $Y$ is complete, so $X^{**}:=\mathcal{B}(X^*,\mathbb{C})$ is complete. Consider natural embedding
$$
i:X\to X^{**}:x\mapsto (f\mapsto f(x))
$$
By corollary of Hahn-Banach theorem $i$ is an isometry. Then $\tilde{X}:=\mathrm{cl}_{X^{**}}(\mathrm{Im}(i))$, is a completion of $X$. Indeed, 


*

*$X$ isometrically embedded into $\tilde{X}$ via $i$ 

*by construction of $X$ is dense in $\tilde{X}$

*$\tilde{X}$ is complete as closed subspace of complete space $X^{**}$

A: You could try to define an isometry from $T: X \rightarrow W$, where $W$ a subspace which is dense in $ \tilde{X}$ and $(\tilde{X}, \tilde{M})$ is complete. Here the elements of $ \tilde{X}$ are equivalence classes. See e.g how $L^p$ is constructed from the space of continuous functions 
