**Question 1** Let $X$ a separable Banach Space and $Y\subset X$ linear subspace. When can we put a norm on $Y$ in such a way so that $Y$ is a Banach space?


Clearly if $Y$ is closed in the norm topology that's sufficient.


So, I'm coming into this with the perspective of Polish spaces(i.e. completely metrizable separable spaces) where a $G\_\delta$ subset of a Polish space is once again Polish and every Polish space is a $G\_\delta$ subset of a Polish space. 


I was wondering if there is some sort of similar condition about complete norms and how the structure of the space matters. 


I know there are linear spaces that can't be Banach spaces, e.g. $c\_{0,0}$ the subspace of $c\_0$ where cofinitely many elements in each sequence are $0$. In this case, there is a trivial Hamel basis, namely $e\_i=(0,\ldots,0, 1, 0, \ldots)$ where $1$ is the $i$th element in the sequence. But this Hamel basis is countable and we know every Hamel basis of an infinite-dimensional Banach space must be uncountable.


We also know that the existence of a complete norm is equivalent to the existence of a complete homogeneous translation-invariant metric $d$, in the sense that


* $d$ is a complete metric
* $d(ax,ay)=|a|d(x,y)$
* $d(x+z,y+z)=d(x,y)$


From here, I believe we can see that $Y$ must at least have a complete metric and so must be a $G\_\delta$ subset of $X$.


I haven't been able to come up with a counterexample of $Y$ a complete metric space with linear structure and $Y$ not a Banach Space, so I'm thinking that the $G\_\delta$ may be a sufficient condition, so I think this can be restated as 


**Question 2**
If $Y$ a linear space and $d$ a complete metric on $Y$. Then there exists a complete homogeneous translation-invariant metric $d'$ on $Y$.


Kind of related to the questions above, I know every vector space has a norm, just take a Hamel basis $(e\_\lambda)$ and $x=\sum a\_i e\_i$. Now let $||x||=\max{|a\_i|}$ or $||x||=\sum |a\_i|$. Either of these forms a norm, but this norm isn't necessarily complete, especially in an infinite-dimensional space.



Under what conditions can we put a complete norm on a linear subspace of a separable Banach space?
The main restriction on putting a complete norm is that Hamel bases in infinite-dimensional separable Banach spaces have cardinality continuum (see Exercise 6.4 on page 191 in Fabian-Habala-Hajek-Montesinos-Pelant-Zizler, Functional Analysis and Infinite-Dimensional Geometry). This gives an answer to your first question: if and only if $Y$ has a Hamel basis of cardinality continuum. (To introduce such norm we use an arbitrary bijection between the Hamel basis of $Y$ and the Hamel bases of an arbitrary separable infinite-dimensional Banach space.)


I think that similar approach can be developed for complete metric spaces.


