Show that all infinite dimensional normed vector space $E$ have a dense hyperplane. Show that all infinite dimensional normed vector space $E$ have a dense hyperplane.
Hint: Consider $\beta$ a convenient Hamel basis of $E$, $S=\mathrm{span}\left\{v_{0},v_{1},\ldots,v_{n},\ldots\right\}$ a countable subset of $\beta$ and let $H=\mathrm{span}\left[\left(\beta\setminus S\right)\cup \left\{\frac{1}{n}v_{n}+v_{0}\:;\:n\geq 1\right\}\right]$.
Remark: I have not been able to build a convenient Hamel base. My attempt was to first build $S$ as a dense, linearly independent and enumerable set, then to extend $S$ to a Hamel basis of $E$, but my attempts have not been successful.
 A: Nobody builds a Hamel basis in an infinite-dimensional space.  They just exist, by Zorn's lemma. By "convenient" the hint means that every element of $\beta$ has norm $1$, which is achieved by taking any Hamel basis and normalizing its elements. 
Since the kernel of an unbounded functional is dense, it's enough to show there is such a functional $f$. To this end, letting $f(v_n)=n$, and also letting $f(v)=0$ for $v\in\beta\setminus S$, and extending by linearity, would work. 
The hyperplane $H$ described in the hint arises in this way if one lets instead $f(v_0)=-1$ and $f(v_n) = n$ for $n\ge 1$ (still $f(v)=0$ for $v\in\beta\setminus S$). I find it a bit more work to do it this way, if one has to show that $H$ is a hyperplane (i.e., the zero set of a linear functional) anyway. 
But in case you want to do it this way, observe that 


*

*$v_0\notin H$, hence $H$ is a proper subspace;

*if $v_0$ is added to the set that span $H$, the new set spans all of $\beta$, hence all of $H$. Therefore, $H$ is of codimension $1$.

