How do you prove that every infinite dimensional normed space contains a dense hyperspace? (Where hyperspace is defined to be maximal proper subspace.)
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Use the existence of an infinite basis to construct an unbounded linear functional. The kernel will be dense (why?) – PhoemueX Sep 07 '14 at 08:48
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@PhoemueX: I suggest you post your answer as an answer, and not as a comment. – Martin Argerami Sep 07 '14 at 14:42
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Using (e.g.) Zorn's Lemma, one can show that each vector space has a basis. So choose a basis $(x_i)_i$.
As your space $X$ is infinite-dimensional, there is an infinite sequence $(i_n)_n$ (consisting of distinct elements). We can assume $\Vert x_i\Vert = 1$ for all $i$. Now define a functional $\varphi$ by
$$ \varphi(x_{i_n}) = n \text{ and } \varphi(x_i) =0 \text{ for } i \notin \{i_n\mid n \}. $$
This implies that $\varphi$ is an unbounded linear functional, hence discontinuous.
Now If a linear functional is not bounded, then it has a non-closed kernel. shows that the kernel of $\varphi$ is a non-closed (hence dense) hyperspace in $X$. It is maximal because of $\Bbb{K} = \varphi(X) \cong X/\text{ker}(\varphi)$.
