Let $E$ be a normed $\mathbb{K}$-vector space ($\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$). I'm interested in the following question:
If every dense subspace is of finite codimension in $E$, then is $E$ finite-dimensional?
As a reminder, the codimension of a subspace $D$ is the dimension of the quotient space $E/D$. It can be shown that it's also the dimension of all algebraic complements of $D$.
This post provides an example of a space in which there exists a dense subspace of infinite codimension, and this post provides a way to construct dense subspaces of any given finite codimension, but I didn't find anything addressing my exact question, though there's probably something I missed here, or maybe it's on MathOverflow I don't know.
If possible, if a non-Banach counterexample is found, it'd be nice to then reconsider the question for $E$ Banach, but let's not get ahead of ourselves. I also do not mind at all even just having partial answers, like "true for $E$ separable Hilbert", or "true for $E$ of the form $\mathcal{C}(X,\mathbb{K})$", or whatever special cases may come up in your minds. I'm also pro-axiom of choice, in case it makes a difference for this question (like, if you need Hamel bases or that kind of things).
Finally, I know that usually question-havers should show what they've tried, but, I'll be honest, I don't really know how to begin… However I thought it was somewhat interesting enough that I'd try asking here anyway, hopefully that is fine? This is not for any homework or project, just my own curiosity.
(Feel free to re-tag or edit this post if needed)
Best Answer
Assume $E$ is an infinite dimensional normed space. Therefore it contains an infinite linearly indepedent set $B_0=\{b_{n,m}\}_{n,m=1}^\infty,$ such that $\|b_{n,m}\|=1.$ This set can be extended to a Hamel basis $B$ of $E.$ Consider the linear operator $A$ defined on $B$ by $$Ab=\begin{cases} mb_{n,1} & b=b_{n,m}\\ 0 & b \in B\setminus B_0 \end{cases} $$ and then extended to $E$ by linearity. Then $\ker A$ is dense in $E.$ Indeed, for $x\in E$ we have $$x=\sum_{b\in B_0} x_bb+\sum_{b\in B\setminus B_0}x_bb$$ where both sums are finite. The second sum belongs to $\ker A.$ The first sum can be represented as $$\sum_{n,m=1}^Kx_{b_{n,m}}b_{n,m} $$ It suffices to approximate that sum by the elements in $\ker A, $ i.e. approximate every term $b_{n,m}$ by the elements in $\ker A.$ We have $$\lim_{k\to\infty}\left [b_{n,m}-{m\over k}b_{n,k}\right ]= b_{n,m}$$ and $$A\left [b_{n,m}-{m\over k}b_{n,k}\right ]=0$$ This completes the proof of density of $\ker A.$
The codimension of $\ker A$ in $E$ is infinite as $E/\ker A$ is isomorphic to ${\rm Im}A$, i.e. the linear span of $\{b_{n,1}\}_{n=1}^\infty.$