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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)

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    $\begingroup$ Every infinite dimensional banach space has uncountable dimension. If the space is separable, there is a dense subspace of countable dimension, which then also has infinite (uncountable) codimension. So separable banach spaces are fine. Nice question, BTW! $\endgroup$
    – PhoemueX
    Commented Jan 22, 2023 at 11:17

2 Answers 2

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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 $\{b_{n,1}\}_{n=1}^\infty.$

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    $\begingroup$ Here you have the same problem as appeared in the answer by kabenyuk. $\ker \phi \cap E$ has codimension $1$ in $E$, whatever non-zero functional $\phi$ you take. So this space is not finite-dimensional. $\endgroup$
    – Matsmir
    Commented Jan 22, 2023 at 13:25
  • $\begingroup$ Moreover, why would $\ker\varphi\cap E$ be finite-dimensional? The assumption would be that it has finite codimension. $\endgroup$
    – Bruno B
    Commented Jan 22, 2023 at 13:25
  • $\begingroup$ Your $\varphi$ seems to be an operator, not a functional. $\endgroup$
    – PhoemueX
    Commented Jan 22, 2023 at 13:29
  • $\begingroup$ @PhoemueX indeed, I missed that $\phi$ is an operator. However, now it is really unclear why its kernel is dense. $\endgroup$
    – Matsmir
    Commented Jan 22, 2023 at 13:58
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    $\begingroup$ @Matsmir I have misread the question. The answer has been re-edited. $\endgroup$ Commented Jan 22, 2023 at 15:12
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This is not a complete answer but I think it is worth to post it. The proof of Proposition 2 essentialy was given above in the comment by PhoemueX.

We say that a normed space satisfies $(D\Rightarrow FCD)$ (short for dense implies finite codimension) if all its dense subspaces have finite codimension.

Proposition 1. If a normed space $E$ satisfies $(D \Rightarrow FCD)$, then so do all its subspaces.

Proof. Indeed, let $L \subset E$ be a linear subspace, and $N \subset L$ be a dense subspace. Take arbitrary algebraic complement $M$ to $L$ in $E$, so $L \oplus M = E$. Then it is clear that $N \oplus M$ is dense in $E$, so it has finite codimension. Finally, $E / (N \oplus M) \sim L / N$ which implies that $N$ has finite codimension in $L$. $\square$

Proposition 2. If $E$ is an infinite dimensional Banach space, then it does not satisfy $(D \Rightarrow FCD)$.

Proof. It is clear that $E$ contains a closed separable infinite dimensional subspace $L$. We prove that $L$ does not satisfy $(D \Rightarrow FCD)$, so, by Proposition 1, $E$ also does not. Consider a countable dense subset $S \subset L$. Then $\mathrm{Span}\:S$ is dense in $L$, but has at most countable dimension. However, $L$ has more than countable dimension (continuum, in fact). Thus, $\mathrm{Span}\:S$ is dense in $L$ but has infinite codimension. $\square$

However, by now I am not able to answer the question for arbitrary normed space $E$. From Proposition 1 it is clear that if an infinite dimensional normed space $E$ satisfying $(D \Rightarrow FCD)$ exists, then there is also a normed space with countable dimension that satisfies $(D \Rightarrow FCD)$. So a proof or a counterexample should be searched in the case of countable dimensional normed spaces.

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