Let $\mathbb{K}$ be a field, $V$ be a $\mathbb{K}$-Banach space and $U<V$ be a dense subspace (with $\overline{U}=V$). I have heard, that: $$V/U\cong\mathbb{K}.$$ Is this result even correct and if not, what is a simple counterexample? If it is, how do you prove it? Unfortunately, it looks a lot like it can simply be done with the fundamental theorem on homomorphisms, but that is not the case. There is no surjective function $f\colon V\twoheadrightarrow\mathbb{K}$ with $\ker(f)=U\Rightarrow f\vert_U=0$ since the latter condition combined with $U$ being dense implies $f=0$.
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2This is not true. You can construct a dense subspace with large codinension. What is true is that if $V$ has codim one then it is either dense or closed. – Evangelopoulos Foivos Jul 13 '22 at 15:52
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1Thanks, this looks pretty similar and is also correct. Looks like the theorem got a bit falsified by being told around. – Samuel Adrian Antz Jul 13 '22 at 16:03
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1When you're thinking about dense subspaces in functional analysis I think it can be helpful to think of these as a subspace that can "approximate" the whole space well. For example, the polynomials are dense in $C[0, 1]$ by the Stone-Weierstrass theorem, which is a very nice example of a dense subspace. But there is no way that this quotient will be one-dimensional - the polynomials are far too small of a subspace. – Izaak van Dongen Jul 13 '22 at 16:59
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Density of a subspace $U$ of a topological vector space $V$ is characterized by the condition that the quotient topology on $V/U$ is trivial, i.e., the only open sets are $\emptyset$ and $V/U$. – Jochen Jul 14 '22 at 08:46
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@IzaakvanDongen Why not an official answer? – Paul Frost Jul 31 '22 at 15:25