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Suppose $X$ is an infinite dimensional topological vector space and $v\in X$ is non-zero. It is then not difficult to construct a vector space $U\subset X$ so that

1) $U$ is dense in $X$.

2) $U+{\mathbb C} v = X$.

In particular $U$ has co-dimension 1 in $X$ but is not closed.

Now, proofs that I can think of uses Zorn: Consider/construct a dense subspace $V$ not containing $v$ (usually not so difficult) and consider the collection of subspaces containing $V$ but not $v$, partially ordered under inclusion. If a vector $w\in X\setminus (V+{\mathbb C} v)$ you may add it to $V$ to get a larger such space. It is strictly inductively ordered so a maximal element $U$ will do the job.

The catch is that this is not very 'visual'. I can not give an explicit description of any such maximal element. So my question is if there is an explicit way to construct such a $U$ without resorting to Zorn/Axiom of Choice?

If possible, best would be an example with $X$ a separable Banach/Hilbert space.

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    $\begingroup$ Even easier, by the way, is to begin by extending $\{v\}$ to a basis, then considering the part of that basis which is not $v$ itself. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 1, 2016 at 16:30

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You are asking for a discontinuous linear functional that is non zero at $v$, but it is consistent with $ZF$ that every linear functional on a Banach space is continuous. However, on some normed spaces you can do what you want in $ZF$. For example, take $X:=c_{00}$, the space of finitely non zero real sequences under the sup norm. Given $0\not= v$ in $X$, you can explicitly construct an element $u$ in $\ell_2 \sim \ell_1$ s.t. $\langle u, v\rangle =1$.

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  • $\begingroup$ Thanks for the reply. If I understand the first part correctly, there is no complete normed space where I can get such an example without Zorn? Without completion I could simply take some $U$ dense in a Banach space $Y$, pick a vector $v \in Y\setminus U$ and let $X=U + {\mathbb C} v$ but the example is a bit trivial. Any reference for the ZF continuity? $\endgroup$
    – H. H. Rugh
    Commented Aug 1, 2016 at 15:29
  • $\begingroup$ @H.H.Rugh: You can find it in Schechter's book about foundation of functional analysis, whose exact name escapes me. Also in papers by Wright, Garnir and a few others. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 1, 2016 at 15:44
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    $\begingroup$ @H.H.Rugh There is no constructive Banach example without Zorn. This is also discussed on StackExchange here. $\endgroup$
    – Chris Gerig
    Commented Aug 2, 2016 at 23:15

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