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Let $X$ be a topological vector space and $V$ be a linear subspace of $X$ such that $\text{dim}(X/V)=1$, then either V is closed or $\overline{V}=X$. In other words if it is not closed then it is dense.

I believe that this is true and TrialAndError's answer to "$\ker f$ is either dense or closed when $f$ is a linear functional on a normed linear space", seems to confirm this. But I would like to make sure that I have understood correctly why this is the case.

My reasoning is simply that if V is a subspace of X, so is $\overline{V}$. If V is not closed then they are distinct subspaces,and my intuition is that there isn't enough room for anything "bigger" than $\overline{V}$.

This is because: $\exists x_0 \in \overline{V} \setminus V$ and so $\text{span}\{x_0\} \subset \overline{V}$ and therefore so does $V+\text{span}\{x_0\}$ but $X=V\oplus\text{span}\{x_0\}$ (I guess I'm using the theorem of incomplete basis to say this) so $X=\overline{V}$. Perhaps there is a better way of seeing this without implicitly invoking the axiom of choice?

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Jack
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1 Answers1

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Perhaps there is a better way of seeing this without implicitly invoking the axiom of choice?

I'm not sure whether it's "better", but we can do it without implicitly using the axiom of choice.

Let $\pi \colon X \to X/V$ be the canonical projection. We know that $\overline{V}$ is a subspace of $X$ with $V \subset \overline{V}\subset X$. Then $W = \pi(\overline{V})$ is a subspace of $X/V$. But since $\dim X/V = 1$, it follows that either $W = \{0\}$, which means $\overline{V} = V$, or $W = X/V$, which means

$$X = \pi^{-1}(W) = \pi^{-1}(\pi(\overline{V})) = \overline{V} + V = \overline{V},$$

since generally for quotient spaces $X/Y$ and subspaces $Z$ of $X$ we have $\pi_Y^{-1}(\pi(Z)) = Z + Y$.

Daniel Fischer
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  • I like the last argument very much, it's much more elegant. Perhaps I should change the requirements in my original post to "Let X a topological vector space", as X Banach seems a bit of an overkill? – Jack Nov 07 '15 at 14:28
  • Yes, $X$ being a Banach space is irrelevant. All we need is a topological vector space. – Daniel Fischer Nov 07 '15 at 14:40
  • Conversely. Is every dense linear subspace of finite co-dimension then? – ABIM Sep 11 '19 at 13:18
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    @AIM_BLB No, in fact already very small subspaces can be dense. Take $X = \ell^p(\mathbb{N})$, with $1 \leqslant p < +\infty$. This is a Banach space of dimension $2^{\aleph_0}$. There, consider the subspace $Y$ of sequences with finite support. This is a dense linear subspace of dimension $\aleph_0$, and hence of codimension $2^{\aleph_0}$. – Daniel Fischer Sep 30 '19 at 14:05
  • Nice example! Thanks Daniel – ABIM Sep 30 '19 at 14:17