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Consider a topological vector space $V$ over $K\in\{\mathbb{R, C}\}$. I ask a simple innocent question: Is the complement of every proper subspace dense?

What if the space is normed? Or has an inner-product? This question popped up while reading Andreas Blass's answer to a previous question of mine.

Atom
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  • It does not have to be. Imagine $\mathbb{R}^2$ with discrete topology. If it is normed it should be true. – Vuk Jovovic Feb 03 '24 at 23:41
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    @VukJovovic A "topological vector space" usually assumes compatibility between the topology and the vector space operations (i.e. continuity of things like addition and multiplication maps). The discrete topology wouldn't meet that requirement, but maybe I'm wrong about the OP's intention for the phrase. – Josh Keneda Feb 03 '24 at 23:53
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    @JoshKeneda You are not wrong about OP's intention. – Atom Feb 04 '24 at 00:00
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    Not the same thing, of course, but a proper subspace of normed a space can be dense -- Does there exist a linearly independent and dense subset? My answer there includes mentioning that a closed proper subspace of a normed space is nowhere dense (hence, the complement is dense in a very strong way) and a Borel proper subspace of a Banach space is first (Baire) category in the Banach space (hence, the complement is dense in a strong way, but not necessarily as strong as if the subspace is closed). – Dave L. Renfro Feb 04 '24 at 00:05

2 Answers2

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To show that a subspace is dense, it is equivalent to show its complement is hollow. Let $F\subset V$ be a proper subspace, $V$ a normed linear space. Let $x\in F$. Let $y\in V\setminus F$. Let $\epsilon>0$. Consider that $z:=x+\frac{\epsilon}{2\|y\|}\cdot y$ is in the ball $B(x,\epsilon)$; moreover, $z\notin F$ because $\frac{2\|y\|}{\epsilon}(z-x)\notin F$, using subspace properties. Therefore $B(x,\epsilon)\not\subset F$ for any $x$ or $\epsilon$; therefore $F$ is hollow, and $V\setminus F$ is dense. Note that this is true even if $F$ is not closed.

In a general TVS I am not sure because I haven't looked at them in yonks but probably if you have a nice TVS with a good basis around the origin you can repeat a variant of this argument.

FShrike
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To finish FShrike's thought, I'd put it this way:

Let $F \subset V$ be your proper subspace. It suffices to show that any $x \in V$ is a limit of a sequence in $V\setminus F$.

If $x \in V\setminus F$, then the constant sequence $x_n := x$ works.

If $x \in F$, then pick any non-zero $y \in V\setminus F$. Then the sequence $x_n := x + \frac{y}{n}$ works.

We know that $\frac{y}{n} \rightarrow 0$ by continuity of the scalar multiplication map, which applies for general topological vector spaces (and therefore also in the normed/inner product setting).

Josh Keneda
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    At first I was worried that there might be some concern about the difference between closure vs. taking sequential limits, because for general topological spaces the closure isn't necessarily the same as the set of sequential limits. Not sure if they're the same for TVS. But even if the two aren't the same, it's always true that the closure contains all sequential limits, which is good enough for us. – Josh Keneda Feb 04 '24 at 00:10