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.
On
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.
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).