Intersection of $\delta$-neighborhoods of lines I think I know how to prove this for inner product spaces, but not how to prove it for the more general normed spaces (assuming it's true).
Let $P$ and $Q$ be two distinct lines through the origin in a real normed vector space. Let $B_\delta(P)$ and $B_\delta(Q)$ be the set of points whose distance to $P$ or $Q$ is less than $\delta$. Similarly, $B_\epsilon(\{0\})$ is a closed ball of radius $\epsilon$ around the origin.
For any $\epsilon$, is there a $\delta$ such that:
$$B_\delta(P)\cap B_\delta(Q)\subseteq B_\epsilon(\{0\})?$$
This came up when I was trying to justify some intuition on the formal definition of the arc-length of a curve.
 A: We shall proceed by contradiction.
Suppose that a certain $\epsilon$ exists, such that $B_\delta(P)\cap B_\delta(Q)\not\subseteq B_\epsilon(\{0\})$ for any positive $\delta$.
This means that for every $\delta>0$ we can find a point $x$ such that $||x||>\epsilon$ and $d(x,P),(x,Q)<\delta$.
This implies that we can find a point $p\in P$ and $q\in Q$ such that $d(x,p)<2\delta$ and $d(x,q)<2\delta$. In turn, this implies again by the triangle inequality that $d(p,q)<4\delta$.
Again, by the triangle inequality we must have $||p||, ||q|| \geq \epsilon-\delta$
Taking any $\delta \leq \epsilon/2$ we conclude we can always find $p\in P$, $q\in Q$ such that $||p-q||<\delta$ and $||p||,||q||\geq \epsilon/2$.
Now, let $p_0$ and $q_0$ be two points in $P$ and $Q$ respectively with norm $\frac{\epsilon}{2}$.
Our previous result can be expressed as follows:
For every $\delta>0$ we can find reals $x,y$ such that $|x|,|y|\geq 1$ and $||xp-yq||<\delta$.
The set $A=\{xp_0+yq_0 | |x|\geq 1, |y|\geq 1\}$ is closed, as it is a closed subset of a finite dimensional subspace of a vector space.
Moreover, when we intersect it with the ball $B_{\epsilon}(\{0\})$ we obtain a compact set. And therefore it contains a vector of minimum norm, a contradiction.
