Is it possible to define a space and/or a distance function such that there is always more than 1 shortest path between any 2 points?

Question

I am in my second semester of university in maths and physics and thought of a question I am unable to answer. I asked my analysis teacher of the last semester if it was possible to define a space and/or a distance function such that there is always more than 1 shortest path between any 2 points. She thought about it and then she told me she didn't know.

Firstly, I thought that maybe it would be possible if you could define the distance between two points in Q'XQ U QXQ' only by going through points (a,b) such that either a is rationnal and b irrational or a irrational and b rationnal, but she told me that there would be no way to connect the points if we use only those points. ({(a,b)|a is rational and b is irrational or a is irational and b is rational})

However, I know that it is possible in special cases. A very simple example of this is that any 2 opposite points on a sphere has infinitely many shortest path between them. However, if the points are not opposite, then there is only one shortest path between them.

It is possible to create as many shortest paths as wanted between some 2 points on a surface, for example you can take the shape of an american football ball, put point 1 on one edge and point 2 on the other edge, and dig into the ball so that there are 2 ways to get from one edge to another. You can repeat this processus to create as many paths as you want between the 2 points on the edges.

These are only special cases though, and my question remain unanswered. So I thought about it and found a possible candidate. Could the distance calculated as the lenght of the circle arc between 2 points work? For example, in R^2, you would get two shortest paths from one point to another.

I believe it respects the usual definition of distance: It's always positive, the distance is equal to zero iff it is the distance from point a to point a, it is symmetric, d(a,b)=d(b,a) and the triangular inequality should be sastified.

The problem I encountered though, is that to create this circle you would need to first find the middle point between the 2 points and then put each point that will form the path from point 1 to point 2 at an equal distance to the middle point, and I believe it does not make sense as we are using normal euclidian distance.

I know this question lacks a lot of rigor and I don't even know if it makes sense asking it. For example I would need to define what I mean by a path. I believe requiring continuity would make sense but I'm not sure about the rest. I would be very grateful if you had any documentation related to this or the answer to my question. Thank you.

Here is a picture of what I meant

Answer 1

Consider the unit circle in $\mathbb{R}^2$ with distance defined by $$d(p_1, p_2) = \frac{1}{2}\left(\text{length of arc on one side of segment p1p2} + \text{length of arc on other side of segment p1p2}\right),\quad p1 \ne p2$$ $$d(p, p) = 0$$.

In this case, the distance metric is valid, and $d(p1,p2)=\pi$ for all $p1 \ne p2$.

Answer 2

This is true of the space $c_0$ of all sequences $(a_n)$ converging to $0$, with the supremum norm $\|(a_n)\|:=\sup_n |a_n|$.

By translation invariance, it suffices to show that every not identically $0$ sequence $(a_n)$ has more than one shortest path between $(a_n)$ and $(0)$.

To see this, let $k$ be an index with $|a_k|<\|(a_n)\|$, and note that if WLOG $a_k\geq 0$, then letting

$$b_i=\begin{cases} \frac{a_i}{2} & i\neq k \\ \frac{\|(a_n)\|}{2} & i= k, \end{cases} $$ we see that $(b_n)$ is a midpoint between $(a_n)$ and $(0)$, hence there are multiple shortest paths joining $(a_n)$ and $(0)$, namely, a linear one, and a piecewise linear one passing through $(b_n)$.

Remarks.

We can replace $\frac{\|(a_n)\|}{2}$ with $-\frac{\|(a_n)\|}{2}$ in the case $a_k<0$, so no generality was lost there.

For a rigorous definition of path length in a metric space, see this answer (thanks to Karl in the comments for linking.)

Also, regarding your idea at the end of your question about using circle-arcs to define the metric, though it's a clever idea, the length of the circle arc joining two points depends on the chosen center point of the circle, and if you choose this point to be the midpoint of the segment joining the points (as suggested in your picture), the distance you define is simply $\frac{\pi}{2}$ times the usual Euclidean distance. Since the Euclidean distance has unique shortest paths, so does your circle-arc distance - despite using circle-arcs to define the distance, the shortest paths are still line segments.

Even if somehow you choose the centers in some more convoluted way, it is still likely not the case that shortest paths coincide with circle arcs (and then you have the additional problem of not necessarily satisfying the triangle inequality), so this approach doesn't really solve the problem.