Can you lay $\aleph_1$ many people on a railway track?

Recently an image was posted to this trolley problem meme page. The image shows two tracks: one with $\aleph_0$ many people and another with $\aleph_1$ many.

A discussion broke out in the comments as to whether it is possible to even have $\aleph_1$ many people on the track. Roughly the two arguments were:

No- You cannot lay uncountably many people in a line.

Yes- If you assume the axiom of choice we can define a well ordering of a set of cardinitality $\aleph_1$ and lay people on the track using this ordering.

• Do people have width? – dbx Apr 26 '18 at 15:03
• I think one can lay aleph-1 people in a long line. – Lord Shark the Unknown Apr 26 '18 at 15:07
• Well, you can. That'd be the reals on the number line. I think the question is can you lay them in order like that. That would require a well-ordering principal (which nobody knows of) which requires axiom of choice. – fleablood Apr 26 '18 at 15:17
• What does "forever" mean here? Is time indexed by ordinals, by real numbers, by surreal numbers? How can the trolley even move, given Zeno's laws of motion??? – Asaf Karagila Apr 26 '18 at 16:45
• Here is a different thought: The image suggests the trolley is propelled by gravity alone. Hence it seems reasonable to assume that the $\aleph_0$ track is steeper than the $\aleph_1$ track. In fact, the angle should be such that without blockage the trolley would travel twice as fast. Running over people would slow down the trolley and may even stop it on the $\aleph_0$ track after finitely many people whereas it may maintain a non-zero speed on the $\aleph_1$ track. So we may able to save all but finitely many people by not pulling the lever. – Stefan Mesken Apr 27 '18 at 10:57

We'll pretend the track is a copy of $[0, \infty)$.

For infinitely thin people you don't even need a well-ordering. You can just put one person at each real number position along the track. Since $|[0,\infty)| = |\mathbb R| \ge \aleph_1$ you can lay at least that many people along the track.

For people of positive finite width this cannot be achieved. For then every person would lie over at least one rational number. Since no two people occupy the same space that would mean there are at least $\aleph_1$ many rational numbers. But we know this is not true.

Of course there are many other possible railways. For example the long line which is obtained by arranging $\aleph_1$ many copies of $[0,1)$ end-to-end. Then we could put one person in each copy of $[0,1)$ easy. The problem with this approach is that no finite speed is sufficient for the train to ever reach the end of the track in finite time. In fact the long line starts with a copy of $[0,\infty)$ and there's no reason to believe the train can ever get beyond that.

Of course there are loads of long railways that are nothing like well-ordered. For example the lexicographically-ordered-square or the product $[0,1) \times [0,\infty)$ with order given by $(a,b) \le (x,y) \iff b \le y$ and $a \le x$. But it's harder to imagine how these would function as railways since once the train gets past $[0,1) \times \{0\}$ there is already infinitely much track behind it. Formally this is saying the following:

Exercise: Prove every continuous increasing function $[0,\infty) \to [0,1) \times [0,\infty)$ with $0 \mapsto (0,0)$ has range contained in $[0,1) \times \{0\}$.

Hint: $[0,\infty)$ does not admit a family of uncountably many pairwise disjoint open sets.

Conclusion: Don't pull the lever. Any reasonable choice of long railway starts with a copy of the short railway and the train never gets beyond that initial copy. So the train will kill people twice as quickly on the upper line to the lower line. Once we reach the end of time the upper train will fall off the end of the track while the lower train will carry on chugging. But we're not going to be around to worry about that.

Addendum: This assumed the people are packed equally densely on both lines. If they are packed three times denser below for example you should pull the lever.

• Note that this assumes the universe behaves like Euclidean space. However, if it behaves like the Alexandroff Line (reference: en.wikipedia.org/wiki/Long_line_(topology) ) the argument does not work. – Joe Apr 26 '18 at 15:09
• Oh yeah there are loads of different linearly ordered sets! Seems anything reasonable would be a Suslin Line though en.wikipedia.org/wiki/Suslin%27s_problem – Daron Apr 26 '18 at 15:17