Is it possible to construct an infinite set of ordered pairs of form S = {(A, B), (B, C), (C, D), (D, x), ..., (y, Z)}?
Every element (B, C...) must appear only once as the first object in one of the pairs (except Z, never), and only once as the second object in one of the pairs (except A, never), and S is infinite.
I saw someone named sets similiar to the S above "R-paths" (the difference being that they were using only finite sets); basically if S (the set above) was a relation on some set A ($S \subseteq A^2$) then S would be an infinite R-path (an example of finite R-path would be {(A, B), (B, C), (C, D)}). My question can be then reformulated: can we construct an infinite R-path (on any set)?
My intuition is no, because at the path-point we got to after n steps, let's call it (N, M), the next path-point is either (M, not-the-end-point) or (M, end-point). The second option can't be right because we only traversed n path-points (and the path is infinite); but if first option is always right (that is, if the next step doesn't get us to the end), then there is no next-to-the-last path-point. This is probably useless since it only shows it is impossible to traverse (going one-by-one) such path (which was obvious from beginning), not that such path is impossible, but it might have to do with the actual annswer...
Thanks!
Yes, it’s possible, assuming that you have infinitely many objects available, but there will be an odd kind of gap in the middle. Suppose that I take $\Bbb N=\{0,1,2,\dots\}$ as my set of objects to chain. Let $A=\{\langle n,n+2\rangle:n\text{ is even}\}$ and $B=\{\langle n+2,n\rangle:n\text{ is odd}\}$. Then I have the chain
$$\langle 0,2\rangle,\langle 2,4\rangle,\langle 4,6\rangle,\dots~\dots,\langle 7,5\rangle,\langle 5,3\rangle,\langle 3,1\rangle\;.$$
Technically it meets your requirements, but it probably isn’t what you really have in mind, since there’s no finite path from the even side to the odd side.