This is a follow-up to another question concerning infinite paths which was admittedly ill-posed. I hope this question is posed better.
The graph $N$ with vertex set $V(N) = \mathbb{N}$ and $(x,y) \in E(N)$ iff $x < y$ contains paths of arbitrary length connecting 0 and an appropriate $n$ .
The graph $Q$ with vertex set $V(Q) = \mathbb{Q}$ and $(x,y) \in E(Q)$ iff $x < y$ contains paths of arbitrary length connecting 0 and 1.
Of course, both graphs contain infinite paths, starting from 0, but ending nowhere.
It's more or less obvious, that $N$ doesn't contain a path of infinite length connecting 0 and an $n\in \mathbb{N}$ (because all $n$ are finite).
But it's hard (for me) to "see" and get a feeling why $Q$ doesn't contain a path of infinite length connecting 0 and 1: each finite path between 0 and 1 is a finite subset of $E(Q)$: $\lbrace (0,q_1),(q_1,q_2),…,(q_n,1)\rbrace$. Why cannot there be an infinite subset of that kind?
Is it impossible to define "that kind", i.e. the
property "being a path
connecting 0 and 1", or can we define
it (in second order logic maybe) but
prove, that no infinite subset of $E(Q)$ with this property exists?
Best Answer
Of what kind? Hans, you are still refusing to be any more precise about what you want this definition to do, and until you start talking with some precision I don't know what there is to say beyond "the standard definitions do not allow you to speak of an infinite path connecting two points because infinite paths do not have two endpoints" or the other things I already said in response to your last question.