Infinite paths that connect two vertices? 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?

 A: 
Why cannot there be an infinite subset of that kind?

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. 
A: I, for one, think I understand what you mean.  To see why there are no infinite sets, try to formulate "that kind of path" (the ones connecting 0 to 1) as a set: We could try letting $P = \{\text{paths }\alpha \text{ in } Q| \alpha \text{ starts at 0 and ends at 1}\}$ but an infinite set never ends, so our vocabulary is lacking.  Let's try something more precise: Let $P=\{(v_0,v_1),(v_1,v_2),\ldots,(v_{n-1},v_n)|v_0=0,v_n=1\}$ but this is inherently assuming that the length of the path is finite, namely, $n$.  I know this isn't a proof, but it seems to me that "connectedness" (at least in graphs) is a finite property.  
In order to distinguish between the two paths brought up by @tomcuchta and @Theo, you would need to assign weights to the edges, and then the definition of minimal path changes drastically. The reason we see the related sequences of real numbers as different is because the distance between the values is shrinking (or staying constant).  But the distance between two vertices in $Q$ is still always 1.  In $Q$, the vertices are just labelled with the elements of $\mathbb{Q}$.  So, since the vertices are not getting closer to one another, you never get any closer to 1, following the path given by tomcuchta.
There may be other ways to describe the possibility of infinite paths, and I'd be interested to hear them, but given the way you've formulated your question above, I don't think you can speak of an infinite path with two endpoints.  The path given by tomcuchta is infinite, but it does not end at 1 (or anywhere else).
A: The inherent problem in this idea is that while you could construct such a thing as an infinite path between two points (using suitable definitions), you'd have to construct it as a sequence of vertices/edges indexed by a totally ordered set where:


*

*There exist infinitely large elements. That is, an element $ω ∈ A$ where the set $\left\{α ∈ A| α < ω\right\}$ has infinite cardinality.

*Every element has a unique successor and predecessor, except for possibly the maximal and minimal elements. The predecessor is important because it makes the path intuitively 'linked'.

*There exist minimal and maximal elements (the start and end of the path).
These conditions preclude indexing the vertices using any well-ordered set (the proof is quite simple, and relies on the requirement for predecessors). This excludes ordinals and natural numbers, but not the set of hyper-natural numbers, which isn't well-ordered but fulfills all of the qualities above (e.g. consider the interval $[1 ... H]$ for some infinite hyper-natural $H$).
In this case, the set of vertices/edges that make up the path would be uncountable, and so you couldn't label the vertices using $ℚ$. I'm not familiar with any countable sets that have these properties, but there probably are.
