# Infinite connectedness and projective graphs

Call two nodes $v$ and $w$ of a graph infinitely connected iff there is an infinite path $P(v)$ starting at $v$ and an infinite path $P(w)$ starting at $w$ such that there is an $x \in P(v) \cap P(w)$ which is not connected (in the standard, i.e. finite sense) to one of $v$ or $w$.

It seems to be a simple matter of fact that there are no graphs with infinitely connected nodes, since every $x$ on a simply infinite path is finitely connected to the starting point.

On the other hand, compare the situation with geometry: In standard (i.e. Euclidean) geometry, no two parallel lines do intersect, opposed to projective geometry where all pairs of lines intersect.

Is there something like a "projective graph theory" with nodes "at infinity"?

The concept of infinitely connected nodes might be vacuous in such a graph theory, since all nodes might be connected, finitely or infinitely. But this would probably depend on the theory.

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Topological graphs with non-isolated vertices that have no edges. (This idea made up on the spot by me.) – Ricky Demer May 20 '11 at 9:35
How can a non-isolated vertex have no edges? – Hans Stricker May 20 '11 at 9:39
topological graphs – Ricky Demer May 20 '11 at 10:10
Sorry, could you point me to a definition of "topological graph"? ("Topological graph theory" doesn't seem to be about topological graphs.) – Hans Stricker May 20 '11 at 10:20
topological graph = graph such that the set of vertices also has a topology – Ricky Demer May 20 '11 at 10:25

Here is an example which could simplified (see below), but I'll leave it this way: Think of the real projective plane as the completion of $\mathbb{R}^2$ by points and one line at infinity. Consider a graph whose vertex set is $\mathbb{Z}^2$ along with the infinite points of lines with rational slope. Use edges in the finite portion from $(a,b)$ to $(a\pm1,b\pm1)$. Then there are two connected components to the induced graph on $\mathbb{Z}^2.$ A path is determined by a starting point and a string using $L,R,\ell,r$ where $L$ and $\ell$ denote a move from $(a,b)$ to $(a-1,b+1)$ and $(a-1,b-1)$ respectively. We might allow exponents so that the path $(5,10)RRRLLrLLrLLrLLr$ could be described by $(5,10)R^3(LLr)^4.$ Finally, allow an exponent of $\infty$ but only as the very rightmost symbol in a string. So eventually periodic paths are (exactly) the finitely nameable infinite paths. I think that it is clear how to add an infinite point to each infinite path. Now every pair of finite vertices is infinitely connected. As it stands the infinite points are not on any edges. We could define some edges in one way or another (I'd think to other infinite points) but that does not affect the example.
A similar idea would use finite decimal expansions ( with $0.120$ connected to the eleven vertices $0.12$ and $0.120j$) So this is an infinite $10$-ary tree. Then ideal points would correspond to rationals and $0.284$ is finitely connected to $0.285$ by a path of length $2$ through $0.28$ as well as infinitely connected through the common point of $0.284999\dots$ and $0.285000\dots.$