Given a path between 2 vertices of a graph, there is only a finite number of vertices in the path? Let us have a weighted graph $G=(V,E)$ with set of vertices $V$ and set of edges $E$, with a function $E\to V\times V$, a weight function $w:E\to\mathbb{R}_{>0}$ etc.
Take the topological realization $|G|$, where if $e\in E$, then $length(e):=w(e)$, and two vertices $v,v'\in V$ such that there is a path $P\subset|G|$ (i.e. homeomorphic to a closed interval in $\mathbb{R}$) joining them. Then I think $V\bigcap P$ should be finite. Why?
For example, it seems to me I could not take as a graph having topological realization the interval $[0,1]$ the one given by vertices $V=\{1,1/2,1/3,1/4,...,1/n,...,0\}$ and edges connecting them with the natural weight, since there is not an edge connecting $0$ with any other vertex.
 A: You can make an open covering of $|G|$ as follows: take the the interior of each edge (that is to say, the edges without their end points) and a small neighbourhood around each vertex (that contains at most a quarter of each adjoining edge). This is an open covering, and since $P$ is compact, it should be covered by finitely many of these open sets. Since each vertex is included in a unique open set in this covering, there can only by finitely many vertices on $P$.
A: By definition a path between two vertices has only finitely many vertices in it, even if the graph the path exists in is infinite.
For a drawing of a graph (that is, your "topological realization") one usually assumes that the set of points that represents vertices is discrete.
If there's a sequence of vertex points that converge towards a different vertex, then this convergence is not a graph-theoretic property, but rather a defect of the drawing. (In other words, the drawing is then not homeomorphic to the abstract realization of the graph as a gluing-together of single points for vertices and copies of $[0,1]$ for edges).
A: You are correct. The paths in an infinite graph are the finite paths, the rays, and the double rays. If you think of the integers as the vertices of a graph $Z$ whose edges are $\{n,n+1\}$ for $n\in\Bbb Z$, the subgraph $N$ induced by the natural numbers is a ray, and the whole graph is a double ray. In an arbitrary infinite graph the rays are the subgraphs isomorphic to $Z$, and the double rays are the subgraphs isomorphic to $N$. Note that only the finite paths have a vertex at each end.
In a topological realization the finite paths are homeomorphic to $[0,1]$, the rays to $[0,\to)$, and the double rays to $\Bbb R$.
