In a graph, connectedness in graph sense and in topological sense
From Wikipedia
Graphs have path connected subsets, namely those subsets for which every pair of points has a path of edges joining them. But it is not always possible to find a topology on the set of points which induces the same connected sets. The 5-cycle graph (and any n-cycle with n>3 odd) is one such example.
As a consequence, a notion of connectedness can be formulated independently of the topology on a space. To wit, there is a category of connective spaces consisting of sets with collections of connected subsets satisfying connectivity axioms; their morphisms are those functions which map connected sets to connected sets (Muscat & Buhagiar 2006). Topological spaces and graphs are special cases of connective spaces; indeed, the finite connective spaces are precisely the finite graphs.
However, every graph can be canonically made into a topological space, by treating vertices as points and edges as copies of the unit interval (see topological graph theory#Graphs as topological spaces). Then one can show that the graph is connected (in the graph theoretical sense) if and only if it is connected as a topological space.
I was wondering if the two bold sentences contradict each other?
Thanks and regards!
Solution 1:
They don't contradict each other as they refer to completely different things.
To make it more precise: If $G$ denotes the $5$-cycle graph then the first statement implies that there is no topology on the vertices of $G$ which has the same connected subspaces.
On the other hand, In the second statement we are identifying the graph $G$ with a pentagon which is not the same as a $5$-element space.
Solution 2:
Actually the 5-cycle (as points) does not admit ANY topology (Hausdorff or not) which gives the same connected sets (ie the edges and paths). This follows by considering that each non-edge induces two open sets. One can prove that each open set will have to be either a set {x} or {y,x,z} where y-x-z are connected, with each type alternating. This implies in turn that any odd cycle cannot have a compatible topology (except 1,3).