If a topological space is a path-connected, it is also connected. However, the converse of this theorem is false.
Can we generalize the notion of a path so that the converse also holds? I was thinking that maybe if the domain of a path was allowed to be an arbitrary connected totally ordered set, this might fix the problem.
EDIT. As Brian M. Scott explains in his answer, this doesn't fix the problem; so here's a last ditch-effort at salvaging the idea. Definition: A connected path in a topological space $X$ is a mapping $\gamma : T \rightarrow X,$ that preserves connectedness under direct images, where $T$ is a connected totally ordered set possessing both a least and a greatest element. Note in particular that $\gamma$ needn't be continuous.
Can anyone see whether this fixes the problem or not? We want to be able to prove that a topological space is path connected iff it is connected.
Anyway, here's the motivation.
The problem, as Brian explains, is that a connected linearly ordered space with endpoints is compact, so the continuous image of such a space is also compact and connected. This is a problem, because:
- No compact, connected subset of the topologist’s sine curve contains the origin and at least one other point of the curve, and
- There exist countably infinite connected Hausdorff spaces, and there are no paths between distinct points in such a space; because if such a path existed, its image would be a compact countably infinite connected Hausdorff space with at least two points, and no such space exist.
So the problem seems to be the compactness of the image of the path, which is implied by the compactness of the domain, because the path is defined as a continuous function. Thus, weakening the requirement that paths need to be continuous might fix the problem.