Connected Linear Graph not Path-Connected

Question

Given a set of vertices $\{x_\alpha\}$ whose cardinality exceeds $\aleph_1$, (assume the axiom of choice) connect each vertex with its successor by an edge, forming a linear graph. Choose two vertices $x_i$ and $x_j$ such that $card\{x|x_i \lt x \lt x_j\} \gt \aleph_1$. On one hand, it is impossible to construct a path corresponding to the edge path from $x_i$ to $x_j$, since there does not exist a continuous, surjective map $\mathcal f: \mathbf I \to \mathbf S$, from the unit interval to the subgraph $\mathbf S$ between $x_i$ and $x_j$. On the other hand, the graph is connected and locally path-connected, and thereby path-connected, so that there must exist a path from $x_i$ to $x_j$, a contradiction.

Answer 1

You’re essentially looking at the lexicographic order topology on $\preceq$ on $X=\alpha\times[0,1)$ for an ordinal $\alpha\ge\omega_2$. This $X$ is not locally path-connected: the points $\langle\eta,0\rangle$ such that $\operatorname{cf}\eta\ge\omega_1$ (and hence in particular the point $\langle\omega_1,0\rangle$) do not have path-connected nbhds. Thus, there is no contradiction.

If you take $\alpha=\omega_1$, so that you have a set of vertices of cardinality $\omega_1$, you’re looking at the closed long ray, which is connected and locally path-connected (and hence path-connected). If, however, you add a righthand endpoint, you have (up to homeomorphism) a subspace of $X$ (above), $\{x\in X:x\preceq\langle\omega_1,0\rangle\}$, with righthand endpoint $\langle\omega_1,0\rangle$. As noted above, this point does not have a path-connected nbhd, and indeed this space is not path-connected: there is no path from and $x\prec\langle\omega_1,0\rangle$ to $\langle\omega_1,0\rangle$. (As Dan Rust suggested in the comments, this answer and the comments below it may be helpful.)