0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Is there a question here? $\endgroup$ – Eric Stucky May 11 '16 at 7:50
  • $\begingroup$ Is it a contradiction that this path-connected space contains points that cannot be connected by a path? -- I didn't state it clearly previously $\endgroup$ – Kevin Yang May 11 '16 at 7:56
  • 1
    $\begingroup$ This looks similar to the long line, or possibly the one-point compactification of the long line. See the comments to this answer. $\endgroup$ – Dan Rust May 11 '16 at 10:05
  • $\begingroup$ @DanRust Thanks a lot $\endgroup$ – Kevin Yang May 11 '16 at 13:27
1
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ @Kevin: You’re very welcome! $\endgroup$ – Brian M. Scott May 11 '16 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.