I know this is a well known result yet I'm having trouble proving it.

The open long line is defined as:

$L = \omega_1 \times [0,1)$ without its minimal element where $\omega_1$ is the first uncountable ordinal

The extended open long line is:

$L$*$ = \omega_2 \times [0,1)$ without its minimal element, where $\omega_2$ is the successor of $\omega_1$.

Can you give me a hint as to how to prove $L$* isn't path connected, without using any notions of compactness? I know I need to choose points like $\{\{\emptyset\},0\}$ and $\{\omega_1,0\}$ and I thought of using the intermediate value theorem - but then a cardinality argument doesn't suffice.

  • $\begingroup$ Be careful. Those two points are (1,0) and (omega_1,0). $\endgroup$ – William Elliot May 10 '17 at 10:05
  • $\begingroup$ Not an answer but I think it is very interesting that the extended long line is "not path-connected", not because it has some gap or something, but just because it is (somehow) too long for any path to reach from one end to the other. This might be more of a flaw of the paths we are considering, than a flaw of $L^*$. $\endgroup$ – M. Winter May 10 '17 at 10:54

Assume there is a path from a = (1,0) to b = (omega_1,0).
Since the long line is Hausdorff, there is an arc from a to b.
Show the arc is increasing and use that to construct an order
embedding of omega_1 + 1 into [0,1], an impossibility.

  • $\begingroup$ Thanks for the answer; what do you mean by an arc? Can you reference why an order embedding as such is impossible? $\endgroup$ – Mariah May 10 '17 at 17:26
  • $\begingroup$ @Mariah An arc within an Hausdorff space is an injective path. If f order imbedds omega_1 into R, then { (f(xi), f(xi+1) | xi < omega_1 } is an uncountable collection of pairwise disjoint sets, each of which contains a rational, an impossibility. $\endgroup$ – William Elliot May 11 '17 at 7:42

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