Long line is connected and compact How to prove that the long line is connected and compact. I was trying  to prove connectedness using contradiction but couldn't.
 A: You can actually show that the long line is path-connected, which shows that it is connected. Pick any two points $x=\langle\alpha,s\rangle$ and $y=\langle\beta,t\rangle$ on the long line, with $x<y$. If $\alpha=\beta,$ then $s<t$ and the long line interval $[x,y]$ is readily homeomorphic to the real interval $[s,t]$, so $x,y$ are connected by a path. Otherwise, the long line interval $[x,y]$ is the union of (in increasing order) $\alpha\times[s,1)$, then (at most) countably-infinitely-many intervals of the form $\gamma\times[0,1)$ with $\alpha<\gamma<\beta,$ then $\beta\times[0,t],$ joined end to end. You should again be able to show an explicit homeomorphism with a closed real interval, so that $x,y$ are connected by a path. As a hint for how to do this, note that $$[0,1)\cup\left(\bigcup_{n\in\Bbb Z^+}\left[\frac{2^{n+1}-1}{2^n},\frac{2^{n+2}-1}{2^{n+1}}\right)\right)\cup[2,3]=[0,3].$$
However, the long line is not compact. For example, let $z$ be the deleted least element of the long line, and for each $\alpha<\omega_1,$ let $z_\alpha=\langle\alpha,0\rangle.$ Then the set of long line intervals $(z,z_\alpha)$ forms an open cover of the long line with no finite subcover (possibly not even a countable subcover, but that is not known). It is, however, locally compact, as every element is contained in a non-degenerate closed long line interval homeomorphic to a closed and bounded real interval.
