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How to prove that the long line is connected and compact. I was trying to prove connectedness using contradiction but couldn't.

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  • $\begingroup$ Page 159 of Munkres. Exercies 12*. Thanks for asking it. $\endgroup$ – Dutta Oct 4 '13 at 16:55
  • $\begingroup$ Long line is really compact? $\endgroup$ – Hanul Jeon Oct 4 '13 at 17:01
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    $\begingroup$ @tetori It's normal, but not paracompact. $\endgroup$ – Michael Hoppe Oct 4 '13 at 17:10
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    $\begingroup$ What is "the long line"? $\endgroup$ – André Caldas Oct 4 '13 at 17:52
  • $\begingroup$ @AndréCaldas: It is the space $\omega_1\times[0,1)$ in the lexicographic order topology, with its lease element removed. $\endgroup$ – Cameron Buie Oct 4 '13 at 18:31
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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.

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    $\begingroup$ $L := \omega_1 \times [0,1)$ in the order topology is path connected, but not compact. $L^*$, the one-point compactification of $L$, is however not path connected. $0$ cannot be joined with $\infty$ by a continuous path. It is "too far away" $\endgroup$ – kahen Oct 4 '13 at 19:32
  • $\begingroup$ That's hilarious! It's worth noting that $L^*$ is connected, however, since $L$ is connected and dense in $L^*$. $\endgroup$ – Cameron Buie Oct 4 '13 at 19:34
  • $\begingroup$ @CameronBuie Thanks... $\endgroup$ – akansha Oct 5 '13 at 15:33
  • $\begingroup$ @manisha: No problem. Let me know if you get stuck along the way of constructing your homeomorphism, and I'll see if I can give you some tips to get unstuck. $\endgroup$ – Cameron Buie Oct 5 '13 at 15:37
  • $\begingroup$ Excuse me, can you explain why the map between $f : [x,y] \to [0,3]$ is homeomorphism? Explicitly, why $f$ is continuous at $(\beta,0)$? (Note that $f(\beta,0)=2)$ and $f^{-1}$ is cuntinuous at $2$? Thank you. $\endgroup$ – Andrews Sep 29 at 6:35

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