Proving that a well-ordered set with the order topology is not homeomorphic to any initial segment of it

Asked Mar 17 '14 at 19:53

Active Mar 17 '14 at 19:58

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Let $\langle X,\leq_X\rangle$ be a well-ordered set. I want to show that $X$ with the topology on $X$ relative to $\leq_X$ (the order topology) is not homeomorphic to any initial segment of $X$. And then I need to conclude that any two well-ordered sets $\langle X,\leq_X\rangle$ and $\langle Y,\leq_Y\rangle$ are homeomorphic iff they are isomorphic. I can feel both seem true, but while the first one I don't know how to proof, the second one I can prove, but not as a conclusion of the first. Any ideas on how can I show the second one as a conclusion of the first? And on how can I prove the first?

Thanks!

edited Mar 17 '14 at 19:58

Michael Hardy

asked Mar 17 '14 at 19:53

Roman Vale

2

This is not true: $\omega+1$ and $\omega+2$ are homeomorphic. See here. – Andrés E. Caicedo Mar 17 '14 at 19:59
$< X,\leq_X>$ is even worse than most instances of the use of that symbol as a substitute for angle brackets. I changed it to $\langle X,\leq_X\rangle$. ${}\qquad{}$ – Michael Hardy Mar 17 '14 at 20:07

Proving that a well-ordered set with the order topology is not homeomorphic to any initial segment of it

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