In the real line in the usual topology the compact sets are the closed and bounded sets. In the left topology, the topology generated by the left-open intervals $(-,a)$, the compact sets are exactly the closed sets, which are the left-closed intervals $(-,a]$ and also it seems the real line itself - since the only open cover of it is itself.

Now the real line is a totally ordered set - a chain - how far can these results be generalised to this context?

In the ordinal $2\omega=\omega+\omega$, with the interval topology; a straight generalisation of the above would say that $[1,\omega+1]$ is compact because it is closed & bounded. But this can't be right since the interval topology just reduces to the discrete topology here; and thus a compact subset is simply a finite subset. The condition appears to be neccessary but not sufficient.

The left topology appears to generalises - it seems that the compact subsets are again exactly the closed subsets.

Finally, do these results generalise to posets?

In the left topology, the compact sets are not exactly the closed subsets (a counter-example being the disjoint union of a countable number of real lines). They seem to be charactised by having a finite number of maximal elements. Is that right? In the interval topology, a neccessary condition is that they have a finite number of maximal & minimal elements - but given the example of the ordinal above it can't be sufficient.

  • 1
    $\begingroup$ Most of the assertions here are wrong. ($1$) A non-empty set in $\Bbb R$ with what you call the left topology is compact if and only if it has a maximum element. In particular, $\Bbb R$ is not compact: $\{(\leftarrow,a):a\in\Bbb R\}$ is an open cover with no finite subcover. ($2$) $\omega+\omega$ is $\omega\cdot 2$,not $2\cdot\omega$. The interval topology on it is not discrete: the point $\omega$ is not isolated. In fact its compact subsets are the closed, bounded subsets. $\endgroup$ – Brian M. Scott Oct 9 '14 at 2:15
  • $\begingroup$ Ok, thanks for the corrections. I'll alter my question accordingly! $\endgroup$ – Mozibur Ullah Oct 9 '14 at 2:20
  • $\begingroup$ If I may ask a couple of questions: what is the standard name for what I call the 'left topology'? And what is the correct definition of 2 x omega? $\endgroup$ – Mozibur Ullah Oct 9 '14 at 2:24
  • $\begingroup$ I’m don’t think that that topology actually has a standard name; you could call it the left open ray topology, just to have a name to use, but you’d still have to define it. // In general, if $\alpha$ and $\beta$ are ordinals, the ordinal product $\alpha\cdot\beta$ is the ordinal with the same order type as the lexicographic (dictionary) order on $\beta\times\alpha$, not $\alpha\times\beta$. Pictorially this is what you get when you string $\beta$ copies of $\alpha$ together. Thus, $2\cdot\omega$ is just $\omega$ back again. $\endgroup$ – Brian M. Scott Oct 9 '14 at 2:28
  • $\begingroup$ Thankyou for the clarification. Is there a good reason for the reversal in the definition of ordinal multiplication, or is it simply conventional? $\endgroup$ – Mozibur Ullah Oct 9 '14 at 2:35

Your Answer

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

Browse other questions tagged or ask your own question.