How can a bounded subspace of the left order topology be compact?

Question

I want to show that every bounded set equipped with the left order topology is compact. This is a statement I found on a wikipedia page and appearently it is lifted from the book Counterexamples in Topology. But how can this be true? Look at this bounded set: $(0,1)$. I can easily find an open covering $A= \{ (0, 1- 1/n) | n \in \mathbb{N} \wedge 2 \leq n\}$, which does not seem to have a finite subcollection covering $(0,1)$.

I have got it from this wikipedia page: http://en.wikipedia.org/wiki/Left_order_topology#Left_and_right_order_topologies. What is wrong with my interpretation?

Thanks!

Answer 1

Wikipedia does appear to be a bit ambiguous, but it is more about the meaning of a "bounded" set. The following is true:

Proposition. The left-order topology on a linearly ordered set $( L , < )$ is compact iff $L$ has a greatest element.

Proof. ($\Rightarrow$) If $L$ does not have a greatest element, then consider the family of all open sets of the form $( - \infty , b )$ for $b \in B$. Given any finitely many $b_1 , \ldots , b_n$ of elements of $L$, one of these, say $b_1$, is the maximum among these. But then $b_1 \notin ( - \infty , b_i )$ for any $i \leq n$. ($\Leftarrow$) If $L$ does have a greatest element, $b$, then the only open set containing $b$ is $L$ itself.\

So it appears that when Wikipedia says "bounded" in this article they mean that the linearly ordered set has either a greatest or least element according to whether you are considering the left- or right-order topology.