Maximum and Minimum of Totally Ordered Compact Sets Let T be a totally ordered compact set. Does this always imply that a maximum and a minimum element of T exist under this total order?
And if not, what about the special case where:


*

*T is a closed and bounded interval of the real line (ergo compact) with the usual total order $\leq$ of the reals

 A: If $T$ is a compact set in a linearly ordered space $X$ with its order topology, then $T$ has both a minimum and a maximum element. 
To see this, merely note that if $T$ has no largest element, then $\{(\leftarrow,x):x\in T\}$ is an open cover of $T$ with no finite subcover, and if $T$ has no smallest element, then $\{(x,\to):x\in T\}$ is an open cover of $T$ with no finite subcover.
In particular, this is true if $X$ is the real line with the usual topology, since that topology is the one induced by the usual linear order on $\Bbb R$. If the linear order is not related to the topology, however, nothing can be said: it is always possible to put a linear order on an infinite set in such a way that the set has neither a least nor a greatest element.
A: First you need some compatibility conditions the reason is the following. Take $T:=[0,1]$ the segment in $\mathbb{R}$ with its usual topology, it is well known that there exists a bijection $f$ between $[0,1]$ and $\mathbb{R}$. 
If you define $\leq_{[0,1]}$ to be the total order induce by $f$ i.e. :
$$a\leq_{[0,1]}b\Leftrightarrow f(a)\leq f(b) $$ 
Then, somehow $T$ is  a totally ordered compact set (i.e. some compact set with a total order) but won't have maximal element nor minimal element...
That is you must have a condition relating both that is (I think) for every $a\in T$ you must have :
$$\{t\in T|a\leq t\}\text{ and } \{t\in T|t\leq a\}\text{ are closed.} $$
i.e. :
$$\{t\in T|t< a\}\text{ and } \{t\in T|a<t\}\text{ are open.} $$
Now if $T$ has no maximal element then for all $t\in T$ you can find $t'\in T$ with $t<t'$ (the order is total) finally :
$$\bigcup_{a\in A}\{t\in T|t< a\}=T $$
But then you can take a finite cover out of it let's say for $a_1,...,a_k$ but then taking the maximal element of those $k$ elements (let's say it is $a_k$) then you have :
$$T=\bigcup_{a\in A}\{t\in T|t< a\}=\bigcup_{i=1}^k\{t\in T|t< a_i\}=\{t\in T|t< a_k\} $$
Finally you have that $a_k\notin T$ which is a contradiction. Of course, the same applies to the minimal element.
