Construct an sequence in order set Assume that we have an uncountable ordered set $\Theta$. 
Can we construct a sequence $(x_n)$ which $x_1<x_2<...<x_n<...$. In addition, for each $x\in \Theta$, there exists $n$ such that $x_n>x$?
 A: You cannot always do so. Let $\Theta$ be $\omega_1$, the first uncountable ordinal, in its natural order; then every strictly increasing sequence in $\Theta$ has an upper bound. $\Theta$ is well-ordered, so if you reverse the order, then every strictly increasing sequence is finite, and you can’t even get an infinite strictly increasing sequence.
Let $\langle\Theta,\preceq\rangle$ be a (non-strict) linear order. A subset $C$ of $\Theta$ is cofinal in $\Theta$ if for each $x\in\Theta$ there is a $y\in C$ such that $x\preceq y$. The cofinality of a linear order is the smallest cardinality of a cofinal subset. Your condition is precisely that $\Theta$ have cofinality $\omega$, i.e., countably infinite cofinality. Some uncountable linear orders do; $\Bbb R$ in its usual order is the obvious example. Some have uncountable cofinality; $\omega_1$ in its usual order is an example, as noted above. And some, like $\omega_1$ in its reversed order, have cofinality $1$, because they have a maximum element.
