Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?



Is $(\omega+1)^\omega/{\cal U}$ complete for ${\cal U}$ free ultrafilter?
No. Every ultraproduct by a free ultrafilter on $\omega$ is $\aleph\_1$-saturated. And infinite $\aleph\_1$-saturated linear orders cannot be complete!


*Proof:* Let $L$ be an infinite $\aleph\_1$-saturated linear order. Since $L$ is infinite, it contains an infinite increasing sequence or an infinite decreasing sequence. Without loss of generality, let's say we have an increasing sequence $(a\_n)\_{n\in\omega}$.


By $\aleph\_1$-saturation, the partial type $\{x>a\_n\mid n\in\omega\}$ is realized in $L$, so the set $\{a\_n\mid n\in\omega\}$ is bounded above. Suppose $b$ is an upper bound. Then the partial type $\{x>a\_n\mid n\in\omega\}\cup \{x<b\}$ is realized in $L$, so $b$ is not a least upper bound. Thus $L$ is not complete.


