Is $(\omega+1)^\omega/{\cal U}$ complete for ${\cal U}$ free ultrafilter? Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?
 A: 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.
A: No. $(\omega+1)^{\omega}/\mathcal{U}$ is not complete whenever $\mathcal{U}$ is a non-principal on $\omega$. Observe that if $(\omega+1)^{\omega}/\mathcal{U}$ is complete, then $(\omega+1)^{\omega}/\mathcal{U}$ is compact in the order topology. In fact, a linearly ordered set is compact in the order topology if and only if it is complete as a linear order.
Let $U=(\omega+1)^{\omega}/\mathcal{U}\setminus\omega$ (i.e. $U$ is the collection of all non-standard natural numbers). Then
$\{U\}\cup\{\{n\}\mid n\in\omega\}$ is an open cover of $(\omega+1)^{\omega}/\mathcal{U}$ with no finite subcover.
More generally, I claim that if $\mathcal{U}$ is a non-$\sigma$-complete ultrafilter on an index set $I$, and $X_{i}$ is a totally ordered set whenever $i\in I$, then the ultraproduct
$\prod_{i\in I}X_{i}/\mathcal{U}$ is only complete if it is finite.
The ultraproduct $\prod_{i\in I}X_{i}/\mathcal{U}$ is necessarily $\aleph_{1}$-saturated. But every $\aleph_{1}$-saturated linearly ordered set is a $P$-space in the order topology. The only compact $P$-spaces are the finite spaces. Furthermore, if $X$ is an $\aleph_{1}$-saturated linearly ordered set or if $X$ is a $P$-space in the order topology, then one can easily show that no strictly increasing sequence $(x_{n})_{n\in\omega}$ in $X$ has a least upper bound, and no strictly decreasing sequence $(x_{n})_{n\in\omega}$ in $X$ has a greatest lower bound.
