Let ${\cal U}$ be a free ultrafilter on $\omega$. Is the linearly ordered set $(\omega+1)^\omega/{\cal U}$ complete?
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asked Dec 12, 2021 at 14:12
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1$\begingroup$ This definition clarifies this problem. en.wikipedia.org/wiki/Complete_lattice $\endgroup$– Joseph Van NameCommented Dec 12, 2021 at 15:43
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5$\begingroup$ You already have two correct answers, answering considerably more than you asked. But to just answer your question negatively, it suffices to notice that the standard natural numbers (the equivalence classes mod $\mathcal U$ of the constant functions $\omega\to\omega$) have no least upper bound. $\endgroup$– Andreas BlassCommented Dec 13, 2021 at 0:41
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2 Answers
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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.
answered Dec 12, 2021 at 16:04
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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.
