Let $\mathbb{R}\_\*=\mathbb{R}^\omega/\mathcal U$ for some ultrafilter $\cal U$. In the definitions of [this question](https://mathoverflow.net/q/72612/118366) and assuming ZFC + CH there are only three types of cuts in $\mathbb{R}\_\*$: $(\omega,\omega\_1),~(\omega\_1,\omega),~(\omega\_1,\omega\_1)$. And only $(\omega\_1,\omega\_1)$ cut could be filled. For an ordered field $\mathbb F$ lets say that a cut $\mathbb{F}=A\coprod B$ is *good* iff for any $c>0$ there are $a\in A$ and $b\in B$ such that $b-a<c$. Any good cut is of type $(\lambda,\lambda)$ where $\lambda$ is cofinality of $\mathbb F$. If any good cut is filled lets call the field *quasi-complete*. Any ordered field can be embedded into quasi-complete field with appropriate universal properties ([see here](https://mathoverflow.net/a/140962/118366)).


Can we proof that (may be for special ultrafilter $\cal U$):


1. $\mathbb{R}\_\*$ is quasi-complete ?
2. If $\mathbb{F}$ is quasi-complete then $\mathbb{F}\_\*=\mathbb{F}^\omega/\mathcal{U}$ is quasi-complete ?
3. Any $(\omega\_1,\omega\_1)$ cut in $\mathbb{R}\_\*$ is good ?



On a completeness property of hyperreals
This is also called Cauchy-completeness, and it coincides for non-Archimedean ordered fields with the natural valuation to the valuation-theoretic notion of completeness. Also, this is the same as having no proper dense ordered field extension.


I will say that an ordered pair $(A,B)$ of subsets of an ordered field $F$ is a *cut generator* if $A<B$ and there is no $x \in F$ with $A<x<B$. If $(A,B)$ is a cut generator, then the pair $(A',B')$ where $A'$ is the set of lower bounds of elements of $A$ and $B'$ is the set of upper bounds of elements of $B$ is a cut in $F$ whose type is $(\operatorname{cf}(A,<),\operatorname{cf}(B,>))$.


The answer to your first question is negative. In ZFC+CH, the field $\mathbb{R}\_\*$ is unique up to isomorphism to be real-closed, of cardinality continuum / $\aleph\_1$ and without $(\omega,\omega)$ type cuts (in particular, the choice of ultrafilter doesn't matter). So it is isomorphic to the field $\mathbf{No}(\omega\_1)$ of surreal numbers with countable birth day. I use the latter because its elements can be represented in a more explicit way, and this helps find good cuts in $\mathbf{No}(\omega\_1)$. For instance, using the sign-sequence presentation of surreal numbers, you have the cut generator $(A,B)$ where $A$ is the set of numbers whose sign-sequence is a concatenation of $(+-)$, and $B$ is the set of numbers obtained by adding $+$ at the end of the sign-sequence of elements of $A$. So $A=\{(),(+-),(+-+-),...\}$ and $B=\{(+),(+-+),...\}$ (the dots hide uncountably many numbers). One can also define $a\_{\gamma}:=\{a\_{\rho} :\rho<\gamma\ | \ b\_{\rho}:\rho<\gamma \}$ and $b\_{\gamma}:= \{a\_{\gamma} \ | \ b\_{\rho}:\rho<\gamma \}$ by induction on $\gamma<\omega\_1$ and obtain $A=\{a\_{\gamma} \ : \ \gamma<\omega\_1\}$ and $B=\{b\_{\gamma} \ : \ \gamma<\omega\_1\}$.


It should be possible to prove the result directly in $\mathbb{R}\_\*$ using $\mathbb{N}\_\*$-indexed sums of fastly growing sequences, but by transfer it's actually easy to obtain a convergent sum, and thus not a cut!


Also, I don't know if ZFC alone proves that $\mathbb{R}\_\*$ is not quasi-complete. In fact I may have asked this exact question on MSE or MO.




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Using the same isomorphism $\mathbb{R}\_\* \cong \mathbf{No}(\omega\_1)$, one can obtain cuts of type $(\omega\_1,\omega\_1)$ that are not good. This again implies some familiarity with surreal numbers, so you can admit the existence of a map $x\mapsto \omega^x: \mathbf{No}(\omega\_1)\rightarrow \mathbf{No}(\omega\_1)^{>0}$, sometimes called the $\omega$-map, which is strictly increasing, with



> 
> $\forall x\in \mathbf{No}(\omega\_1)^{>0},\exists ! d\_x \in
> > \mathbf{No}(\omega\_1), \exists r \in \mathbb{R}^{>0}, r^{-1} \omega^{d\_x}<x<r\omega^{d\_x}$
> 
> 
> 


Then $(\omega^A,\omega^B)$ (where $\omega^X=\{\omega^x \ : \ x \in X\}$ is a cut generator, whose corresponding cut has type $(\omega\_1,\omega\_1)$. Indeed if there were a number $c \in \mathbf{No}(\omega\_1)$ between $\omega^A$ and $\omega^B$, then $d\_c$ would have to lie between $A$ and $B$. The corresponding cut is not good, because we have $x+1<y$ for all $(x,y) \in \omega^A\times \omega^B$.


Everything I wrote requires some justification so feel free to ask if you want me to elaborate.


