On a completeness property of hyperreals Let $\mathbb{R}_*=\mathbb{R}^\omega/\mathcal U$ for some ultrafilter $\cal U$. In the definitions of this question 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).
Can we proof that (may be for special ultrafilter $\cal U$):

*

*$\mathbb{R}_*$ is quasi-complete ?

*If $\mathbb{F}$ is quasi-complete then $\mathbb{F}_*=\mathbb{F}^\omega/\mathcal{U}$ is quasi-complete ?

*Any $(\omega_1,\omega_1)$ cut in $\mathbb{R}_*$ is good ?

 A: 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.

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.
