Cardinality of Dedekind Completion of Hyperreals Let $^*\mathbb{R}$ denote the hyperreal field constructed as an ultra power of $\mathbb{R}$. At the expense of losing the field properties, we may take the Dedekind completion of $^*\mathbb{R}$ to get a new totally ordered set. Denote the Dedekind completion by $\overline{^*\mathbb{R}}$.
I’m curious what the cardinality of $\overline{^*\mathbb{R}}$ is. It is easy to see that $|^*\mathbb{R}| = |\mathbb{R}|$ from the ultra power construction. Since $^*\mathbb{R}$ is dense in $\overline{^*\mathbb{R}}$, it follows that $|\mathbb{R}| \leq |\overline{^*\mathbb{R}}| \leq 2^{|\mathbb{R}|}$.
There are some intuitive arguments I can make for $|\overline{^*\mathbb{R}}|$ living at either extreme. I have thus far been unsuccessful in proving anything of substance though.
 A: Here is a partial answer.
For any infinite cardinal $\kappa$, we write $\mathrm{ded}(\kappa)$ for the supremum, over all linear orders $X$ of size $\kappa$, of the cardinality of the set of Dedekind cuts in $X$. We always have $\kappa<\mathrm{ded}(\kappa)\leq 2^\kappa$.
Now suppose $S$ is a saturated dense linear order of cardinality $\kappa$. I claim that $S$ has the maximal number of Dedekind cuts, i.e., the number of Dedekind cuts in $S$ is $\mathrm{ded}(\kappa)$.
Since $\mathrm{ded}(\kappa)$ is defined as a supremum, and $\kappa<\mathrm{ded}(\kappa)$, it suffices to show that for any $\lambda$ with $\kappa<\lambda\leq \mathrm{ded}(\kappa)$, if there is a linear order $X$ with $\lambda$-many Dedekind cuts, then $S$ has at least $\lambda$-many Dedekind cuts.
Now since $S$ is a saturated dense linear order and $|X|\leq |S|$, there is an order embedding $e\colon X\to S$. For each Dedekind cut $(L,R)$ in $X$, $(e(L),e(R))$ generates a Dedekind cut in $S$, unless it is "filled", i.e., there are elements of $S$ between $e(L)$ and $e(R)$. But since $|S| = \kappa$, at most $\kappa$-many of the images of the cuts in $X$ are filled in $S$. Since $\lambda>\kappa$, $\lambda$-many cuts in $X$ have images that are cuts in $S$. This completes the proof that $S$ has $\mathrm{ded}(\kappa)$-many Dedekind cuts.

How is this relevant to the hyperreals? Let $^*\mathbb{R}$ be an ultrapower of $\mathbb{R}$ by a non-principal ultrafilter on $\omega$. Then $^*\mathbb{R}$ is $\aleph_1$-saturated. If we assume CH, $|^*\mathbb{R}| = \aleph_1$, so the underlying linear order of $^*\mathbb{R}$ is $\aleph_1$-saturated and dense. By the result above, the number of Dedekind cuts in $^*\mathbb{R}$ is $\mathrm{ded}(\aleph_1)>\aleph_1$.
Thus, assuming CH, $|\overline{^*\mathbb{R}}|>\aleph_1$. And if we also assume $2^{\aleph_1} = \aleph_2$, then $|\overline{^*\mathbb{R}}|\leq 2^{|\mathbb{R}|} = \aleph_2$, so we obtain that $|\overline{^*\mathbb{R}}| = \aleph_2 = 2^{|\mathbb{R}|}$.
I suspect it is consistent that CH holds but $\mathrm{ded}(\aleph_1)<2^{\aleph_1}$ (can anyone confirm?). If so, then there is a model of ZFC in which $2^{\aleph_0} = \aleph_1 < \mathrm{ded}(\aleph_1) < 2^{2^{\aleph_0}}$, so $|\mathbb{R}| <|\overline{^*\mathbb{R}}| < 2^{|\mathbb{R}|}$.
I have no idea what happens if CH fails. It seems likely that without CH, the cardinality of $\overline{^*\mathbb{R}}$ depends on the ultrafilter you use to take the ultrapower.
