Consider $(\mathbb{Q}; \leq)$ and let $T$ be the theory of dense linear orderings without endpoints.

Let $\mathfrak{A}$ be the $\omega_1$ saturated model of $T$. Note that $|\mathfrak{A}|=2^{\aleph_0}$ since $\mathbb{R} \subset A$. However, $\mathbb{R} \neq A$ since $\mathfrak{A}$ has elements of the form "I am smaller than every positive rational, but larger than 0". Since $\mathbb{Q}$ is dense in $\mathbb{R}$, this element is not in $\mathbb{R}$. However, since $\mathfrak{A}$ is only $\omega_1$ saturated, we don't immediately get elements of the form "I am smaller than every positive real number, but larger than zero". For this to be the case, $\mathfrak{A}$ must be ${2^{\aleph_0}}^{+}$- saturated. Therefore, in some model which is ${2^{\aleph_0}}^+$- saturated, it is clear that $\mathbb{R}$ is not dense in that model.

Question: Is $\mathbb{R}$ dense in $\mathfrak{A}$ (the $\omega_1$ saturated model)?

  • $\begingroup$ But if you are smaller than every positive rational you are also smaller that every positive real. $\endgroup$ – Rene Schipperus Aug 11 '14 at 5:24
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    $\begingroup$ The question is a bit funny, since there's not a canonical embedding of $\mathbb{R}$ in $\mathfrak{A}$. But the point is that no embedding of $\mathbb{R}$ in $\mathfrak{A}$ is dense. $\endgroup$ – Alex Kruckman Aug 11 '14 at 5:30
  • $\begingroup$ @AlexKruckman: Do you have any advice in rephrasing my question? $\endgroup$ – Kyle Aug 11 '14 at 5:36
  • $\begingroup$ Sure: Observe that $\mathbb{R}$ embeds in $\mathfrak{A}$, since certainly $\mathbb{Q}$ embeds in $\mathfrak{A}$, and we can extend the embedding to $\mathbb{R}$ using $\aleph_1$-saturation, since the type of each element of $\mathbb{R}$ is determined by its type over $\mathbb{Q}$. Is any embedding of $\mathbb{R}$ in $\mathfrak{A}$ dense? The answer is no, as per Rene Schipperus's answer. $\endgroup$ – Alex Kruckman Aug 11 '14 at 5:41

If $\mathbb{R}$ were dense then so to would be $\mathbb{Q}$.


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