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By previous question, if there is a elementary embedding from $\mathfrak A$ into $\mathfrak B$, then $\mathfrak A \equiv \mathfrak B$.

Now it is naturally to ask conversely, if $\mathfrak A \equiv \mathfrak B$, is there a elementary embedding to link them? Or there are $\mathfrak A,\mathfrak B$ such that $\mathfrak A \equiv \mathfrak B$ but none can be embedded to the other.


For example, real field $\mathbb R$ and hyperreal field $\mathbb R^*$ are elementary equivalent but $\mathbb R \prec \mathbb R^*$.

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Not in general. However, if $\left| \mathfrak{A} \right| < \lambda$ and $\mathfrak{B}$ is $\lambda$-universal, then there does exist an elementary embedding $\mathfrak{A} \to \mathfrak{B}$. (This is essentially the definition of $\lambda$-universality.)

For example, let $\Sigma$ be a signature with two unary relation symbols $X$ and $Y$, and let $\mathfrak{A}$ and $\mathfrak{B}$ be the $\Sigma$-structures where $\left| X^\mathfrak{A} \right| = \aleph_0$, $\left| Y^\mathfrak{A} \right| = \aleph_3$, $X^\mathfrak{A} \cap Y^\mathfrak{A} = \emptyset$, $X^\mathfrak{A} \cup Y^\mathfrak{A} = \mathfrak{A}$; $\left| X^\mathfrak{B} \right| = \aleph_1$, $\left| Y^\mathfrak{B} \right| = \aleph_2$, $X^\mathfrak{B} \cap Y^\mathfrak{B} = \emptyset$, $X^\mathfrak{B} \cup Y^\mathfrak{B} = \mathfrak{B}$. Obviously $\mathfrak{A}$ and $\mathfrak{B}$ are elementarily equivalent, but for cardinality reasons there cannot exist any elementary embedding of one into the other.

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  • $\begingroup$ I understand, they can be checked elementarily equivalent by EF-Game. Thank you very much. $\endgroup$ – Popopo Dec 8 '12 at 16:04
  • $\begingroup$ Is there any other (fairly basic) definition of $\lambda$-universality? $\endgroup$ – tomasz Dec 9 '12 at 14:41

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