Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb R \prec \mathbb R^*$.

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.