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I corrected the spelling of Keisler.

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edited Aug 27, 2013 at 2:09

Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for each first order structure $\mathcal{A}$, the ultrapower $\mathcal{A}^{\mathcal{U}}$ is elementarily embeddable in $\mathcal{A}^{\mathcal{V}}$, and $\mathcal{U}=_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-KieslerKeisler equivalent to $\mathcal{V}$) iff $\mathcal{A}^{\mathcal{U}}$ is isomorphic to $\mathcal{A}^{\mathcal{V}}$ for each first order structure $\mathcal{A}$. If $\mathcal{U},\mathcal{V}$ are ultrafilters, then there is a unique up-to Rudin-KieslerKeisler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ is isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ for all first order structures $\mathcal{A}$. I have a few questions about the product operation on ultrafilters.

If $\mathcal{U}\cdot\mathcal{V}=_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{V}\leq_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}=_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}\leq_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{U}=_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}=_{RK}\mathcal{V}$?

If $\mathcal{U}\cdot\mathcal{U}\leq_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}\leq_{RK}\mathcal{V}$?

I would also be interested in any proof or reference of similar results about the product of ultrafilters.

Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for each first order structure $\mathcal{A}$, the ultrapower $\mathcal{A}^{\mathcal{U}}$ is elementarily embeddable in $\mathcal{A}^{\mathcal{V}}$, and $\mathcal{U}=_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Kiesler equivalent to $\mathcal{V}$) iff $\mathcal{A}^{\mathcal{U}}$ is isomorphic to $\mathcal{A}^{\mathcal{V}}$ for each first order structure $\mathcal{A}$. If $\mathcal{U},\mathcal{V}$ are ultrafilters, then there is a unique up-to Rudin-Kiesler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ is isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ for all first order structures $\mathcal{A}$. I have a few questions about the product operation on ultrafilters.

If $\mathcal{U}\cdot\mathcal{V}=_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{V}\leq_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}=_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}\leq_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{U}=_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}=_{RK}\mathcal{V}$?

If $\mathcal{U}\cdot\mathcal{U}\leq_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}\leq_{RK}\mathcal{V}$?

I would also be interested in any proof or reference of similar results about the product of ultrafilters.

Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for each first order structure $\mathcal{A}$, the ultrapower $\mathcal{A}^{\mathcal{U}}$ is elementarily embeddable in $\mathcal{A}^{\mathcal{V}}$, and $\mathcal{U}=_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler equivalent to $\mathcal{V}$) iff $\mathcal{A}^{\mathcal{U}}$ is isomorphic to $\mathcal{A}^{\mathcal{V}}$ for each first order structure $\mathcal{A}$. If $\mathcal{U},\mathcal{V}$ are ultrafilters, then there is a unique up-to Rudin-Keisler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ is isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ for all first order structures $\mathcal{A}$. I have a few questions about the product operation on ultrafilters.

If $\mathcal{U}\cdot\mathcal{V}=_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{V}\leq_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}=_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}\leq_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{U}=_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}=_{RK}\mathcal{V}$?

If $\mathcal{U}\cdot\mathcal{U}\leq_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}\leq_{RK}\mathcal{V}$?

I would also be interested in any proof or reference of similar results about the product of ultrafilters.

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asked Aug 26, 2013 at 18:06

Joseph Van Name

Is the product of ultrafilters cancellative?

Suppose that $\mathcal{U},\mathcal{V}$ are ultrafilters on sets. Recall that $\mathcal{U}\leq_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Keisler less than or equal to $\mathcal{V}$) iff for each first order structure $\mathcal{A}$, the ultrapower $\mathcal{A}^{\mathcal{U}}$ is elementarily embeddable in $\mathcal{A}^{\mathcal{V}}$, and $\mathcal{U}=_{RK}\mathcal{V}$ (here we say $\mathcal{U}$ is Rudin-Kiesler equivalent to $\mathcal{V}$) iff $\mathcal{A}^{\mathcal{U}}$ is isomorphic to $\mathcal{A}^{\mathcal{V}}$ for each first order structure $\mathcal{A}$. If $\mathcal{U},\mathcal{V}$ are ultrafilters, then there is a unique up-to Rudin-Kiesler equivalence ultrafilter $\mathcal{U}\cdot\mathcal{V}$ such that $\mathcal{A}^{\mathcal{U}\cdot\mathcal{V}}$ is isomorphic to $(\mathcal{A}^{\mathcal{U}})^{\mathcal{V}}$ for all first order structures $\mathcal{A}$. I have a few questions about the product operation on ultrafilters.

If $\mathcal{U}\cdot\mathcal{V}=_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{V}\leq_{RK}\mathcal{U}\cdot\mathcal{W}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}=_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}=_{RK}\mathcal{W}$?

If $\mathcal{V}\cdot\mathcal{U}\leq_{RK}\mathcal{W}\cdot\mathcal{U}$, then do we necessarily have $\mathcal{V}\leq_{RK}\mathcal{W}$?

If $\mathcal{U}\cdot\mathcal{U}=_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}=_{RK}\mathcal{V}$?

If $\mathcal{U}\cdot\mathcal{U}\leq_{RK}\mathcal{V}\cdot\mathcal{V}$, then do we necessarily have $\mathcal{U}\leq_{RK}\mathcal{V}$?

I would also be interested in any proof or reference of similar results about the product of ultrafilters.

set-theory ultrafilters