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Let $U$ is an ultrafilter on a set $X$, and $V$ an ultrafilter on a set $Y$.

Wikipedia says: Ultrafilters $U$ and $V$ are Rudin–Keisler equivalent, $U\equiv_{RK}V$, if there exist sets $A\in U$, $B\in V$, and a bijection $f: A → B$ which satisfies the condition above. (If $X$ and $Y$ have the same cardinality, the definition can be simplified by fixing $A = X$, $B = Y$.)

where "the condition above" is:

$C\in V\iff f^{-1}[C]\in U.$

How to prove that the special case of the same cardinality is equivalent to the case of arbitrary $X$ and $Y$?

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    @Andres, I see what you mean and thought about such notion when I was working on that question about normal ultrafilters. I was reading the definition on wikipedia before posting the last two comments. I was being petty and pointed out how porton possibly misunderstood the definition and posed the question in a bad way. :-)2011-03-04

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Assume that we have $|X|=|Y|$, let $\mathcal{U}$, $\mathcal{V}$ ultarfilters on $X$ and $Y$ respectively and let $A\in\mathcal{U}$, $B\in\mathcal{V}$ and a bijection $f:A\to B$ such that $C\in\mathcal{V}\iff f_{-1}[C]\in\mathcal{U}$.

Observe that if $|X-A|=|Y-B|$ you are done since both these sets are not in $\mathcal{U}$ and $\mathcal{V}$ (respectively) and you can extend $f$ arbitrarily at these points. Now assume that $|X-A|<|Y-B|$. Let B'\supset B such that |X-A|=|Y-B'| while |B|=|B'| (such B' exists; to see this observe that $|B|=|Y|$). We need to find a bijection g:B\to B' that satisfies the condition of RK. Then we would be done, since we would be able to extend $g\circ f$ as I described above. Let D=B'\setminus B and notice that $D\notin\mathcal{V}$. Let $E\subset B$ an infinite set of size greater than or equal to that of $D$ such that $E\notin\mathcal{V}$. Let $h:E\to D\cup E$ some arbitrary bijection and let $g$ be the identity for every $x\in B\setminus E$ and equal to $h$ for the elements of $E$. This is the $g$ we are looking for.

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    I have followed your proof and found it correct (not mentioning little omissions) writing more detailed version of the proof in my draft. If you want to be acknowledged in the article I am writing, you probably should tell me your real full name.2011-04-26
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I hope the following works: Let $U, V$ be ultrafilters on $X$ and $Y$ respectively. One side is clear. If there is a bijection $f\colon X\rightarrow Y$ satisfying RK-condition then the restriction of $f$ to any $A\in U$ satisfies RK-condition and it is a bijection.

Now the other side: Assume that there is $A,B$ and a bijection $f\colon A\rightarrow B$ satisying RK-condition as given in the wikipedia page. Hence, we know that $f$ is defined at every point of $U,$ because $f$ satisfies RK-condition by assumption. Hence, we have a function $f\colon X\rightarrow Y$ and we want to show that it is 1-1 and onto. If $f$ is not onto, then $\exists C\subset Y$ such that $f(X)\cap C=\emptyset.$ Then trivially $f^{-1}(C)=\emptyset\in U$ by the RK-condition. This certainly not possible. Hence $f$ is onto.

Now, suppose that $f\colon X\rightarrow Y$ is not 1-1. This means there are $x_1,x_2\in X$ $x_1\neq x_2$ but $f(x_1)=f(x_2).$ Let $K:=\left\{x_1,x_2\right\}\subset X.$ If $K\cap A=\emptyset$ and $K\notin U$ then consider the set $A\cup K\in U$ since $U$ is an ultrafilter. The set $f(A\cup K)=f(A)\cup f(K)$ where $f(K)$ is a one point set. But since $V$ is an ultrafilter and $B\in B$ we necessarily have $f(K)\subset B$ which contradicts to $A\cap K=\emptyset.$ Hence, $A\cap K\neq \emptyset.$ $K$ contains two points. But these two points cannot be in $A$ simultaneously, because $f$ is a bijection on $A$. Therefore, WLOG say $x_1\in A$ and $x_2\notin A.$ But then consider $f(A)\in V.$ $f(A)$ contains $f(x_1).$ By RK-condition again inverse image $f^{-1}f(A)=A$ should contain $x_2$ too. This is again a contradiction.

Hence, $f\colon X\rightarrow Y$ is 1-1 and onto.

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    Your proof is clearly wrong: "Hence, we know that $f$ is defined at every point of $U$, because $f$ satisfies RK-condition by assumption." is a nonsense because the condition need to be satisfied only by subsets $C$ of $Y$. Please consult Wikipedia for details.2011-04-26