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I've found some time to read a little more on set theory, and I've come across the following question.

Suppose I have four sets $X$, $Y$, $Z$, and $W$ such that $Y\subseteq W$ and $Z\subseteq X$. Suppose also that $X\cup Y\sim Y$, where by $\sim$ I mean that the two sets $X\cup Y$ and $Y$ are equinumerous. How can I show that $Z\cup W\sim W$?

I thought the Bernstein-Schroeder theorem might be applicable. The identity function maps $W$ into $Z\cup W$ injectively, so I figured it suffices to show that there is an injection from $Z\cup W$ into $W$. From $X\cup Y\sim Y$, there is an injection $f\colon X\cup Y\to Y$, and thus $f|_X$ is an injection from $X$ into $Y$. Since $Z\subseteq X$ there is an injection from $Z$ to $X$, and likewise from $Y$ into $W$. Composing all these would give an injection from $Z$ into $W$. Those were my thoughts, but I don't think I can use them to show that $Z\cup W$ maps injectively into $W$. There must be a better way. Thanks for any help.

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    What you need id to make $Z$ and $W$ disjoint, isn't it.2011-02-18

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$Z \cup W \subseteq X \cup W = X \cup (Y \cup (W \;\backslash Y)) = (X \cup Y) \cup (W \;\backslash Y) \sim Y \cup (W \;\backslash Y) = W \subseteq Z \cup W$

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    To be specific, it requires the Zorn's lemma, which is obviously equivalent to The Axiom of Choice.2011-02-18
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If $f\colon X\cup Y\to Y$ is an injection and and $Z\subseteq X$, then the restriction to $f\colon Z\cup Y\to Y$ is also an injection. With $Y\subseteq W$ and using the identity injection $i\colon W \backslash Y \to W \backslash Y$ restricted to the injection $i\colon W \backslash (Z \cup Y) \to W \backslash Y$, you can combine $f|_{Z\cup Y}$ and $i|_{W \backslash (Z \cup Y)}$ to provide an injection from $Z\cup W$ to $W$

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    Things may be easier using the product of sets, the set-theoretical sum of sets [here](http://en.wikipedia.org/wiki/Disjoint_union#Set_theory_definition) and your question will be totally settled up by basic theorems of set theory.2011-02-18