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Let $A_1$, $A_2$, $B_1$ and $B_2$ be sets such that $$ |A_1|=|A_2|,\quad|B_1|=|B_2|,\quad B_1\subset A_1,\quad B_2\subset A_2. $$ Where $|\ \ |$ denotes the cardinality of a set.

How to show that $|A_1\setminus B_1|=|A_2\setminus B_2|$ ?

We know that there is a bijection from $A_1$ to $A_2$ and one from $B_1$ to $B_2$, and we just have to find one from $A_1\setminus B_1$ to $A_2\setminus B_2$, but how to prove its existence?

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This is wrong. Let, for example, $A_1 = A_2 = B_1 = \mathbb Z$, $B_2 = \mathbb N$. –  martini Sep 9 '12 at 15:18
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up vote 5 down vote accepted

But this is not true at all...

Consider the following example:

$A_1=A_2=\mathbb N$ and $B_1=\{n\in\mathbb N\mid n\text{ is even}\}$, $B_2=\{n\in\mathbb N\mid n>3\}$.

Clearly all sets involved have the same size but $A_1\setminus B_1$ is infinite where as $A_2\setminus B_2$ is finite.

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As posted by others, the statement need not be true if all four sets have the same infinite cardinality. On the other hand, it is true if the $A_i$ are finite. It is also true if $|B_{1,2}|<|A_{1,2}|$ and $|A_{1,2}|$ infinite because then $|A_i\setminus B_i|=|A_i|$.

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