Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer

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 $|A_i\setminus B_i|=|A_i|$.

share|cite|improve this answer

Your Answer


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.