# If $B$ is an infinite set and $A\subset B$ is finite then $|B|=|B \setminus A|$

I'm having a hard time with this proof.

The definition of cardinality that I'm using is: Two sets have the same cardinality if there is a bijection between them.

($$\star$$) I already prove that $$|B\setminus A|$$ is an infinite set.

I'm doing a proof by contradiction. So I'm assuming that $$|B|\neq|B\setminus A|$$ wich means that there is no bijection between $$B$$ and $$A\setminus B$$, what I'm trying to do is that this assumption leads me to say that $$|A\setminus B|$$ is finite which will contradict $$(\star)$$.

But I don't know how the assumption will let me say that $$|A\setminus B|$$ is finite.

I already try using the fact that each infinite set has a countable subset, but I didn't succeed.

Any ideas for this?

Thank you.

• @NajKamp $n \mapsto n+1$? – It'sNotALie. Sep 28 '20 at 17:51
• Could you build such a bijection explicitly? – It'sNotALie. Sep 28 '20 at 17:52
• Are you assuming the axiom of choice ? If so, you can choose a countable subset $A'$ such that $A \subset A' \subset B$. You then can find a bijection $f$ between $A'$ and $A'\setminus A$. The map from $B$ to $B\setminus A$ that coincides with $f$ on $A'$ and is the identity map outside is a bijection. – Didier Sep 28 '20 at 17:52
• @DIdier_ Thank you, this direct proof is better than my idea. – Choxom Sep 28 '20 at 18:15
• – halrankard2 Sep 28 '20 at 19:01

Every infinite set has an infinite countable subset (assuming AC). So $$B\setminus A=C\uplus D$$ where $$C$$ is countable and infinite and the union is disjoint. Then $$B\setminus A=|C|+|D|=\aleph_0+|D|$$. Also $$B=A\uplus C\uplus D$$, so $$|B|=|A|+\aleph_0+|D|$$. As $$A$$ is finite, $$|A|+\aleph_0=\aleph_0$$.