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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.

Arctic Char
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Choxom
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  • @NajKamp $n \mapsto n+1$? – It'sNotALie. Sep 28 '20 at 17:51
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    Could you build such a bijection explicitly? – It'sNotALie. Sep 28 '20 at 17:52
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    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
  • See: https://math.stackexchange.com/questions/2912030/suppose-x-is-infinite-and-a-is-a-finite-subset-of-x-then-x-and-x-setm?rq=1 – halrankard2 Sep 28 '20 at 19:01

2 Answers2

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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, $|A|+\aleph_0=\aleph_0$.

Angina Seng
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Instead of writing a bijection between both sets, try to find a pair of injective functions $f:B \rightarrow (B\setminus A)$ and $g:(B\setminus A) \rightarrow B$. Then, by Schröder–Bernstein theorem you have the same cardinality (no matter what cardinality it really are).

For $g$ take the identity fucntion which is clearly an injective function from $B\setminus A$ to $B$.

Now you just have to find some injective function from $B $ to $B\setminus A$.

Hope this works for you.

Sergio Ivan
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