Cardinality does not change when substracting a countable subset from an uncountable set

cardinalsset-theory

This comes from Exercise 11 of Section 2.1 of Kaplansky's "Set Theory and Metric Spaces".

Let $A$ be an uncountable set. Let $B \subseteq A$ be a countable set and let $B^c$ be its complement in $A$. Prove that there exists a bijection between $A$ and $B^c$.

I've found some related questions here: removing a countable subset from an infinite set doesn't change cardinality, and here: Altering an Infinite Set does not change cardinality, but none seems to be exactly the one I'm looking for.

Suposedly this problem can be solved without the axiom of choice, due to it being proposed before the axiom's introduction in the book.

It is quite easy to show that $B^c$ must be uncountable, but moreover I need to prove that it has the same cardinality as $A$. That doesn't look so simple. Here are some thoughts I've been considering:

  • Since $B^c$ can be split into some countable $C$ and uncountable $C^c$, and the same for $C^c$, this problem is equivalent to partitioning $A$ into a countable number of countable sets and its complement, which must be uncountable. And if we prove that it has indeed the same cardinality as $A$, we would've proved that any partition of $A$ must include a proper subset. It is a fact that proper subsets exist in $A$ due to it being infinite, but I don't think that's enough to say that it will be present on every partition. ¿Is there a way to prove this?

  • As a particular case, we obtain from this result that irrationals and trascendents have both the same cardinality as $\mathbb{R}$.

EDIT

So, following Berci's answer, I wrote this proof.

Since $B$ is countable and $A = B \cup B^c$ is uncountable it follows that $B^c$ is uncountable. Then it contains an uncountable infinite subset; call it $B_1$. By the same argument $B^c \backslash B_1$ must be uncountable. On the other hand, we have that $B \cup B_1$ is countable, so there exists a one to one correspondence between $B$, $B_1$ and $B \cup B_1$. And since $A \backslash B \cup B_1$ and $B \backslash B_1$ both have the same elements, it follows that there must exist a bijection between $A$ and $B^c$.

I think that's how it is suposed to be solved. I have some doubts about the conclusion though.

Best Answer

Since $B^\complement$ is uncountable, it contains a countably infinite subset $C$.

Now simply apply a bijection between $B\cup C$ and $C$, and extend it identically on $B^\complement\setminus C$.

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