# Proof there is a 1-1 correspondence between an uncountable set and itself minus a countable part of it

Problem statement:

Let A be an uncountable set, B a countable subset of A, and C the complement of B in A. Prove that there exists a one-to-one correspondence between A and C.

My thoughts:

There's a bijection between A and A (the identity function). There's a bijection between C and C (the identity function). There's a bijection between B and $\mathbb{N}$. That's all I know.

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I take it you don't have cardinal arithmetic on hand yet? – Arturo Magidin Mar 6 '11 at 3:07
Cardinal numbers are in the next chapter, so I guess not.  This chapter proves that a countable union of countable sets is countable, $\mathbb{N}$ is countable, $\mathbb{Q}$ is countable, the algebraic numbers are countable, and $\mathbb{R}$ is not countable. – Matt Gregory Mar 6 '11 at 3:08

## 1 Answer

Since $A\setminus B$ is uncountable, assuming countable choice it has a countably infinite subset $B'$. Then $B'\cup B$ is countable, so there is a bijection $g:B'\cup B\to B'$. Define $f:A\to A\setminus B$ by $f\vert_{B'\cup B}=g$ and $f(a)=a$ for $a\in A\setminus(B'\cup B)$.

Basically, you just take chunks off of $A$ and $A\setminus B$ that have equal size, respectively $B'\cup B$ and $B'$, leaving the remaining sets equal to $A\setminus(B'\cup B)$, then piece together the two bijections.

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