2

Given $A\subseteq B$ such that $A$ is countable and $B\setminus A$ is infinite. I want to show that the cardinality of $B\setminus A$ is the same as the cardinality of $B$.

I know that since $B \setminus A$ is infinite, there is a function $g: \mathbb{N} \rightarrow B \setminus A$ injective. But how does this help me find a bijective function $f: B \setminus A \rightarrow B$ ?

Peter Sampodiras
  • 478
  • Can you assume that any set can be well ordered? – Robert Shore Apr 08 '23 at 00:48
  • $A$ is countable, which means it is bijective to a subset of $\Bbb N$. $g$ is also a bijection between $\Bbb N$ and $g(\Bbb N)$. So if you can think of a way to map the disjoint union of a subset of $\Bbb N$ and $\Bbb N$ to $\Bbb N$ bijectively, you have a ready-made bijection between $A\cup g(\Bbb N)$ and $g(\Bbb N)$. – Paul Sinclair Apr 09 '23 at 01:54
  • Not really @Robert Shore – Peter Sampodiras Apr 09 '23 at 14:36
  • @Paul Sinclair If I understand what you say the idea is to find a mapping $h: C \cup \mathbb{N} \rightarrow \mathbb{N}$ bijective where $C \subset \mathbb{N}$? – Peter Sampodiras Apr 09 '23 at 14:45
  • Almost. Note that I emphasized the disjoint union. You don't get to ignore that part. This turns the problem from being about sets of which you know almost nothing to a set which you've been studying since you were a child yourself, and for which a vast body of knowledge exists. – Paul Sinclair Apr 09 '23 at 14:59
  • But if $C \in \mathbb{N}$ how come $C \cup \mathbb{N}$ can be disjoint? @Paul Sinclair – Peter Sampodiras Apr 09 '23 at 15:07
  • See Disjoint union. What is important here is that both sets can be treated as natural numbers, so that you can use whatever operations are handy to meet the challenge of fitting these together. – Paul Sinclair Apr 09 '23 at 16:07

1 Answers1

1

Let $h: \Bbb N \to A$ be a bijection and denote $a_n=h(n)$.

Let $g: \Bbb N \to B \setminus A$ be an injection and denote $b_n=g(n)$.

$$\text{Define } f: B \to B \setminus A \text{ via }f(x)= \begin{cases} b_{2n}, & \text{if }x=a_n \\ b_{2n+1}, & \text{if } x=b_n \\ x & \text{otherwise} \end{cases}$$

Then $f$ is a bijection.

Robert Shore
  • 26,056
  • 3
  • 21
  • 49