1

I know that the set of all surjective mappings of ℕ onto ℕ (lets name this set as F) should have cardinality |ℝ|.

How to strictly prove that?

From the fact that cardinality of every possible function is |ℝ|, |F| <= |ℝ|. Saw similar question on this site, but I need strictly defined function that shows that |F|>=|ℝ|.

Thank you for your time.

kushtibargo
  • 99

1 Answers1

1

HINT: Let $M=\{2n+1:n\in\Bbb N\}$. For each $A\subseteq M$ define

$$f_A:\Bbb N\to\Bbb N:n\mapsto\begin{cases} n/2,&\text{if }n\in\Bbb N\setminus M\\ 1,&\text{if }n\in A\\ 0,&\text{if }n\in M\setminus A\;. \end{cases}$$

Brian M. Scott
  • 631,399
  • Understood this beautiful solution, but my teacher wants injective function that would transfer subset of N into surjective function.

    Could you help me?

     – kushtibargo May 28 '13 at 16:40
  • @kushtibargo: Just modify this one slightly. For each $A\in\Bbb N$ let $\widehat A={2n+1:n\in A}$; the map $A\mapsto\widehat A$ is a bijection from $\wp(\Bbb N)$ to $\wp(M)$. Now just use the injective function $A\mapsto f_{\widehat A}$. – Brian M. Scott May 28 '13 at 17:54
  • So we make from N set of odd numbers and work on them? Did I understand it right? – kushtibargo May 28 '13 at 17:59
  • 1
    @kushtibargo: Yes, that’s right: first convert the subset of $\Bbb N$ to a subset of the odd numbers, and then use my original trick. – Brian M. Scott May 28 '13 at 18:01
  • Why this function will be injective? Different domains? How can we strictly prove injectivity? – kushtibargo May 28 '13 at 19:05
  • @kushtibargo: If $A,B\subseteq\Bbb N$ with $A\ne B$, then $\widehat A\ne\widehat B$, and therefore $f_{\widehat A}\ne f_{\widehat B}$. – Brian M. Scott May 28 '13 at 19:07
  • @BrianM.Scott I know this is unrelated to this question, but would you mind taking a look at this question? I'm trying to prove that $\dim_{\mathbb{Q}}\mathbb{R} = |\mathbb{R}|$ without using the continuum hypothesis or using the Baire category theorem. – Gord452 Dec 31 '20 at 00:10