2
$\begingroup$

I know $|\mathbb{N^{N}}| = |\mathbb{R}|$ , but I just cant find an injection from $|\mathbb{N^{N}}|$ to $|\mathbb{R}|$...

I tried to use this one: Given $\ f:\mathbb{N} \to \mathbb{N}$, define $F: F(f) = 2^{f(0)}3^{f(1)}... = \prod_{i=1}^{\infty}p_{i}^{f(i-1)}$ then by the prime factorization theorem, $F$ is injective..

But my professor said this function is not well defined since I may get $\infty$.

And he said I can consider this one: $F(f) = f(0) + \frac{1}{f(1)+\frac{1}{f(2)+...}}$

But what if $f(n) = 0 \ for\ all \ n \in \mathbb{N}?$

And can someone give me more injections form $\mathbb{N^{N}}$ to $\mathbb{R}$ ?

  • 2
    If your problem is $0$, find an injection from $\mathbb N^\mathbb N$ to $(\mathbb N\setminus\{0\})^\mathbb N$ first. – 2017-02-27
  • 0
    Why is $f: n\mapsto 0$ a problem? The continued fraction you get just equals $1$, doesn't it? ($0$ *does* pose a problem, but this isn't it . . .) – 2017-02-27
  • 3
    For the continued fraction example to work you need to exclude $0$. But there's a simple injection from $\mathbb{N}$ to $\mathbb{Z}^+$ that will fix this issue. – 2017-02-27

2 Answers 2

7

Note that your proposed argument actually maps $\mathbb{N}^\mathbb{N}$ into $\mathbb{N}$, so it can't possibly work - as the former is uncountable. And indeed your proposed map is not defined on any sequence $f\in\mathbb{N}^\mathbb{N}$ with infinitely many terms $>1$.

One way to do this is to view the terms of your sequence $f\in\mathbb{N}^\mathbb{N}$ as measuring the gaps between $1$s in a binary expansion - e.g. $03204...$ gets translated into $$0.10001001100001...$$ It's not hard to show that this is injective.

0

Interpreting functions as sets of ordered pairs, $\mathbb{N}^\mathbb{N}\subseteq 2^{\mathbb{N}^2}$. Your favourite injection of $\mathbb{N}^2$ into $\mathbb{N}$ injects $\mathbb{N}^\mathbb{N}$ into $2^\mathbb{N}$, which I'm sure you can inject into $\mathbb{R}$ easily.