Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A=\{f\in\{0,1\}^{\mathbb{N}}\,|\, f(0)=0,f(1)=0\}$, i.e. all the infinite binary vectors, that start from $0,0$. Need to proof that $A\sim\mathbb{N}^{\mathbb{N}}$. Any ideas or hint?

share|cite|improve this question
What knowledge do you have to attack this problem? What have you tried? – inactive... for now Dec 3 '12 at 15:13
up vote 2 down vote accepted

The condition $f(0)=f(1)=0$ does not change the cardinality of the set, hence $\#A=\# \{0,1\}^\mathbb N$. We will show that $\#(\mathbb N^\mathbb N)\leq\#\bigl(\{0,1\}^\mathbb N\bigr)$. To each infinite sequence $n_0, n_1, \dotsc \in\mathbb N$ we assign a word $0^{n_0}10^{n_1}10^{n_2}1\dotsm\in\{0,1\}^\mathbb N$. Where $0^n$ means "zero repeated $n$ times". Example: $0,1,2,3,4,5,\dotsc$ maps to $101001000100001000001\dotsm$ This map is not a bijection, because the image always contains infinitely many $1$'s. But clearly it is an injection (e.g. to two sequences of integers two different strings are assigned).

Since $\{0,1\}^\mathbb N\subseteq \mathbb N^\mathbb N$, the inequality $\#(\mathbb N^\mathbb N)\geq\#\bigl(\{0,1\}^\mathbb N\bigr)$ is obvious. Cantor-Bernstein Theorem gives that the sets have the same cardinality.

share|cite|improve this answer
You mean $b(n)=n\, mod\:2$ ? – Denis Turov Dec 3 '12 at 15:35
Don't mind the question, I get it. But this way can't I immediately prove that $\mathbb{N}^{\mathbb{N}}\sim2^{\mathbb{N}}$ ? – Denis Turov Dec 3 '12 at 15:44
Well, the truth is that you can ;) I'll edit the answer. – yo' Dec 3 '12 at 15:45
I think that the vector $0,0,0,0,0,...$ maps into the empty word? may be to stay with the previous version of binary representation of each number? – Denis Turov Dec 3 '12 at 15:56
Oh, sorry, the "1"s still remain. But still is there a problem with a binary representation? – Denis Turov Dec 3 '12 at 15:58


Show that $\mathbb{2^N\sim N^N}$ first, then show that $A\sim 2^\mathbb N$.

The first part is the more difficult part, but recall that $\mathbb{N^N}\subseteq\mathcal P(\mathbb{N\times N})$ and that $\mathbb{N\times N\sim N}$.

share|cite|improve this answer
Thanks Asaf, to show that $2^{\mathbb{N}}\sim\mathbb{N^{\mathbb{N}}}$ I used the way that tohecz proposed. But in the proof that you mentioned: why $$\mathbb{N}^{\mathbb{N}}\subseteq\mathcal{P}(\mathbb{N}\times\mathbb{N})$$ $\mathbb{N}^{\mathbb{N}}$ is the set of all functions from $\mathbb{N}$ to $\mathbb{N}$ , and $\mathcal{P}\left(\mathbb{N}\times\mathbb{N}\right)$ is set of all subsets of pairs $\left(n_{1},n_{2}\right)$ ? – Denis Turov Dec 3 '12 at 19:13
@DenisTurov Because for $f:\mathbb N\to\mathbb N$ we have $\{(n,f(n)):n\in\mathbb N\}\subset N\times N$. – yo' Dec 3 '12 at 22:27
@Denis: I apologize, I missed your comment in my inbox until tohecz wrote his. As tohecz wrote, every $f\in\mathbb{N^N}$ is a set of ordered pairs from $\mathbb{N\times N}$. Therefore $f\colon\mathbb{N\to N}$ implies that $f\in\mathcal P(\mathbb{N\times N})$. – Asaf Karagila Dec 3 '12 at 22:30
Thanks, I missed that point. – Denis Turov Dec 4 '12 at 16:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.