# prove that: $|\{x\in\mathbb{\mathbb{R}}:\text{ }\exists r\in S \frac{x}{r}\in\mathbb{N}\}|= |\mathbb{N}|$

1.let $$S = \{r_{1},.......,r_{n}\}\subset\mathbb{R}$$ as $S$ is an infinite set of objects different than $0$.

I need to prove that $$|\{x\in\mathbb{\mathbb{R}}:\text{ }\exists r\in S \frac{x}{r}\in\mathbb{N}\}|= |\mathbb{N}|$$

1. prove that $$|\{0,1\}^{\mathbb{N}}\times\{0,1\}^{\mathbb{N}}| =|\{0,1\}^{\mathbb{N}}|$$

I know i can prove it if i'll show that an Inverse function from one set to another exist. but i don't know how to define such function, and defeintly don't what the Inverse function will be.

also, I don't know how to use CBS at this matter. I know how to prove that: $$|\{x\in\mathbb{\mathbb{R}}:\text{ }\exists r\in S \frac{x}{r}\in\mathbb{N}\}|\leq|\mathbb{N}|$$ by defining a function between the sets and prove that it's one-to-one. but how to prove that this function is onto? (I know the definition of onto and on this case, it doesn't dircetly follows by the definition)

about Q.2, I know that it can be proved by making a function that will compose an ordered pair of functions to the original function, but again, does only one function exist of the form $$f:n\longrightarrow\{0,1\}^{\mathbb{N}}?$$

this is obviously a true verse, but I can't find a way using function to prove that.

## 2 Answers

Answering question (1):

Let $A=\{x\in\mathbb{R}:\exists r\in S, \frac{x}{r}\in\mathbb{N}\}$ and $A_i=\{x\in \mathbb{R}: \frac{x}{r_i} \in \mathbb{N}\}=\{x\in \mathbb{R}: x = 0\mod r_i\}$. $A_i$ is the ideal made of the multiples of $r_i$ it is thus countable. Then, $A=\cup_{i\in\mathbb{N}}A_i$. It is a countable union of countable sets, it is thus countable.

See Gödel's answer for question (2).

• Your second answer is wrong. The cardinal of $\{0,1\}^\mathbb{N}$ is $2^{\aleph_0}$ and by Cantor theorem $2^{\aleph_0}>\aleph_0$ – Gödel May 3 '18 at 16:33
• I do not doubt what you say but then I fail to see why the usual writing of any number $n\in \mathbb{N}$ as $n=\sum_{i=0}^\infty b_i2^i$ where $(b_i)_{i\in \mathbb{N}}\in \{0,1\}^\mathbb{N}$ is not a suitable bijection. – Bill O'Haran May 3 '18 at 16:38
• Because this sums are finite, How you represent the sequence $f(n)=1$ for all $n$? – Gödel May 3 '18 at 16:42
• Thank you! Makes much more sense now. – Bill O'Haran May 3 '18 at 16:44

For the second question consider $$F:\{0,1\}^\mathbb{N}\times\{0,1\}^\mathbb{N}\rightarrow\{0,1\}^\mathbb{N}$$ Such that $$F(f,g)(n):=\left\{\begin{array}{ccc} f(k)&if&n=2k\\ g(k)&if&n=2k+1 \end{array}\right.$$ Is easy to show that $F$ is a bijection so $|\{0,1\}^\mathbb{N}\times\{0,1\}^\mathbb{N}|=|\{0,1\}^\mathbb{N}|$