Do we have $|\mathbb{Q}\times\mathbb{Q} | = |\mathbb{Q}|?$ or $|]a,b[|= |\mathbb{R}|$?

Question

I'm stuck into two proof about cardinality and countable set :

I have to prove that |$\mathbb{Q} \times \mathbb{Q}$| = |$\mathbb{Q}$|, i have a hint in my lessons which is $|\mathbb{N}|= |\mathbb{Q}|$, but I don't know how to proceed. This is what I did for the moment : Since $\mathbb{Q}$ is countable, it exist a $f: \mathbb{Q} \to \mathbb{N}$ which is a bijection. I define g: $\mathbb{N} \times \mathbb{N} \to \mathbb{N}$ by $$g(m,n)=2^m(2n+1)-1$$ which is bijective. then $g \circ (f \times f)$ is a bijection $\mathbb{Q} \times \mathbb{Q} \to \mathbb{N}$

Another question is to prove that $|]a,b[|= |\mathbb{R}|$, for $a<b.$ I have a hint which is to find a monotone continuous function $f:(a,b) \to \mathbb{R}$

Answer 1

So you have proved that there is a bijection between $\mathbb{Q}\times\mathbb{Q}$ and $\mathbb N$. And you know that there is a bijection between $\mathbb N$ and $\mathbb Q$. Therefore, there is a bijection, between $\mathbb{Q}\times\mathbb{Q}$ and $\mathbb Q$. So, they have the same cardinal.

Answer 2

the first question is easier since we have $$ f^{-1}\circ g \circ (f \times f): \mathbb{Q} \times \mathbb{Q} \to \mathbb{N}\to \Bbb Q$$

is a bijection. the first arrow map is what you proved and the second was your hint.

For the last question it suffices to prove for $a=-1 $ and $b=1$. Consider $$x\mapsto \frac{1}{e^{-x}+1}~~~or ~~~~ x\mapsto \frac{x}{|x|+1}~~~~~or~~~ x\mapsto \frac{2}{\pi}\arctan x.$$ which are bijections from $\Bbb R$ to $(-1,1)$.