Prove that there is a bijection $\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ in form of $f(x,y)=a(x)+b(y)$ In this question tells how to build bijection $\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ in form of $f(x,y)=g(a(x)+b(y))$
We doing something with $x$ and $y$ on $[0,1]$, add up them and after mapping that in $\mathbb{R}$ (this is a function $g$).
How to do it without $g$? I think it will be difficult to come up with a constructive proof, so that proof of existence will be enough.
 A: A more elementary approach, but still requiring Zorn's lemma:
Let $\phi:\mathbb R\times\mathbb R\to\mathbb R$ send $(x,y)\mapsto x+y\sqrt 2$.
Call $A\subset R$ "nice" if $A$ is:


*

*Closed under addition: $a_1+a_2\in A$ if $a_1,a_2\in A$.

*Closed under rational multiplication: $qa\in A$ if $q\in\mathbb Q$ and $a\in A$.

*$\phi_{\mid A\times A}$ is one-to-one.


For example, $\mathbb Q$ is "nice."
It is pretty easy to see that "nice" subsets satisfy Zorn's lemma, so there must be a maximal $A$.
Now, for maximal $A$, show that $\phi_{\mid A\times A}$ is onto by assuming some $x$ is not in the image.
Then $x\not\in A$, since $A$ is in the image. Since $A$ is maximal, we must have that $A+x\mathbb Q$ is not "nice." But from that we can show that $x=a_1+a_2\sqrt{2}$ for some $a_1,a_2\in A$. (It's a tedious calculation, but it essentially follows because when $p+q\sqrt{2}$ is non-zero, with $p,q$ rational, then $\frac{1}{p+q\sqrt{2}}=\frac{p}{p^2-2q^2}-\frac{q}{p^2-2q^2}\sqrt{2}$.)
Now, let $B=\sqrt{2}\cdot A$. Then any real number can be written uniquely as $a+b$ for some $a\in A,b\in B$.
Now, we have $|A|=|B|$, trivially, so since $A\times B=\mathbb R$, we have that $A=|R|.$ Letting $a:\mathbb R\to A$ and $b:\mathbb R\to B$ be bijections, then $h(x,y)=a(x)+b(y)$ is a bijection.
In this case, we can actually use $b(y)=a(y)\sqrt{2}$.

My original, more advanced answer.
$\mathbb R$ is a vector space over $\mathbb Q$. By a result equivalent to the axiom of choice, any vector space has a basis $B$ such that every element $x\in\mathbb R$ can be written essentially uniquely as the a finite sum:
$$x=r_1b_1+r_2b_2+\cdots+r_kb_k$$ where the $b_i$ are distinct, and the $r_i$ are rational.
Now, necessarily, $B$ is infinite, or $\mathbb R$ would be countable.
Now you need another theorem: For infinite set $B$, there is a bijection between $B$ and $B\times\{0,1\}$.
Now, $\mathbb R\times\mathbb R$ is also a vector space over $\mathbb Q$, and we have a basis, $B_2=B\times\{0\}\cup \{0\}\times B$.
So we have an easy bijection $B_2 \leftrightarrow B\times\{0,1\}$, and thus a bijection $B_2\to B$.
For $(b,0)\in B_2$, let $\phi_1(b)$ be the corresponding element of $B$, and for $(0,b)\in B_2$, let $\phi_2(b)$ be the corresponding element of $B$.
Then define:
$$a(r_1b_1+r_2b_2+\cdots r_nb_n)=r_1\phi_1(b_1)+r_2\phi_1(b_2)+\cdots r_n\phi_1(b_n)\\
b(r_1b_1+r_2b_2+\cdots r_nb_n)=r_1\phi_2(b_1)+r_2\phi_2(b_2)+\cdots r_n\phi_2(b_n)$$
Then you get $(x,y)\to a(x)+b(y)$ is a bijection.
This is just a change basis, essentially. The left side $\mathbb R^2$ and the right side $\mathbb R$ are $\mathbb Q$-vector spaces with bases of the same cardinality, and thus there is a linear isomorphism.
