Bijection between countable ordinals and reals The set of all countable ordinals is $\omega_1$, which has a cardinality of $\aleph_1$. When accepting the continuum hypothesis, $2^{\aleph_0} = \aleph_1$, so a bijection between countable ordinals and real numbers exists. Is there any known bijection of this type?
 A: The use of the term "known bijection" is unclear. If you mean an explicit formula in the language of set theory that $\sf ZFC$ proves it to be a bijection, then the answer is no. If that would have existed then the continuum hypothesis would have been a theorem, rather than a consistent sentence.
Under certain assumptions, for example $V=L$, there is a definable bijection, and one can say, if so, that there is a statement in the language of set theory which under the additional axiom $V=L$, defines a bijection between $\omega_1$ and $\mathcal P(\omega)$.
However, in general there is no parameter free $\varphi$ such that $\sf ZFC+CH$ proves that $\varphi$ defines a bijection. Assume by contradiction that $\varphi(x,y)$ defines such a bijection (so $x\subseteq\omega$ and $y<\omega_1$). $\newcommand{\forces}{\mathrel{\Vdash}}$
Now consider $V\models\sf ZFC+CH$, and let $\Bbb P=2^{<\omega}$ in $V$ be the Cohen forcing adding a single real, whose canonical name is $\dot r$. Let $G$ is a generic filter, then $V[G]$ is also a model of $\sf ZFC+CH$. 
Therefore in $V[G]$ the formula $\varphi(x,y)$ still defines a bijection, therefore there is some $\alpha<\omega_1$ and $p\in G$ such that $p\forces\varphi(\dot r,\check\alpha)$. Let $m,n$ two integers not in the domain of $p$ (recall that a condition is a partial function from $\omega$ into $2$ with a finite domain) and consider the $2$-cycle $(m\ n)$ as a permutation of $\omega$. By standard arguments this permutation extends to the forcing and the names, and we have, $$p\forces\varphi(\dot r,\check\alpha)\iff\pi p\forces\varphi(\pi\dot r,\pi\check\alpha)=p\forces\varphi(\pi\dot r,\check\alpha).$$
Therefore $p$ forces that $\varphi$ does not define a bijection. This is a contradiction to the assumption that it does.

Note: If we allow parameters, then of course that such formula exists, since we just need to write as a formula in $\varphi(x,y,u)$ that $u$ is a bijection between $\mathcal P(\omega)$ and $\omega_1$ and that $u(x)=y$. That clearly creates circularity in the question, so I only treated the parameter free case.
Of course one can define a whole wide spectrum of allowed parameters, real numbers, countable sets of real numbers, and so on and so forth. I have only dealt with the simplest case in my above answer, and there might be a positive answer if we use strong enough parameterization; but my suspicion is that any such answer would amount to encoding the bijection in one way or another.
