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Is it possible to find a bijective map from $ \{ (x,y) \in \mathbb{R} ^{2} : y \geq 0 \} $ to the whole of $\mathbb{R} ^{2} $ (or consider it $ \mathbb{C} $ if needed)?

John Doe
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Sure, you can just pick your favourite bijection from $\mathbb R$ to $\mathbb R^{\ge 0}$, to apply to the $y$ bit of your coordinates. If you'll accept the Schröder-Bernstein theorem, there exists such a bijection since $x \mapsto e^x : \mathbb R \to \mathbb R^{\ge 0}$ is an injection, and the inclusion map the other way is an injection, so there exists $f: \mathbb R^{\ge 0} \to \mathbb R$ bijective, and then $(x, y) \mapsto (x, f(y))$ is such a bijection.

We could explicitly define a bijection $f$ as something like \begin{align*} f: \mathbb R \times \mathbb R^{\ge 0} &\to \mathbb R^2 \\ (x, y) &\mapsto \begin{cases} (x, \ln(y + 1)) & y \in \{0, 1, 2, \dotsc\} \\ (x, \ln y) & \text{otherwise} \end{cases} \end{align*} but it's just a bit messy.

Izaak van Dongen
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    Thanks, a bijection like that is fine. I had found another one that was a lot more complex to write down that involved splitting the upper half plane into strips and mapping half of the strips to the lower half plane. It wasn't easy to use though. – John Doe Jan 19 '20 at 14:57