# A plane has the same cardinality as a half-plane: how to map all values in polar coordinates? Also needs verification/refuting of 2 alternative proofs

Prove that a plane has the same cardinality as a half-plane.

1. I decide to find a bijection in polar coordinates. The plane consists of points $(r, \phi)$ such that $r\in\mathbb{R}, \phi\in [0^{\circ}, 360^{\circ})$. ($)$ because we don't want same point on $OX$ to have two different representations.) By mapping $(r, \phi) \mapsto (r, \phi/2)$, we almost get the half-plane -- other than the left half of the dividing line. I tried quite a few things to circumvent this (mapping four arbitrary half-lines of the half-plane to the four half-lines of $OXY$ and constructing from there, for example), but nothing yielded a bijection.

My Q1, is it possible to find a polar bijection from half-plane to whole plane? How?

1. $(x, y) \mapsto (x, \frac{1}{2^y})$. That is, when we define half-plane as not including the separating line.

2. The third attempt seems valid to me, but not sure if rigourous enough.

My Q2 and Q3 are, are the attempts 2 and 3 correct? If no, where? How can I improve them?

Alternative proofs are also very welcome.

Thank you.

• Remember that the polar coordinates of the origin are not unique. Commented Jun 2, 2018 at 15:28
• To just show that cardinalities are the same, one usually wouldn’t construct a bijection explicitly, one would just show the existence of in injection (the inclusion map) and a subjection from the half plane to the plane. For the surjection just double the angle. Commented Aug 15, 2022 at 11:16

As noted, the origin has no unique polar coordinates so we'll define a bijection from $\mathbb{R}^2 \setminus \{0,0\}$ to $\mathbb{R} \times (\mathbb{R} \setminus \{0\})$.

Basically you're looking for a bijection between $[0, \pi\rangle$ and $\left[0, \frac\pi2\right]$.

First construct a bijection $f : [0, \pi\rangle \to [0,\pi]$ with

$$f(x) = \begin{cases} \frac1{n-1}, & \text{if x = \frac1n for some n \in \mathbb{N}, n \ge 2} \\ \pi, & \text{if x = 1}\\ x, & \text{otherwise} \end{cases}$$

Then $\frac12 f : [0, \pi] \to \left[0, \frac\pi2\right]$ is a bijection.

Now define $$(r, \phi) \mapsto \left(r, \frac12f(\phi)\right)$$

and verify that this is a bijection between $\mathbb{R}^2 \setminus \{0,0\}$ and $\mathbb{R} \times (\mathbb{R} \setminus \{0\})$.

Your second approach is correct, but yes, it doesn't include the $x$-axis.

Your third approach is also correct, if you use rays of the form $\mathbb{R} \times [k, k+1\rangle$.