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I took my first class in Discrete Math this semester and I was given a problem to explain why there is a bijection between the points on the unit circle and the real numbers. I initially thought the points on the unit circle were a subset of the real numbers , which my Professor told me was not true but gave me no explanation as to why. Which brought me here with a few questions I hope to get answered.

1) What are the points on the unit circle if not a subset of the reals?

2) What approach would I take in explaining why there is a bijection between the unit circle and the real numbers and how would I prove it? (I don't want the solution I would just appreciate some direction so I know how to approach the problem).

Sal Moe
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  • Recall that points are ordered pairs of real numbers $(a,b)$ and are in $\mathbb{R}^2$, not the real numbers themselves. Thus a set of ordered pairs is a subset of $\mathbb{R}^2$, but not $\mathbb{R}$ – jl00 Dec 02 '19 at 06:01
  • Equivalently you can find a bijection between $[0,1)$ and $\mathbb{R}$. The fact that one is a subset of the other does not significantly help. – Michael Dec 02 '19 at 06:04
  • Alternatively, the unit circle can be visualized in $\Bbb C$ as the set of points $e^{i \theta}$ for $\theta \in [0, 2\pi)$. Of course, $\Bbb C \cong \Bbb R^2$ if I remember correctly, so there's a sort of equivalence to the "$x^2+y^2=1$ in $\Bbb R^2$" formulation. Though I feel the complex case makes establishing bijections easier for this. – PrincessEev Dec 02 '19 at 06:04
  • Do you mean there's a bijection between the real numbers and the unit circle minus a point? – JDZ Dec 02 '19 at 06:07
  • Yes, the point (0,1) @JDZ – Sal Moe Dec 02 '19 at 06:10
  • I feel foolish for not coming to that realization on my own haha it makes perfect sense. Thank you! @jl00 – Sal Moe Dec 02 '19 at 06:11
  • Also, notice that any point on the unit circle can be represented by just one real number, $\theta\in [0,2\pi)$. So it is sufficient to find a bijection $f:\mathbb{R} \to [0,2\pi)$. Can you do it? – jl00 Dec 02 '19 at 06:13
  • It's not clear to me whether you want a constructive proof. – Peter Taylor Dec 02 '19 at 06:28

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Points on a unit circle are points - they're geometric objects. And even if you embed the circle in a Cartesian space, they have co-ordinates with two real components, so you haven't put them into a subset of the reals, but instead of $\mathbb{R}^2$, the set of pairs of real numbers.

To motivate the idea of how you might create a bijection from the unit circle to the reals, here's a suggestion that is flawed but which suggests a direction - imagine taking the unit circle as embedded in a Cartesian $xy$-plane, and moving it up so that its base is lying on the $x$-axis. Now, take a point on the circle, and draw the tangent to the circle at that point. Let your mapping be the function that takes the point, and maps it to the point where that tangent meets the $x$-axis.

Is it a bijection? No. In fact, it's not even a function, but it's really close. For almost every point on the circle, that tangent will meet the $x$-axis at a single point, and vice versa. Unfortunately, there are two problem points on our circle - the top and the bottom. The tangent at the bottom is the $x$-axis, so it meets that axis everywhere. And the tangent at the top never meets the $x$-axis, so it doesn't map to any point on it.

That said, if your domain is infinite, failing to create a bijection at finitely many points is usually as good as having a perfect bijection. Can you see if there's any way to adjust this mapping to make it work? Or is there another way that you could do it?

ConMan
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  • That's similar to the the question that was posed to me. I was given a projection of the unit circle onto the real number line with the center of the circle at the origin and the point (0,1) undefined. Initially I tried to think of the unit circle as a line sectioned off from the North Pole which, if I was were to visualize a mapping f: UC \to\ R it would not be a proper mapping because then the NP would have two unique elements it maps to, which a function can't have therefore that is the reason it is left undefined. – Sal Moe Dec 02 '19 at 21:25