Prove the unit circle is uncountable

Question

This is a homework problem, so avoid giving the answer. I think a discussion of my attempt at a proof would be more appropriate. The problem goes as follows:

Let $S$ be the circle of unit radius in the Euclidean plane: $$S = \{ (x,y) \in \mathbb{R}^2: x^{2} + y^{2}=1 \}$$ Prove that $S$ is uncountable.

This is my attempt at a proof. I don't know if it is valid, or if my logic, and for that matter my approach to the proof, is correct. Feedback/comments/thoughts of any kind are welcome.

Let $G^{+} = \{(x,y)\in G: y \geq0\}$ and $G^{-} = \{(x,y)\in G: y \leq 0 \}.$These are the upper and lower segments of the unit circle.

Notice that $G^{+}\subset S$ and $ \hspace{1mm}G^{-}\subset S$, so $S=G^{+} \cup G^{-}.$

Let $f: G^{+} \rightarrow [-1,1],$ where $f(x,y)=x$. This can be thought of as a projection of the semi-circle onto the $x$-axis.

Since the image of $G^{+}$ under $f$ is equal to the codomain; i.e., $f(G^{+})= [-1,1]$, then $f$ is surjective.

Now since for every $(x,y) \in G^{+}$ we have the cardinality $|f(x,y)|=1$, there exists an inverse function $f^{-1}:[-1,1] \rightarrow G^{+}$ defined by $f^{-1}(x) = (x,\sqrt{1-x^{2})}$.

Thus, there exists a bijection between $G^{+}$ and $[-1,1].$

Similarly we have the same argument for $G^{-}.$ Let $g:G^{-} \rightarrow [-1,1]$ then

$ g(G^{-})=[-1,1]$ (surjective)
$|g(x,y)|=1 \hspace{4mm} \forall (x,y)\in G^{-}$ (one-to-one)
$g^{-1}: [-1,1] \rightarrow G^{-}, \hspace{4mm} g^{-1}(x)= (x,-\sqrt{1-x^{2}})$ (inverse)

which shows the bijection.

Since the set of real numbers $[-1,1]$ is uncountable as can be shown by Cantor diagonalization, and we have $G^{+} {\raise.17ex\hbox{$\scriptstyle\sim$}} [-1,1]$ and $G^{-} {\raise.17ex\hbox{$\scriptstyle\sim$}} [-1,1]$, then that implies that $G^{+}$ and $G^{-}$ are uncountable. Therefore $S= G^{+} \cup G^{-}$ must also be uncountable. q.e.d.

One other approach I thought about was to think of the semicircles as intervals in their own right, where the length of the upper semicircle would be $[0,\pi]$ and the interval of the lower semicircle would be $[\pi,2\pi],$ and I suppose the metric would be the arc length. So essentially you take the arc length and straighten it out, but I didn't know how to approach it or even formalize it. However, I think it is essentially the same thing as what I did in my proof.

This certainly works, but it seems easier just to note that $f(x) = (\cos(2\pi x), \sin(2\pi x))$ is a bijection from $[0,1)$ to $S$. — eyeballfrog, Sep 22 '17 at 20:39
Looks good - though you could shorten it a bit once you know that $x \mapsto (x, \sqrt{1-x^2})$ is an injective map $[-1, 1] \to S$. And similarly, for the other approach, $t \mapsto (\cos t, \sin t)$ is an injective map (and in fact a bijective map) $[0, 2 \pi) \to S$. — Daniel Schepler, Sep 22 '17 at 20:39
Seems like the easiest thing is to note that the projection of the circle to a line (say, the x-axis) is an interval, which is uncountable: $\pi_x(S)=[-1,1]$. — MPW, Sep 22 '17 at 20:42
So this was a bit overkill? Thanks for the suggestions. The idea about using sine and cosine maps crossed my mind, but I didn't know exactly how to approach it. — CrypticParadigm, Sep 22 '17 at 20:43

score 4 · Answer 1 · answered Sep 22 '17 at 20:40

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I think the proof is fine.

You might also like to note that a unit circle can be charactherized by using the corresponding angle, $\theta \in [0, 2\pi)$.

Since $[0, 2\pi)$ is uncountable, you obtain your result as well.

answered Sep 22 '17 at 20:40

Siong Thye Goh

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Can you elaborate? I get that $[0,2\pi)$ is uncountable, but what would the function $f$ look like? For $\theta \in [0,\pi],$ we could have $f(x) = cos^{-1}(x) = \theta$, but $2\pi$? I think what you are saying is what I was talking about towards the end in my original post. – CrypticParadigm Sep 22 '17 at 20:59
https://en.wikipedia.org/wiki/Atan2 check out atan2 function which map $(x,y)$ to $(-\pi, \pi]$. – Siong Thye Goh Sep 22 '17 at 21:08
One could also establish a surjection from the unit circle to $[-1,1]$ by projecting onto the $x$-axis (or $y$-axis). – Oiler Sep 22 '17 at 21:14
nice, in fact, showing surjection is sufficient to conclude that it is uncountable. – Siong Thye Goh Sep 22 '17 at 21:16

score 1 · Answer 2 · answered Sep 22 '17 at 20:40

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Your solution is fine, but as a refinement, note that to prove a set is uncountable, it suffices to show that it has uncountable subset. Since you prove $G^+$ is uncountable, you're done, and don't need to consider $G^-$ at all.

e: It's also worth nothing that you also don't need to prove that $g$ is injective; proving that it's surjective is enough; this proves that the cardinality of the domain is at least the cardinality of the codomain.

answered Sep 22 '17 at 20:40

Duncan Ramage

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for e, you mean if I only considered $G^{-}$ and not $G^{+}$, then I don't need to show $g$ is injective. So I could have stopped when I showed $f$ was surjective-- meaning $G^{+}$ was mapped onto $[-1,1]$ under $f$? – CrypticParadigm Sep 22 '17 at 20:57
Yes. You only need to consider one of $G^-$ and $G^+$, and you would only need to prove your function to be surjective. – Duncan Ramage Sep 22 '17 at 20:59

Prove the unit circle is uncountable

2 Answers2