prove $S^2=\{(x,y,z)\in R^{3}:x^2+y^2+z^2=1\}\sim\mathbb{R}$

Question

attempt:

Find two injective functions $f:S^2\to\mathbb{R}$ and $g:\mathbb{R}\to S^2$ sufficent to prove equinumerous.

$$f:S^2\to\mathbb{R}\\(x,y,z)\to e^x+e^y+e^z$$

Verify: if $(x,y,z)\neq(\hat{x},\hat{y},\hat{z}),$ then $f((x,y,z))\neq f((\hat{x},\hat{y},\hat{z}))$ by observation.

re-edited For simplicity, just consider the projection function $$f:S^2\to\mathbb{R}\\(x,y,z)\to e^x$$

$$g:\mathbb{R}\to S^2\\r\to \begin{cases} x =\sin(\tan^{-1}(r)) \\ y =\cos(\tan^{-1}(r))\\z=0 \\ \end{cases}$$

Verify: $\tan^{-1}(\cdot)$ is injective function, sign any real numbers to $(\frac{-\pi}{2},\frac{\pi}{2})$, and $\sin(\cdot)$ and $\cos(\cdot)$ are injective in the domain $(\frac{-\pi}{2},\frac{\pi}{2})$.

I spent plenty of time coming up with function $g$, really appreciate it if you can let me know if this is correct

Your function $f$ is not injective, for example check the poles (the points having all coordinates $0$ except one which is $1$). — Wiktor Vacca, Mar 02 '22 at 10:32
@WiktorVacca Thank you so much for your reply. You are right, I believe my re-edited function is correct now. — LJNG, Mar 02 '22 at 10:50

Chris Sanders · Accepted Answer · 2022-03-02T11:09:19.980

1

It's best to think simplistically.

Yes, your example of $g$ is correct, but coming up with $g$ shouldn't take a long time. $\mathbb{R}$ has the same cardinality as the open interval $(0,1)$. The set $S^2$ is just the sphere in three-dimensional real space. If you think to yourself, "Hmm, what can I draw on the sphere that resembles an open interval?", you'll realise the answer before too long.

As for $f$, if you can find an injective map from $\mathbb{R}^3$ to $\mathbb{R}$, then the restriction of such a map to $S^2\subset\mathbb{R}^3$ will also be injective.

This link has more info: Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$

edited Mar 02 '22 at 11:09

answered Mar 02 '22 at 10:58

Chris Sanders

7,137
10
23

Thank you for your reply. I glanced at you were asking if I see open interval and sth have the same cardinality, would you please post that question again? – LJNG Mar 02 '22 at 11:08
$\mathbb{R}$ and $(0,1)$ have the same cardinality. If you're learning about cardinality, you'll probably have come across this fact – Chris Sanders Mar 02 '22 at 11:11
Hi, it's been a while since I retrieved the comment. First, Could you please give an example of how to come up with a good $g$, is example $g$ overcomplicated? I am not confident in thinking of this example in the exam. What would you propose for $g$? Second, would you please elaborate "Hmm, what can I draw on the sphere that resembles an open interval?" . I've seen the link that you shared. In that example, what I see is a tuple rather than an interval, what if $(a,b)$ such that $b<a$. Thank you so much – LJNG Feb 26 '24 at 14:28

prove $S^2=\{(x,y,z)\in R^{3}:x^2+y^2+z^2=1\}\sim\mathbb{R}$

1 Answers1