Prove $\{(x,y,z)\in \mathbb{R^3}:x^2+y^2+z^2=3\}$ is compact set

Question

@SeverinSchraven I have tried C is compact set and find Contact results but failed — Peter, Nov 23 '20 at 23:21
Hint: If you could find a surjective continuous map $f:[0,1]\rightarrow [0,1]^2$, how would that help you? Do you know such a map? — Severin Schraven, Nov 23 '20 at 23:26
@SeverinSchraven I found the $f$ mapping but still don't understand what you mean — Peter, Nov 23 '20 at 23:47
The beautiful answer of Robert Israel shows you what to do :) — Severin Schraven, Nov 24 '20 at 08:02
Just try to find a surjective continuous map $g:[0,1]^2 \rightarrow S^2$ (where by $S^2$ I mean the unit sphere in $\mathbb{R}^3$). Then $g\circ f$ produces the unit sphere and its complement is not connected. — Severin Schraven, Nov 24 '20 at 09:12
@SeverinSchraven Why "the unit sphere and its complement is not connected" ? — Peter, Nov 24 '20 at 10:40
The complement of the unit sphere $\mathbb{R}^3 \S^2$ is clearly not connected. Use the very definition of being connected and write it as the disjoint union of open sets (there is one obvious choice). — Severin Schraven, Nov 24 '20 at 11:33

score 3 · Answer 1 · answered Nov 23 '20 at 23:26

3

Not necessarily. The unit sphere, for example, is the image of $[0,1]$ under a continuous mapping. You can construct it using a space-filling curve.

answered Nov 23 '20 at 23:26

Robert Israel

I still don't understand what you mean. I tried to prove it $\mathbb{R^3}\setminus$ unit sphers is it connected – Peter Nov 25 '20 at 11:59
It is not connected, because it is the union of two disjoint open sets: the inside of the sphere and the outside. – Robert Israel Nov 25 '20 at 15:54
One more problem I have is finding a f continuous mapping: $[0,1] to$ unit sphers. Your link is $[0,1]\to$ unit ball, so i don't know mean you – Peter Nov 25 '20 at 18:48
You can wrap a ball using a square sheet of paper. That is, there is a continuous map from $[0,1] \times [0,1]$ onto the unit sphere. So map $[0,1]$ onto $[0,1] \times [0,1]$ by a space-filling curve, then map that onto the unit sphere. – Robert Israel Nov 25 '20 at 19:22
I have an idea as follows: Theorem: for all compact set is continuous image of Cantor set. Unit sphere is compact set, so have f continous mapping: Cantor to unit sphere. You think my idea is true or not. I meet a problem is f just from Cantor to unit sphere but i don't sure can extend f from [0,1] to unit sphere. Hope you help – Peter Nov 26 '20 at 07:40
If you want an explicit map from $[0,1]\times[0,1]$ onto the unit sphere, use $(x,y) \mapsto (\sin(\pi x)\cos(2\pi y), \sin(\pi x) \sin(2\pi y), \cos(\pi x))$. – Robert Israel Nov 26 '20 at 15:59

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