Find a continuous bijection with an noncontinuous inverse in metric spaces

continuitygeneral-topologyinverse function

I am asked to find a continuous bijection $f$ from $[0,1)$ to $S$, when $S = ${$(x,y) | x^2 + y^2 \le 1$}, so the it inverse function would be noncontinuous.

In other words, $S$ is the boundaries of the unit circle.

I've tried somehow stretching the interval $[0,1)$ and i managed o find a bijection like this: if $0 \le x \le 1/4$ than $f(x) = (x,\sqrt{1-x^2})$ , and so on for all the parts.

However I'm pretty sure it is not continuous and I got nothing on the inverse.

So I might need some help.

Then there is the same question but with $f$ from $[0,\infty)$ to $S$, with which I also couldn't manage.

Best Answer

I think your question is stated incorrectly, as $S$ is supposed to be the boundary of the unit circle, so should be $\{(x,y) \,:\, x^2+y^2=1\}$.

A bijection from $[0,1)$ to $S$ is $x\mapsto (\cos 2\pi x, \sin 2\pi x)$; the inverse is discontinous at $(1,0)$.

For $[0,\infty)$, you can map that smoothly onto $[0,1)$, e.g. by $x\mapsto \frac{x}{x+1}$, and use the first part.