I was asked to prove that the unit circle is homeomorphic to the unit square(both closed), so I came up with this function: $$f:\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{2}}\rightarrow\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{\infty}}, \left(x,y\right)\mapsto\left(\frac{\left\Vert \left(x,y\right)\right\Vert _{2}}{\left\Vert \left(x,y\right)\right\Vert _{\infty}}x,\frac{\left\Vert \left(x,y\right)\right\Vert _{2}}{\left\Vert \left(x,y\right)\right\Vert _{\infty}}y\right)$$which seems right to me(it's like sending lines from (0,0) passing trough (x,y) if it makes any sense). I proved it's a bijection, but i am having a hard time proving it's a continuous function (with the definition that in a continuous function the pre-image of an open set is open).
What I tried to do is taking $x\in\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{2}}$ and U an open neighborhood of f(x), so I know that $U=B_{r}\left(y\right)\cap\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{\infty}}$ for some y and I can take some radius of f(x) so I can get $$f\left(x\right)\in B_{r'}\left(f\left(x\right)\right)\cap\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{\infty}}\subseteq U$$ and I want to know there is an open neighberhood E of x such that $f\left(E\right)\subseteq B_{r'}\left(f\left(x\right)\right)\cap\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{\infty}}$ so I know that for every $x\in\mathbb{R}^{2}$ i get: $$\left\Vert x\right\Vert _{\infty}\leq\left\Vert x\right\Vert _{2}\leq\sqrt{2}\left\Vert x\right\Vert _{\infty}$$ so I took $E=B_{\frac{r'}{\sqrt{2}}}\left(x\right)\cap\overline{B}_{1}^{\left\Vert \cdot\right\Vert _{2}}$ but I didn't manage to go any further, when I take $x'\in E$ I get:$$\left\Vert f\left(x'\right)-f\left(x\right)\right\Vert _{\infty}=\left\Vert \frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x\right\Vert _{\infty}\leq\left\Vert \frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x\right\Vert _{2}$$ but I can't manage to continue because the norms inside are different, if they were the same I could get $$\left\Vert \frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x\right\Vert _{2}=\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}\left\Vert x'-x\right\Vert _{2}\leq\frac{\left\Vert x\right\Vert _{2}}{\frac{\left\Vert x\right\Vert _{2}}{\sqrt{2}}}\left\Vert x'-x\right\Vert _{2}<\sqrt{2}\cdot\frac{r'}{\sqrt{2}}$$ I also tried using triangle inequality and got $$\left\Vert \frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x\right\Vert _{\infty}\leq\left\Vert \frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x'\right\Vert _{2}+\left\Vert \frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x\right\Vert$$ but I didn't manage to bound $\left\Vert \frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}x'-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}x'\right\Vert _{2}\leq\left(\frac{\left\Vert x'\right\Vert _{2}}{\left\Vert x'\right\Vert _{\infty}}-\frac{\left\Vert x\right\Vert _{2}}{\left\Vert x\right\Vert _{\infty}}\right)$, so I feel like I am going in circles right now, any help is welcomed
This is easy if you divide your problem into a few subproblems. Your function is: $$f\colon (\mathbb R^n\setminus\{0\},\|\cdot\|_2)\to (\mathbb R^n\setminus\{0\},\|\cdot\|_\infty),\quad f(p) =\frac{\|p\|_2}{\|p\|_\infty}p.$$ First prove that $\|\cdot\|_2,\|\cdot\|_\infty$ are continuous from $(\mathbb R^n\setminus\{0\},\|\cdot\|_2)$ to $\mathbb R\setminus\{0\}$. Then use the fact that the quotient of two continuous functions is continuous.
Then prove that $$\mathrm{id}\colon (\mathbb R^n,\|\cdot\|_2)\to (\mathbb R^n,\|\cdot\|_\infty),\quad \mathrm{id}(p) =p,$$ is continuous.
Finally, show that if $$g\colon (X,d)\to \mathbb R,\ h\colon (X,d)\to (\mathbb R^n,\|\cdot\|_\infty)$$ are continuous then $g\cdot h$ is also continuous.
It can be helpful if you observe that Cauchy definition is more or less the same as the definition involving preimages of open sets (for metric spaces).