Consider the set $$A=\{(x.y) \in \Bbb{R}^2: x^\frac{2}{3}+y^\frac{2}{3}=1\} \subset \Bbb{R}^2$$
How to prove $A$ is compact with respect to the usual metric on $\Bbb{R}^2$ ?
My try:
It is enough to check $A$ is closed and bounded.
Since $A$ is a inverse image of $\{1\}$ under the continuous map $(x,y) \mapsto x^\frac{2}{3}+y^\frac{2}{3}$, it is closed.
It is bounded, since $\sqrt[3]{x^2}+\sqrt[3]{y^2}=1$ implies $\vert x \vert \leq 1$ and $\vert y \vert \leq 1$.
My Questions are:
1) Is my approach correct? If not, what is the mistake in my answer and how to prove further ?
2) How to draw this graph (generally this type of graphs) in a plane ?
Thanks in advance!
Your argument is correct. I'd describe the intended set as $$A:=\bigl\{(x,y)\in{\mathbb R}^2\bigm||x|^{2/3}+|y|^{2/3}=1\bigr\}\ .$$ Note that the set $A$ is symmetric with respect to both axes. For a plot of the part in the first quadrant you can plot the function $$y=f(x):=\sqrt{(1-x^{2/3})^3}\qquad(0\leq x\leq 1)\ ,$$ or you can use the parametric representation $$t\mapsto\bigl(x(t),y(t)\bigr):=\bigl(\cos^3 t,\sin^3 t\bigr)\qquad\left(0\leq t\leq{\pi\over2}\right)\ .$$