Let $D \subseteq \mathbb{R}^2$ be a non-empty subset that is compact in the Euclidian metric space $(\mathbb{R}^2, d)$ and let $E = \{x^2 + y^2: (x, y) \in D\} \subseteq \mathbb{R}$ be another subset. Prove that $E$ is a compact subset of $(\mathbb{R}, | \cdot |)$.
Note: Such a set is called compact if it is closed and bounded.
My attempt:
Bounded:
Since $D$ is bounded, there is some $M$ so that $d(x, y) < M$ for all $x, y \in D$. Suppose $x = (a_1, a_2)$ and $y = (b_1, b_2)$ then
$$d(x, y) = d((a_1, a_2), (b_1, b_2)) = \sqrt{(a_1 – b_1)^2 + (a_2 – b_2)^2} < M$$
From this we need to show that the usual metric on $\mathbb{R}$, denoted by $d_1(x, y) = |x – y|$ is bounded too.
Not sure how to proceed from here.
(Does bounded here mean bounded above and below? Or does what I've done suffice?)
Closed:
A set is closed if either all its limit points are contained in the space or if its complement is open. Finding the limit points of $E$ doesn't seem doable. More so, looking at $E$, I can't seem to figure out its complement.
Another approach could be to show that $E$ is complete. Then, $E$ would be closed but I can't think of how to show that all Cauchy sequences converge in $E$.
Not sure how to do this one.
Any assistance is much appreciated.
Best Answer
The function $f(x,y)=x^2+y^2$ is continuous and $f(D)=E$. Since D is compact, and since f is continuous, E must be compact. Here we are using the fact that the continuous image of a compact set is compact.
Edit: I'll try to give another proof without using topological machinery.
Bounded: If $D$ is bounded then it is contained in some disk $||(x,y)|| \leq R$. Since $x^2+y^2=||(x,y)||^2$ we have $f(x,y) \leq R^2$ for each $(x,y) \in D$ (here we used $f(D)=E$).
Closed: This follows from the reverse triangle inequality $| ||x||-||y|| | \leq ||x-y||$ where $||x-y||=d(x,y)$