# Compact set proof

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.

• Have you proved that if $f:X\to Y$ is continuous, and $K\subseteq X$ is compact, then $f[K]$ is compact? Mar 17 at 21:10
• Can you use the fact that every sequence of elements of a compact set has a subsequence which converges to an element of that set? Mar 17 at 21:11
• @BrianM.Scott No I have not Mar 17 at 21:17
• @JoséCarlosSantos How would that work? Mar 17 at 21:17
• @Kraftsman: This problem is sufficiently similar that you can use the technique in my answer to prove your result. Mar 17 at 21:22

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 $$||x-y||=d(x,y)$$

• Literally an Orange Mar 17 at 21:14
• Unfortunately, the OP doesn’t have that tool available. Mar 17 at 21:18
• Would that follow from this proof? math.stackexchange.com/a/874059/812835 Mar 17 at 21:21
• @kraftsman I gave another proof using your definitions Mar 17 at 21:34
• Could you elaborate on your proof? I'm still a little lost. Mar 17 at 21:38

In order to prove that $$E$$ is bounded take $$z\in E$$. Then $$z=x^2+y^2$$ for some $$(x,y)\in D$$. Since $$D$$ is bounded, there is some $$M\in\Bbb R_+$$ such that the distance from each element of $$D$$ to $$(0,0)$$ is smaller than $$M$$. But then\begin{align}z&=x^2+y^2\\&=(\text{distance from (x,y) to (0,0)})^2\\&

I shall prove now that it is closed. Let $$(e_n)_{n\in\Bbb N}$$ be a sequence of elements of $$E$$ which converges to some $$l\in\Bbb R$$. For each $$n\in\Bbb N$$, let $$(x_n,y_n)\in D$$ be such that $$x_n^{\,2}+y_n^{\,2}=e_n$$. The sequence $$\bigl((x_n,y_n)\bigr)_{n\in\Bbb N}$$ has a subsequence $$\bigl((x_{n_k},y_{n_k})\bigr)_{k\in\Bbb N}$$ which converges to some $$(x_0,y_0)\in D$$. But then, since the map$$\begin{array}{ccc}D&\longrightarrow&E\\(x,y)&\mapsto&x^2+y^2\end{array}$$is continuous,$$l=\lim_{k\to\infty}e_{n_k}=\lim_{k\to\infty}x_{n_k}^{\,2}+y_{n_k}^{\,2}=x_0^{\,2}+y_0^{\,2}\in E.$$Since every convergent sequence of elements of $$E$$ converges to an element of $$E$$, $$E$$ is closed.

• I have not shown that it is bounded though... Mar 17 at 21:30
• Now, I have done it. Mar 17 at 21:41

The easier way to do it, as others have pointed out, is to invoke the fact that $$E=f(D)$$ is compact if $$f$$ is continuous and $$D$$ is compact. However, it may be useful for you to try to follow your reasoning, to grapple with the concepts of boundedness and closedness.

First, I want to clarify that you can only prove that $$d_1$$ is bounded from above since, being a distance, it's non-negative, and $$d(x,x)=0$$, so it doesn't make sense to bound it from below. This being said, take two points $$e_1,e_2\in E$$, write them as $$e_1=x_1^2+y_1^2, \quad e_2=x_2^2+y_2^2$$ for some $$(x_1,y_1),(x_2,y_2)\in D$$. Now, try to compute $$d_1(e_1,e_2)$$ and see if you can bound it by the distance $$d((x_1,y_1),(x_2,y_2))$$, which you know is itself bounded. Then, you will have boundedness.

$$Tip:$$ Since our distance comes from a norm, we can actually avoid bounding the distance between points, and just bound the norm of each point. By the triangle inequality, having a bounded distance is equivalent to having a bounded norm.

Now, to prove closedness, pick a sequence $$\{e_n\}$$ in $$E$$ that converges to a point $$e\in\mathbb{R}$$. Again, write each point in the sequence as $$e_n=x_n^2+y_n^2.$$ Since $$D$$ is closed, the sequence $$\left\{(x_n,y_n)\right\}$$ has a limit point in $$D$$, shall we call it $$(x,y)$$. Since the function $$f(x,y)=x^2+y^2$$ is continuous, we have that $$e=\lim e_n=\lim f(x_n,y_n)=f(\lim(x_n,y_n))=f(x,y)=x^2+y^2.$$ So $$e$$ is in fact an element of $$E$$.