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:


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?)


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.

  • $\begingroup$ Have you proved that if $f:X\to Y$ is continuous, and $K\subseteq X$ is compact, then $f[K]$ is compact? $\endgroup$ Mar 17 at 21:10
  • $\begingroup$ 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? $\endgroup$ Mar 17 at 21:11
  • $\begingroup$ @BrianM.Scott No I have not $\endgroup$ Mar 17 at 21:17
  • $\begingroup$ @JoséCarlosSantos How would that work? $\endgroup$ Mar 17 at 21:17
  • $\begingroup$ @Kraftsman: This problem is sufficiently similar that you can use the technique in my answer to prove your result. $\endgroup$ 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 $||x-y||=d(x,y)$

  • $\begingroup$ Literally an Orange $\endgroup$
    – C Squared
    Mar 17 at 21:14
  • $\begingroup$ Unfortunately, the OP doesn’t have that tool available. $\endgroup$ Mar 17 at 21:18
  • $\begingroup$ Would that follow from this proof? math.stackexchange.com/a/874059/812835 $\endgroup$ Mar 17 at 21:21
  • $\begingroup$ @kraftsman I gave another proof using your definitions $\endgroup$ Mar 17 at 21:34
  • $\begingroup$ Could you elaborate on your proof? I'm still a little lost. $\endgroup$ 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\\&<M^2.\end{align}

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.

  • $\begingroup$ I have not shown that it is bounded though... $\endgroup$ Mar 17 at 21:30
  • 1
    $\begingroup$ Now, I have done it. $\endgroup$ 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 $E$.


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