0

My claim is that the set $\lbrace x: x=\cos(y), y\in [0,1]\rbrace$ is compact. Here is my solution.

Attemption:

First plug the bounds of the set and observe that: $$x=\cos(0)=1 \text{ and } x=\cos(1) \rightarrow \cos(1)\leq x \leq 1$$

So the set has the form: $$S= \lbrace x: \cos(1) \leq x \leq 1 \rbrace$$

It is clear that the set $S$ is closed and bounded so by Heine-Borel we can say S is compact.

Am I right? Thanks in advance!

beingmathematician
  • 687
  • 3
    Yes, should be right. You can also say that $\cos$ is continuous and $[0,1]$ compact and the set is the image of a continuous map: $\cos([0,1])$. But you should fill in more details in your proof. – psl2Z Jun 11 '22 at 15:14
  • 2
    This is indeed a theorem, "the continuous image of a compact set is compact", there are many references on the site, for instance https://math.stackexchange.com/q/26514/399263 – zwim Jun 11 '22 at 15:58

1 Answers1

0

Hint, give a look at the definition of a compact set: Set $A \subset \mathbb{R}^m$ is called compact, if for every sequence $x_n \in A$ there exists a subsequence with a limit belonging to A.

aveline de grandpre
  • 21