Is the set $\{ x \in [0,1] | e^x \cos(\sqrt{x^2+1}) \leq 1 \}$ compact?

Question

Is the set $S = \{ x \in [0,1] | e^x \cos(\sqrt{x^2+1}) \leq 1 \}$ compact?

I already proved that continuous image of compact sets is a compact set and also that $f(x)=e^x \cos(\sqrt{x^2+1})$ is a continuous function, but I don't know how to analyse this set because of the condition $f(x) \leq 1$.

I think (by graphic analysis) that it is compact since the image of every point in $[0,1]$ lies below the line $y=1$, so it would be the continuous image of a compact set. But I need to properly justify this, and I'm failing to do this.

Any help would be appreciated.

Answer 1

Unless I am misunderstanding what you mean, in order to show that $S$ is compact you should not be interested in the image of a function. You should be interested in the preimage of some set under a function. If I am not getting your meaning, please let me know.

Specifically, we should look at $f: [0,1] \to \mathbb{R}$ given by $f(x) = e^x \cos(\sqrt{x^2+1})$. With this definition, we should have $ S = \{ x \mid f(x) \leq 1 \}$. This is the same as saying $S = f^{-1}((-\infty, 1])$. Since $f$ is continuous, and $(-\infty, 1]$ is closed, $S$ must also be closed. Since $[0,1]$ is compact, and $S \subseteq [0,1]$, S must be compact - since a closed subset of a compact set is compact.

Answer 2

Yes, a continuous function defined on a compact set maps this set into also a compact set. The condition in the set definition includes boundary, that means that should it ever be $f(x) = 1$, it will be for $x = x_0$ so that $f(x_0) = 1$ and $x_0$ will be part of this set. In this particular case, it looks like $e ^ x \cos(\sqrt {x^2 + 1}) \leq 1$ for all $x \in [0, 1]$ (see figure below) and this set narrows down to $x \in [0, 1]$.

Answer 3

The image of a compact set by a continuous function is compact. Hence : $f([0,1])$ is compact.

Now suppose there is an $x$ such that : $f(x) > 1$, then by the intermediate value theorem there are two points : $0 \geq x_1 < x < x_2 \leq 1$ such that : $f(x_1) = f(x_2) = 1$ since $f(0),f(1) \leq 1$. Hence in this case we have :

$$S = \{x \in [0,1] \mid e^x \cos(\sqrt{x^2+1}) \leq 1 \} = [0,x_1] \cup [x_2,1]$$

which is the union of two compact sets so it is compact.

You can now easily generalise the above argument in order to prove that $S$ is compact. Note that here was using that the inequality is $\leq$ and not just $<1$.

If you want any clarifications don't hesitate to ask.