Sequential Compactness: Show that there exists a number $\alpha$ and a sequence of positive integers $a_1, a_2, a_3,...$

Question

Here's the problem:

Consider the function $f(x)=\text{cos}(\sqrt{x}e^x)$. Show that there exists a number $\alpha$ and a sequence of positive integers $a_1, a_2, a_3,...$ such that

$$ \lvert f(a_1)-\alpha \rvert \lt 1, \lvert f(a_2)-\alpha \rvert \lt \frac{1}2,\lvert f(a_3)-\alpha \rvert \lt \frac{1}3,... $$

(Hint: remember the definition of sequential compactness and apply it to $[-1,1]$.)

Here is the definition of sequential compactness from my lecture notes: "A set $S \subset \Bbb R$ is sequentially compact if ANY sequence in $S$ has a subsequence converging to a point in $S$."

So, I know that a set is sequentially compact iff the set is closed and bounded. I also know that the outputs of the given function are all, obviously, within the interval $[-1,1]$. There's also this idea that a subsequence can be created that converges to a point in the set, but I don't think that that applies here. Basically, I'm pretty lost on this one, so any help would be appreciated.

Answer 1

Hint 1: what do you know about the sequence $(f(n))_{n=1}^{\infty}$?

Hint 2: note it says the definition of sequential compactness.

Hint 3: Recall the definition of convergence: a sequence $(f(n))$ converges to $\alpha$ if for any given $\varepsilon>0$, there is an $N$ such that $$ \left| f(n) - \alpha \right| < \varepsilon. $$

So, we have a convergent subsequence from compactness. First take $\varepsilon=1$, and choose the first term of your sequence that satisfies $$ \left| f(n) - \alpha \right| < 1, $$ and call this $a_1$. Now put $\varepsilon=1/2$...

Answer 2

As you know, sequence $\cos n$ is dense in $[-1,1]$. The same we can say about the sequence $\cos \sqrt{n}e^{n}$. Let $A$ be an image of $\mathbb{N}$ with function $f$. Let's apply the definition of dense set

$$ \forall \alpha \in [-1;1] \forall \varepsilon>0 \quad \exists a\in A: |\alpha-a|<\varepsilon $$ and remember that $$\forall a\in A \quad \exists n\in \mathbb{N}: a=f(n). $$

Consider a sequence $b_k=\frac{1}{k}$ instead of $\varepsilon$. Then for each $b_k$ we find $a_k\in A$ which is also denoted as $f(n_k)$. Now it's left to substitute these variables to the definition of density and make sure that everything is true.