Proving the supremum of the set {$\cos (1/x)$} is 1

Question

I am trying to prove that the sup { $\cos (1/x): x\in\mathbb{N}$ } = $1$ using the Archimedean property.

Here is my proof so far:

Suppose $y\in$ { $\cos (1/x): x\in\mathbb{N}$ }. Then, since $\cos(1/x)$ is bounded by $1$, we have that $y< 1$ for all $y$ in the set so $1$ is an upper bound.

Now, suppose $z$ is an upper bound such that $z<1$. Then by the Archimedean Property, we have that $\frac {1}{n}$ $< 1-z$ where $n\in\mathbb{N}$. Here is where I am not sure how to continue. I know that $cos(x)>1-\frac{x^2}{2}$ for all $x\neq 0$, so I think I may need to modify the way I have chosen by upper bound.

Answer 1

From $cos(x)>1−\frac{x^2}{2}$ it follows that $cos(\frac{1}{n})>1−\frac{1}{2n^2}$.

Therefore, if you find $n \in \mathbb{N}$, so that

$$ 1−\frac{1}{2n^2}>z \qquad(1) $$

that would imply that $cos\left(\frac{1}{n}\right)>z$

The (1) is equivalent to $\frac{1}{2n^2}<1-z$, оr $\frac{1}{n} < \sqrt{2(1-z)}$. Such a number $n$ exists due to Archimedian property.