Derived sets of $\Big\{\frac{n}{\sqrt2}+\frac{\sqrt2}{n}:n\in\mathbb N\Big\}$ and $\Big\{\frac{n}{\sqrt2}+\frac{\sqrt2}{m}:n,m\in\mathbb N\Big\}$

Question

Find the set of all limit points of these sets: $$A:=\left\{\frac{n}{\sqrt2}+ \frac{\sqrt2}{n}:n\in \mathbb N\right\}$$ $$B:=\left\{\frac{n}{\sqrt2}+ \frac{\sqrt2}{m}:n,m\in \mathbb N\right\}$$ The sequence in $A$ diverges to $\infty$ so I don't see any limit points and in $B$ every point $\frac{n}{\sqrt2}$ seems to be a limit point but there could be more.

Answer 1

You are right (twice).

If $A_n=\frac n{\sqrt2}+\frac{\sqrt2}n$, then $\lim_{n\to\infty}A_n=\infty$ and therefore any bounded set of real numbers only has finitely many $A_n$'s. So, no real number is a limit point of the set $A$.

On the other hand, let $l\in\mathbb R$ and let $(n_k,m_k)_{k\in\mathbb N}$ be an injective sequence of pairs of natural numbers such that $\lim_{k\to\infty}\frac{n_k}{\sqrt2}+\frac{\sqrt2}{m_k}=x$. This implies that $x$ is an upper bound of the set of all numbers of the form $\frac{n_k}{\sqrt2}$. But then $\sqrt2x$ is an upper bound of the set $\{n_k\,|\,k\in\mathbb N\}$ and, since this is a set of natural numbers, it must be finite. So, there is natural number $N$ such that $n_k=N$ infinitely often. So, for a subsequence of the sequence $(n_k)_{k\in\mathbb N}$, every $n_k$ is equal to $N$. Let $(n_k^\star,m_k^\star)_{k\in\mathbb N}$ be a subsequence of $(n_k,m_k)_{k\in\mathbb N}$ such that $n_k^\star$ is always equal to $N$. Then we still have $\lim_{k\to\infty}\frac{n_k^\star}{\sqrt2}+\frac{\sqrt2}{m_k^\star}=x$ but now, since the sequence is injective, the $m_k^\star$'s are all distinct. Therefore, $\lim_{k\to\infty}m_k^\star=\infty$ and so $x=\frac N{\sqrt2}$.