$M=\lbrace \frac{(-1)^n}{n}:n \in \mathbb N \rbrace$ not orderly countable

Question

A non-empty set $M$ is called orderly countable if there is an surjective, strictly increasing $f: \mathbb N \to M$.

Proof: $M=\left\{ \frac{(-1)^n}{n}:n \in \mathbb N \right\}$ is not orderly countable.

Answer 1

Assume that such an $f$ exists. Since $f$ is surjective, there is a $k\in\mathbb{N}$, such that $f(k)=\frac{1}{2}$.

The set $A=\{\frac{1}{4},\frac{1}{6},\ldots\}$ is a subset of $M$, and since $f$ is increasing, we must have that $f^{-1}(A) \subseteq \{1,2,\ldots,k-1\}$, which is impossible, because $f$ is surjective and $A$ is infinite.

Answer 2

Let $M = M_-\cup M_+$ where $M_-$ (resp. $M_+$) contains negative (resp. positive) elements of $M$. Each contains infinitely many elements. But there is no $N$ such that $f(N) < 0 < f(N+1)$ (because no matter $n$ we choose, there is $m$ such that $\frac{-1}{2n+1} < \frac{-1}{2m+1}<0$).

Answer 3

Assume such a function $f$ exists. Distinguish two cases:

• there is $m$ with $f(m)> 0$.

• there is no $m$ with $f(m)> 0$.

The second case, contradicts surjectivity quite directly. For the first, note that then there can only be finitely many negative values of $f$, which also contradicts surjectivity.