How to prove the set $A = \left\{ \frac 1{2^n} \mid n \in \mathbb N\right\}$ is an infinite set?

Question

I have started to learn in university, and the way they teach us is very simpled... not the way they tought as geometry. The question is simple, but I want an concrete simple mathematic proof(verbal would be enough) something that would help me grasp the concept:

I have this set $$A = \left\{ \frac 1{2^n} \mid n \in \mathbb N\right\}$$ and I need to prove it to be infinite.

I know that $\mathbb N$, the set of natural numbers, is an infinite set, but they don't want me to rely on that, since the second part of the question is to prove their correlation.

So the question is, how do I prove that a set is infinite using simple concepts of sets?

Answer 1

It's a bit hard to work around the fact you've been given the set of elements which is infinite. Almost any proof you can come up with kind of uses that. However, how about a proof by contradiction. Suppose $A$ were finite of order $x$. Then you can write down an element of $A$ which generates a finite subgroup of order more than $x$, a clear contradiction.

Answer 2

HINT

Assume $A$ is finite. Certainly $1/2 \in A$. Then $A$ has a minimal element, say $a$. Then, $a^2 \in A$. Since $a$ is minimal, $a \le 1/2$.

Find a contradiction.

Answer 3

Note $\mathbf N$ is not a group (natural numbers have no additive inverse). Your group $A$ is in bijection with $\mathbf N$, by the map $n\longmapsto\frac1{2^n}$.

Indeed this map is injective, because if $n\ne n'$, say $n<n'$, then $$\frac1{2^{n'}}=\frac1{2^n}\cdot\frac1{2^{\mkern1mu\smash[b]{\underbrace{n'-n}_{>0}}}}<\frac1{2^n}.$$

Also it is surjective by the very definition of $A$. Hence $A$ is infinite (countable, may we add).