Two definitions of convergence?

Question

I remember that I've ever seen these two definitions of convergence somewhere:

1. A sequence $S_n$ converges to $s$ iff $\forall m\in\mathbb{N}, \exists N$ such that $\forall n>N,|S_n-s|<\frac{1}{m}$
2. A sequence $S_n$ converges to $s$ iff $\forall r\in\mathbb{R}>0, \exists N$ such that $\forall n>N,|S_n-s|<r$.

My question is whether in reals those two definitions are equivalent or not? I suppose they are, but I am just not sure about the proof. Cheers!

Answer 1

2) to 1) is clear. Now assume that 1) holds, for every $r>0$, we find a large $m\in{\bf{N}}$ such that $1/m<r$, and by 1), we have something like $|S_{n}-s|<1/m<r$, so $|S_{n}-s|<r$ is then established.

Answer 2

Yes, they are equivalent, basically because $\frac{1}{n}\to 0$. For a rigorous justification of this, see the Archimidean property of the reals.

Answer 3

Let's show that

i) $\forall m\in\mathbb{N^*}, \exists N \in \mathbb{N^*},\forall n>N,|S_n-s|<\frac{1}{m}$

and

ii) $\forall r >0, \exists N \in \mathbb{N^*}, \forall n>N,|S_n-s|<r$

are equivalent.

No problem showing ii) implies i).

Now suppose i). Let $r>0$. There exists an $m \in \mathbb{N^*}$ such that $\frac{1}{m} < r$. There also exists $N$ such that $\forall n>N,|S_n-s|<\frac{1}{m}$.

Hence $\forall n>N,|S_n-s|< r$.

Hence i).