The Sequence $(\frac{1}{n})_{n \in \mathbb{N}}$ Does Not Converge in $\mathbb{R}$ w. discrete metric

Question

Show that, in $\mathbb{R}$ with the discrete metric, the sequence $(\frac{1}{n})_{n \in \mathbb{N}}$ does not converge.

I'd appreciate a proof verification/help. This is how I started my proof:

Our sequence $S := (\frac{1}{n})_{n \in \mathbb{N}}$ is $1, \frac{1}{2}, ... , \frac{1}{n}$ in $\mathbb{R}$. Since none of the terms in $S$ are equal, i.e. $s_i \ne s_j$ for $i \ne j$, with the discrete metric the distance between all the terms of $S$ is $1$. (Since wrt. discrete metric, $d(x, y) = 1$ if $x \ne y$). This means the distance between all our terms is $1$. This means that $s_n$ is not getting arbitrarily getting close to any term. So, $S$ cannot converge.

I can't really put the bolded part of the proof into "mathematical language." Should I have started this with proof by contradiction? I'd appreciate any correction/guidance in proving the statement. Thank you.

Answer 1

Suppose $(s_n)$ would converge. Then it is a Cauchy sequence (convergence always implies Cauchy). However, as you pointed out, it is not Cauchy, which contradicts the convergence assumption.

Answer 2

Assume that the sequence converges to $x\in\mathbb{R}$. Then, any open set containing $x$ must contain all but finitely many terms of the sequence (definition of limit in a topological space). Consider the open set $\{x\}$. Obviously, this contains $x$. Then, all but finitely many terms of the sequence are in the set $\{x\}$. But, this means all but finitely many terms of the sequence are equal to $x$ and yet at the same time, $\frac{1}{n}\neq \frac{1}{m}$ for $m\neq n$ so we have a contradiction.