I found this proof of the sequential criterion online:
And I understand everything up to the second part where they prove that the convergence of the sequences implies the convergence of the function.
I understand their logical arguments yet I'm having trouble understanding how they proved that their $\{x_n\}$ sequence converges to $c$ because $\forall n \in \mathbb{N}: \; 0 < \vert x_n – c \vert < \frac{1}{n}$ but this doesn't necessarily mean that $\forall \varepsilon > 0$ the condition for convergence holds, only for $\varepsilon$ of the form $\frac{1}{n}$.
Is what I'm saying wrong? Can anyone please explain to me how I might prove that $\displaystyle\lim_{n \to \infty} x_n = c$?
Best Answer
Assume you have the condition for every $n$, and take $\varepsilon>0$. By the Archimedean Principle, there exists $n_0\in\mathbb N$ with $\tfrac1{n_0}<\varepsilon$. So if $|x_n-c|<\tfrac1n$ for all $n$, for any $n<n_0$ you have $\tfrac1n<\tfrac1{n_0}<\varepsilon$. Thus $x_n\to c$.
Note that it looks that you are getting confused by the presence of $\varepsilon$ in the definition of limit. That "$x_n\to c$" should mean that as $n$ grows, $x_n$ should be closer and closer to $c$. And that's definitely guaranteed if the distance between $x_n$ and $c$ is less than $\tfrac1n$.