The theorem of sequential continuity states that:
f: $\mathbb{R} \rightarrow \mathbb{R}$ is continuous at $a \in \mathbb{R}$ $\Longleftrightarrow$ $f(x_n) \rightarrow f(a)$ $\forall$ sequences $x_n \rightarrow a$.
In the proof of the converse statement, i.e. suppose we know $f(x_n) \rightarrow f(a)$ and we want to prove $f$ is continuous, it is written that
(1) Suppose for a contradiction $f$ is not continuous, and suppose $f(x_n) \rightarrow f(a)$ for all sequences $x_n \rightarrow a$.
(2) Then $\exists \epsilon > 0$ s.t. $\forall \delta>0, \exists x \in
(a-\delta, a+\delta)$ s.t. $|f(x)-f(a)| \geq \epsilon$.(3) Choose $\delta = \dfrac{1}{n}$. Then $\exists x_n ∈ (a – \dfrac{1}{n}, a + \dfrac{1}{n})$ such that $|f(x_n) − f(a)| \geq \epsilon$. So $|x_n – a|<\dfrac{1}{n}$ $\forall n$ and therefore $x_n \rightarrow a$. But $f(x_n) \not\rightarrow f(a)$.
I can understand that from (2) is the negation of the definition of continuity which we assumed for a contradiction but I cannot understand how we get to $x_n \rightarrow a$ from $|x_n – a|<\dfrac{1}{n}$ $\forall n$.
Best Answer
The definition of $x_n \to a$ is
$$\forall \varepsilon >0: \exists N \in \mathbb{N}: \forall n \ge N: |x_n -a| < \varepsilon $$
So pick $\varepsilon>0$ and note that there is some $N \in \mathbb{N}$ such that $\frac{1}{N}< \varepsilon$. This is a consequence of the Archimedean property of the real numbers, if you want to be precise, or a consequence of the fact that $\frac{1}{n} \to 0$ in $\Bbb R$, if you covered that. I claim that this $N$ is as required, because for all $n \ge N$ we have $\frac{1}{n} \le \frac{1}{N}$ so
$$|x_n -a| < \frac{1}{n} \le \frac{1}{N} < \varepsilon$$
So the fact that $x_n \to a$ is an almost immediate consequence of $\frac{1}{n} \to 0$, really.