Reasoning why the implication $t - \epsilon \le x \le t + \epsilon$ for $\epsilon \ge 0 \Rightarrow x = t$ holds using sequences. In texts I've seen the following reasoning used several times:
Suppose $t - \epsilon \le x \le t + \epsilon$ holds for $\epsilon \ge 0$. Then it in particular holds for $t - \frac 1 n \le x \le t + \frac 1 n$ $n = 1, 2, ..$
Because it holds for any $n \in \mathbb N$ we can apply the theory of sequences to conlude that $\lim_{n \rightarrow \infty}t - \frac 1 n \le \lim_{n \rightarrow \infty} x \le \lim_{n \rightarrow \infty} t + \frac 1 n$ to conclude that $t \le x \le t  \Rightarrow t = x$.
Why is it allowed to use the theory of sequences here ? The inequlity is not exactly a sequence. Indeed it can be transformed into several real sequences, but wouldn't it be more accurate to conclude that $x = t$, because any other value of $x$ doesn't satisfy the inequality ?
Looking forward hearing your opinion.
 A: Yes, the three statements are equivalent :


*

*$x\neq t$

*$\exists \epsilon > 0$ such that $|x-t| > \epsilon$

*$\exists n\in \mathbb{N}$ such that $|x-t| > 1/n$
The last two are equivalent by the Archimedean property of $\mathbb{R}$.
The reason for working with sequences instead of any $\epsilon > 0$ might have something to do with the context.
A: I don't like what you described either. In my opinion it is very wrong unless properly interpreted. As you said, the entities involved aren't sequences.
This is how I reinterpret that kind of reasoning when I see it:
So given $x,t\in \Bbb R$, you know what $\forall n\in \Bbb N\left(t-\dfrac 1 n\leq x \leq t+\dfrac 1 n\right)$ or equivalently $\color{blue}{\forall n\in \Bbb N\left(t-\dfrac 1 n\leq x\right)}$ and $\color{green}{\forall n\in \Bbb N\left(x \leq t+\dfrac 1 n\right)}$.
Blue tells you that $x$ is an upper bound of $\left\{t-\dfrac 1 n\colon n\in \Bbb N\right\}$, therefore it is greater than it's supremum, that is, $t=\color{grey}{\sup\left(\left\{t-\dfrac 1 n\colon n\in \Bbb N\right\}\right)}\leq x$.
Similarly green and infima  gives you $x\leq t$.
