Symmetrically Bounded?

Consider the first inequality, that is, $a - \dfrac{1}{n} \leq b$

Taking $n \to \infty$ , and noting that $\lim_{n \to \infty}\dfrac{1}{n} = 0$ (as a consequence of the Archimedian property), we deduce that $a \leq b$

Similarly, the other inequality gives $b \leq a$

Hence, we conclude that $a = b$

One of the most fundamental properties of the real numbers is the Archimedean property : if a number $x \ge 0$ satisfies that $x \le \frac{1}{n}$ for all positive integer $n$ , then $x = 0$ . In some constructions of real numbers, this is a theorem that follows from the basic axioms; in other constructions, this is taken as an axiom (so the set of real numbers is an object that satisfies certain properties, including Archimedean property).

With this, the proof is rather straightforward. First, we need to prove a few of the most basic properties, like commutativity/associativity/inverse of addition and stuff. Next, we need to show that if $x \le y$ for some reals $x, y$ , then $x + z \le y + z$ for all real $z$ ; also, if $x \le y$ , then $-y \le -x$ . With this, we can now represent $b = a + c$ , so the inequalities become $a - \frac{1}{n} \le a + c \le a + \frac{1}{n}$ for all positive integer $n$ ; if we can show $-\frac{1}{n} \le c \le \frac{1}{n}$ implies $c = 0$ , we're done. But this final part can be done by Archimedean property as well; $c \le \frac{1}{n}$ means $c \le 0$ , and $-\frac{1}{n} \le c$ means $-c \le \frac{1}{n}$ , so $-c \le 0$ , so $0 \le c$ . Thus $0 \le c \le 0$ , and we can show that $c = 0$ . This shows that $a = b$ .

Relevant wiki: Squeeze Theorem

Since the inequality $a - \dfrac 1n \le b \le a + \dfrac 1n$ is true for all positive integers $n$ , then it is true for:

$\begin{aligned} \lim_{n \to \infty} \left(a - \dfrac 1n\right) & \le b \le \lim_{n \to \infty} \left(a + \dfrac 1n\right) \\ a & \le b \le a & \small \color{#3D99F6} \text{By squeeze theorem} \\ \implies & \boxed{a=b} \end{aligned}$

let say that $a>b$ so we must have a fix real number $\epsilon$ such that $a=b+\epsilon$ . Define $N=\lceil \frac{1}{\epsilon} \rceil$ if the statement is true how about when $n>N$ ? a contradiction now let say $a<b$ so $a=b-\epsilon$ . Define $M=\lfloor \frac{1}{\epsilon} \rfloor$ how about $n<M$ ? a contradiction so $a=b$

If a=b we've got: $a-\frac{1}{n}\leq a\leq a+\frac{1}{n}$ By decreasing with $a$ the entire inequality we will get: $\frac{-1}{n}\leq 0\leq \frac{1}{n}$ which is true.

$a - \frac { 1 } { n } \leq b \leq a + \frac { 1 } { n } \rightarrow an - 1 \leq bn \leq an + 1 \rightarrow \boxed { a = b }$ we get.