What is a rigorous way of deducing $ A \le B \le A$ from $ A-\varepsilon \le B \le A+\varepsilon$?

Question

Deduction. $ \ $ Let $A, B\in \mathbb{R}$, and suppose that, $\forall \varepsilon >0$,

$$ A-\varepsilon \le B \le A+\varepsilon.$$

Then, since $\varepsilon$ can be chosen to be arbitrarily small $ (*)$, we may conclude that,

$$ A \le B \le A.$$

Hence $$ A=B.$$

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I have used the argument $(*)$ above (in italic) several times in proofs, but it has always felt non-rigorous.

What is a more rigorous way of deducing $ A \le B \le A$ from $ A-\varepsilon \le B \le A+\varepsilon$?

I thought of simply taking the limit $\ \lim_{\varepsilon \to 0} \ $ across $ \ A-\varepsilon \le B \le A+\varepsilon$, $ $ then the result follows. But again, I don't have any good reason why taking limits across this inequality is justified.

Answer 1

From the given inequality, $ -\varepsilon \leq B - A \leq \varepsilon $, so $|B-A| \leq \varepsilon$. By properties of absolute values, either $|B-A| = 0$, forcing $B = A$, or $|B-A| > 0$, a contradiction of the given inequality for any positive $\varepsilon$ less than $|B - A|$.

Answer 2

Suppose that $B>A$ and let $\varepsilon=\frac12(B-A)$. Then $B>A+\varepsilon$ (since $B=A+2\varepsilon$ and $\varepsilon>0$), which is impossible.

By the same argument, you can't have $B<A$. Therefore, $B=A$.

Answer 3

Suppose $A\neq B$. Then there is $\epsilon>0$ such that $|B-A|=2\epsilon$. But then $|B-A|\not\leq\epsilon$.