Prove: If real numbers $a$ and $b$ satisfy the inequality $a \leq b+\varepsilon$ for any $\varepsilon > 0$, then $a \leq b$.
2026-04-25 00:09:31.1777075771
Proving the Epsilon Trick
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If $a>b$ then there is $n > 1/(a-b)$. Take $\epsilon=(1/n) \lt (a-b)$. Therefore, $b+\epsilon < a$. Contradiction.