Real Numbers Between Epsilons

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Theorem

Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$.

Then $a = b$.


Proof

From Real Plus Epsilon, $b < a + \epsilon \implies b \le a$.

From Real Number Ordering is Compatible with Addition, $a - \epsilon < b \implies a < b + \epsilon$.

Then from Real Plus Epsilon, $a < b + \epsilon \implies a \le b$.

The result follows.

$\blacksquare$


Sources