Real Numbers Between Epsilons

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

  • 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: Exercise $\S 1.8 \ (5)$
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