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