Real Plus Epsilon
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Theorem
Let $a, b \in \R$, such that:
- $\forall \epsilon \in \R_{>0}: a < b + \epsilon$
where $\R_{>0}$ is the set of strictly positive real numbers.
That is:
- $\epsilon > 0$
Then:
- $a \le b$
Proof
Aiming for a contradiction, suppose $a > b$.
Then:
- $a - b > 0$
By hypothesis, we have:
- $\forall \epsilon \in \R_{>0}: a < b + \epsilon$
Let $\epsilon = a - b$.
Then:
- $a < b + \paren {a - b} \implies a < a$
The result follows by Proof by Contradiction.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 1$: Real Numbers: $\S 1.7$: Example