Real Plus Epsilon

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
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