How Does the Epsilon Argument Prove \( a \leq b \)?

[HMPARTICLE]

Homework Statement

Suppose that , for any $$ \epsilon > 0, a < b + \epsilon $$ . Then $$ a\le b $$

The Attempt at a Solution

I have the proof, its not a question that was assigned to me, it was an example used.
According to the proof i can choose ANY epsilon greater than 0, so let's choose 10. then 2 < 1 + 10 = 11. Right?
Then $$ 2 \le 1 $$ . NOT right.

it's late here in the UK, is it me?

 

[RUber]

It isn't saying for anyone epsilon, but for any epsilon you could possibly choose. Therefore, you can take epsilon as small (positive) as you like and the relation will still be true.

 

[RUber]

$2 < 1 + \epsilon$ for $1 < \epsilon$ but, you can choose any epsilon, so choose $\epsilon = 1/2$, then $2 < 1+ \epsilon$ is false.

 

[HMPARTICLE]

Thanks for clearing that one up! The penny/cent/nickel/kuna has dropped!