Theorem 1.1. Given real numbers $a$ and $b$ such that $$a \leq b + \epsilon, \text{ for every } \epsilon \gt 0,\text{ then } a \leq b \tag 1$$
Proof. If $$b \lt a,$$ then inequality $(1)$ is violated for $$\epsilon = (a-b)/2 $$because
$$b + \epsilon = b + (a-b)/2 = (a+b)/2 \lt (a+a)/2 = a $$
I understand Theorem, but I didn't understand the proof part.
why $$\epsilon = (a-b)/2$$ comes out, and what $$b + \epsilon = b + (a-b)/2 = (a+b)/2 \lt (a+a)/2 = a$$ stands for?
The line $$ b + \epsilon = b + (a-b)/2 = (a+b)/2 \lt (a+a)/2 = a $$ stands for a proof of $b+\epsilon < a$, done in several steps: $$ b + \epsilon = b + (a-b)/2, \\ b + (a-b)/2 = (a+b)/2 \\ (a+b)/2 \lt (a+a)/2 \\ (a+a)/2 = a $$
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Why $\epsilon = (a-b)/2$?
In fact, we see in the statement of the theorem, that any positive number could be used for $\epsilon$. This particular choice for $\epsilon$ was made because the calculation (see above) is simple for it.