how does $m \leqslant a+\epsilon$ bring out $m \leqslant a$

Question

I have seen something like below in a proof :

$m \leqslant a+\epsilon$ for every $\epsilon>0$ and it bring out $m \leqslant a$

the explanation is that:

as $\epsilon \to 0$ the $a+\epsilon \to a$, so $m \leqslant a$

I can't fully understand it. I know this is something related to limit. Can someone give me more explanation?

Answer 1

Suppose $m>a$ but the above mentioned conditon holds for all $\epsilon$.

Let $m=a+\delta$ ,$\delta>0$ . Now choose any $\epsilon < \delta$ and see that $m$ is always greater than such $a +\epsilon$ .So this leads to a contradiction and we must have $m \leq a$