Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? I am reading Tom Apostol's Analysis and come across this theorem. 
Should $a \leq b$ if $a\leq b+\epsilon$ for all $\epsilon >0$? 
I don't doubt the proof in the book but I don't understand the intuition or geometric explanation behind this. Could somebody shed some light on this equation? I just started studying analysis on my own.$\ \ $
 A: What is the possible alternative to $a≤b$ ?
Obviously, it is $b<a$.
Is it possible that at the same time  $a≤b+ϵ$ for all $ϵ>0$ and $b<a$.
OK, let us consider that possibility — in naive geometric sense it means that $b$ is to the left of $a$.
But real numbers have that great property — if we have two different numbers, there exists a number "between" them. So, for some small $ϵ$, for example, the $ϵ$ is equal to half of distance between $a$ and $b$, it is true that $b+ϵ<a$.
A: Draw a number line. Mark the point $b$. Where can you mark $a$? Every number greater than $b$ may be written as $b+\varepsilon$ for some $\varepsilon >0$. Then $a\leqslant b+\varepsilon$ says every number greater than $b$ is also greater than $a$. Thus, you erase all what comes after $b$. The only remaining choices are the numbers to the left or $b$ itself.
A: The contrapostive of this statement says if $a>b$ then there exist $\epsilon>0$ such that $a>b+\epsilon$, take $\epsilon = (a-b)/2$.
A: Assume for contradiction $a > b$. However we can pick $\epsilon = (a-b)/2 > 0$ so that $a > b + \epsilon$, a contradiction of our original assumption $a \le b + \epsilon$. Imagine the real line:
$$
....\underbrace{b ......... b+\epsilon}_{\text{dist of } (a-b)/2} ......... a ....
$$
The same trick works for $a - \epsilon \le b + \epsilon$ using $\epsilon = (a-b)/3$. 
