# 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.$\ \$

• It means that if $a$ isn't bigger than $b$, then $a \le b$.
– user61527
Feb 17 '14 at 3:34
• I suggest you concentrate on the words for all. As often happens in basic analysis, it isn't the algebra that really matters, it's the logic. Hope this helps. Feb 17 '14 at 3:42
• For any real number we have either $a>b$ or $a\le b$ (but not both at the same time) and since the former yields a contradiction for a particular $\varepsilon$ as the book have shown what is the only alternative? And for the intuition behind think about the meaning of the term "for all". Feb 17 '14 at 3:54
• Just take the limit $\epsilon \to 0$
– Hawk
Feb 18 '14 at 0:42
• @sidht, why limits preserve $\le$? BTW limits don't preserve $<$. May 11 '14 at 13:47

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 choices left are the numbers to the left or $b$ itself.

• +Upvote. consummate intuition. Apr 29 '14 at 11:31

The contrapostive of this statement says if $a>b$ then there exist $\epsilon>0$ such that $a>b+\epsilon$, take $\epsilon = (a-b)/2$.

• does this expatiate on the intuition? Mar 12 '14 at 12:36
• I don't remember to explain "intuition" when the question was asked.
– TTY
Mar 13 '14 at 23:54
• @Tucker Rapu nonetheless, the intuition really should really be a sense of "approximation": if something is true with $\epsilon$ arbitrarily small, then the statement probably works when $\epsilon=0$. But I myself don't find it that helpful to talk about intuition in this case.
– TTY
Mar 14 '14 at 0:01
• but question questioned for 'intuition'? I don't understand your first comment. Apr 29 '14 at 11:30
• I don't understand it either now, lol.
– TTY
Apr 29 '14 at 16:32

What is the possible alternative to $$a≤b$$ ?

Obviously, it is $$b. Is it possible that at the same time $$a≤b+ϵ$$ for all $$ϵ>0$$ and $$b.

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.

• Your last sentence makes no sense. Feb 17 '19 at 19:47

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 $$\epsilon = (a-b)/3$$.