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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.$\ \ $

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    $\begingroup$ It means that if $a$ isn't bigger than $b$, then $a \le b$. $\endgroup$
    – user61527
    Feb 17 '14 at 3:34
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    $\begingroup$ 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. $\endgroup$
    – David
    Feb 17 '14 at 3:42
  • $\begingroup$ 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". $\endgroup$ Feb 17 '14 at 3:54
  • $\begingroup$ Just take the limit $\epsilon \to 0$ $\endgroup$
    – Hawk
    Feb 18 '14 at 0:42
  • $\begingroup$ @sidht, why limits preserve $\le$? BTW limits don't preserve $<$. $\endgroup$ May 11 '14 at 13:47
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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.

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    $\begingroup$ +Upvote. consummate intuition. $\endgroup$
    – NNOX Apps
    Apr 29 '14 at 11:31
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The contrapostive of this statement says if $a>b$ then there exist $\epsilon>0$ such that $a>b+\epsilon$, take $\epsilon = (a-b)/2$.

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    $\begingroup$ does this expatiate on the intuition? $\endgroup$
    – NNOX Apps
    Mar 12 '14 at 12:36
  • $\begingroup$ I don't remember to explain "intuition" when the question was asked. $\endgroup$
    – TTY
    Mar 13 '14 at 23:54
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    $\begingroup$ @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. $\endgroup$
    – TTY
    Mar 14 '14 at 0:01
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    $\begingroup$ but question questioned for 'intuition'? I don't understand your first comment. $\endgroup$
    – NNOX Apps
    Apr 29 '14 at 11:30
  • $\begingroup$ I don't understand it either now, lol. $\endgroup$
    – TTY
    Apr 29 '14 at 16:32
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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 $b+ϵ<a$.

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  • $\begingroup$ Your last sentence makes no sense. $\endgroup$ Feb 17 '19 at 19:47
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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 $\epsilon = (a-b)/3$.

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