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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.$\ \ $

Martin Sleziak
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MathNoobe
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    It means that if $a$ isn't bigger than $b$, then $a \le b$. –  Feb 17 '14 at 03:34
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    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. – David Feb 17 '14 at 03:42
  • 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". – Jose Antonio Feb 17 '14 at 03:54
  • Just take the limit $\epsilon \to 0$ – Lemon Feb 18 '14 at 00:42
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    @sidht, why limits preserve $\le$? BTW limits don't preserve $<$. – Martín-Blas Pérez Pinilla May 11 '14 at 13:47
  • @Martin But he has $\leq$ not $<$ – Lemon May 12 '14 at 03:26
  • @sidht, yes. My point is that the preservation of inequalities by limits isn't obvious (and in the strict case is false). – Martín-Blas Pérez Pinilla May 12 '14 at 06:13
  • @Martín-BlasPérezPinilla, I agree it is false for $<$, but isn't it true for $\leq$? – Lemon May 12 '14 at 21:09
  • @sidht, yes, is true. But not obvious. Proving it requires the same or more work than proving the original question. – Martín-Blas Pérez Pinilla May 13 '14 at 06:15
  • @Martín-BlasPérezPinilla, but we know that if $g(x) \leq f(x) + \epsilon$, it certainly is true when you take limit. This is just a special case. I am still not seeing the problem…; in fact I thought since (as a weaker statement), if $a < b + \epsilon$ for any $\epsilon >0$, then it is safe to take limit to prove the result. – Lemon May 13 '14 at 06:55
  • @sidht, one more time: my point is that the preservation of $\le$ by limits isn't obvious, and the proof of it isn't significantly easier than the proof of the OP question: http://math.stackexchange.com/questions/418499/limits-preserve-weak-inequalities. – Martín-Blas Pérez Pinilla May 13 '14 at 07:22
  • @Martín-BlasPérezPinilla, oh okay that's what you meant. The link helped clarified. – Lemon May 13 '14 at 07:33

3 Answers3

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

user2820579
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Pedro
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The contrapostive of this statement says if $a>b$ then there exist $\epsilon>0$ such that $a>b+\epsilon$, take $\epsilon = (a-b)/2$.

TTY
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    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
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    @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 00:01
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    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
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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 $b+ϵ<a$.

sas
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