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This is Theorem 1.2.6 in Abbott. Understanding Analysis (2016 2 edn). pp 9 - 10. I'm NOT asking about proof that the author proves. Please don't prove. I'm longing just for intuition.

  1. I can't intuit how, God willing, an equality (on LHS) can be equivalent to a conjunction of strict inequalities (on RHS).

  2. Is there any picture that can assist?

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    If $a\neq b$, then $|a-b|$ is positive, so it cannot be smaller than all positive numbers. This is both the intuition, but also the proof. – Wojowu Nov 15 '19 at 09:46
  • How about this. Take a strip of paper. Tell yourself "this strip is too long", tear a part off and throw it away. Then tell yourself again "this strip is too long", tear a part off and throw it away. What will you have left over if you repeat this to infinity – WafflesTasty Nov 15 '19 at 10:06
  • Think of a microscope with resolution $\epsilon$, that is, all objects that differ less than $\epsilon$ are not distinguishable. Now what would you say about two objects you can't distinguish regardless which resolution you use? – Michael Hoppe Nov 15 '19 at 10:14
  • The issue becomes trivial by translating it into natural language: the only non-negative number which is smaller than all positive numbers is $0$. The word "number" may refer to either real numbers or rationals. This is equivalent to "Given any positive number we can find a smaller positive number". And this can be compared with "given any positive integer, we can a greater positive integer (just by adding one to it)". – Paramanand Singh Apr 22 '20 at 04:58

4 Answers4

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Identity of Indiscernibles :

If we cannot find the slightest difference between two objects $a$ and $b$, then necessarily the two must be identical.

And vice versa.

Mauro ALLEGRANZA
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For $x$ non-negative, $\forall \epsilon>0:x<\epsilon$ is a contrived way to say $x=0$ without saying it.

Because if $x$ is positive, you will find some $\epsilon$ that contradicts the inequality $x<\epsilon$ (for example $\epsilon=\frac x2$).

Note that $\forall \epsilon>0:x\le\epsilon$ also works. But not $\forall \epsilon\ge0:x<\epsilon$.

1

Short answer:

Smaller than anything can only be zero.

0

Suppose $|a-b| < \epsilon$ for all $\epsilon > 0$. Thus the distance between $a$ and $b$ is smaller than every strictly positive number.

Suppose to the contrary that this distance is non-zero, then $|a-b| > 0$ and then your claim says that $|a-b| < |a-b|$, which is a contradiction.

Thus we must have that the distance $|a-b| = 0$, which only happens when $a=b$.

Drawing a picture to see what's going on will definitely help. Take two distinct points $a\neq b$. On the picture, argue why $|a-b| < \epsilon$ for all $\epsilon > 0$ can never happen (my proof above contains the answer). So I would argue that the proof is actually the intuition.

J. De Ro
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  • Sorry, but I ruled out proofs like your first 3 paras. that I already understand. –  Nov 15 '19 at 09:57
  • See last lines. The proof is basically the intuition. – J. De Ro Nov 15 '19 at 09:57
  • I just drew the picture, but I'm not seeing the intuition. –  Nov 15 '19 at 09:58
  • Did you draw two different distinct points? Consider their distance $d$. Can the distance $d$ ever be smaller than every positive number? No, you have a problem when your positive number that you chose becomes smaller than this distance. – J. De Ro Nov 15 '19 at 10:00
  • If I would say to you: this object is smaller than $\epsilon > 0$ meters for every $\epsilon > 0$, how large is the object? It must be smaller than $1$ meter, because $1/2 < 1$. It must also be smaller than $1/2$, because $1/4$. It must also be smaller than $1/4$ and keep doing this. – J. De Ro Nov 15 '19 at 10:04