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So for the supremum of a set, we say:

$S=sup(A) \Leftarrow \Rightarrow S - \epsilon < x$ for some $x \in S$ and $\epsilon > 0$.

I don't understand this for specific examples where the supremum is not in set A, such as:

$A=\{1-1/n:n \in \mathbb{N} \}$

Here the supremum would be $S = 1$, but if $\epsilon = 0.00000... 0001$, it would be greater than 0, but $S - \epsilon \not < x$ for some $x \in S$ (They would be the same).

Arctic Char
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Roland Killian
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  • There isn't any real number that behaves like your $\epsilon$. If there were, what would $\frac \epsilon{10}$ be? What would $10\epsilon$ be? This post discusses your $\epsilon$ in more detail. – MJD Sep 23 '20 at 05:08

1 Answers1

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Your definition of $sup(A)$ is not quite right.

$S = sup(A) \iff \forall \epsilon > 0 ~\exists x ~\in ~A ~\text{such that}~ (S - \epsilon) < x.$

This means, per the definition of $sup(A)$,
first any (small positive) $\epsilon$ is chosen,
then a satisfying $x ~\in A$ must be found.

From the definition of $sup$ that I have given you, you can see that the example that you offered is not in fact a counter example.

That is, in your example, you offered a small $\epsilon$. Per the definition of $sup(A)$, now an $x ~\in A$ must be selected
that satisfies this particular value of $\epsilon$.

Per the definition of $sup(A)$, at no time must an $x ~\in A$ be found that
would satisfy all possible (small positive) values of $\epsilon.$

user2661923
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    Your definition of $\sup(A)$ is still wrong. What means $x \in S$?? – J.-E. Pin Sep 23 '20 at 03:12
  • @J.-E.Pin good catch - thanks. my blind spot in editing. Answer corrected. – user2661923 Sep 23 '20 at 03:15
  • You should also remove the symbol $\ni$. – J.-E. Pin Sep 23 '20 at 03:18
  • @J.-E.Pin Interesting - good suggestion. Never occurred to me that $\ni$ might not be recognized. Editing done. – user2661923 Sep 23 '20 at 03:20
  • I think what is confusing me is, why can't $\epsilon = \lim_{x\to \infty} 1/x$, because then there would not exist an $X \in A$ such that $S - \epsilon < x$. So basically, why can't we set $\epsilon$ to the smallest irrational number after 0. – Roland Killian Sep 23 '20 at 13:02
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    @RolandKillian "why can't we set $\epsilon$ to the smallest irrational number after $0$?" Answer: there is no such animal. That is, for any $\epsilon > 0$ that you choose, no matter how small, after you choose $\epsilon$, I will simply choose some positive integer $M > \frac{1}{\epsilon}.$ Then, I will have that $\epsilon > \frac{1}{M}.$ – user2661923 Sep 23 '20 at 13:23
  • Ok, that makes sense. Thanks for the help! – Roland Killian Sep 23 '20 at 14:06