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I was reading a real analysis text that gave the following lemma after introducing the axiom of completeness and least upper bounds, it reads:

Assume $s \in \mathbb{R}$ is an upper bound for a set $A \subseteq \mathbb{R}$. Then, $s = sup A$ if and only if, for every choice of $\epsilon \gt 0$, there exists an element $a \in A$ satisfying $s - \epsilon \lt a$.

When thinking about this lemma I was testing it with some example sets, and I stumbled across one that seemed to disprove it. The set is simply the set of real numbers on the interval $[0,1)$, our supremum is $s = 1$, and I set $\epsilon = 1 - a$. Putting it all together we get

$$1 - (1 - a) \lt a$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a \lt a$$ As far as I can tell this set fulfills all of the requirements for A, 1 is the supremum, and $1-a$ is a valid choice for $\epsilon$, as no choice of $a \in [0,1)$ will make it less than or equal to zero. But it also feels very unlikely this lemma would be disproved so easily, so I'm sure I went wrong somewhere, but I'm not sure where. If anyone has some insight into this it would be greatly appreciated, I can provide further context if necessary.

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    $\begingroup$ That should say "for every choice of $\epsilon > 0$". $\endgroup$ – aschepler Mar 7 at 16:56
  • $\begingroup$ Oh, thanks for catching that. $\endgroup$ – dancingvulture Mar 7 at 16:58
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    $\begingroup$ $\forall \epsilon \exists a$ ... we allow $a$ to depend on $\epsilon$, but we do not allow $\epsilon$ to depend on $a$. So $\epsilon = 1-a$ is not valid for a counterexample. $\endgroup$ – GEdgar Mar 7 at 17:05
  • $\begingroup$ I'm not entirely sure what $\epsilon$ depending on $a$ or vice versa has to do with this. Doesn't the lemma state that there must be some $a \in A$ that works for every possible choice of $\epsilon$? Isn't $\epsilon = 1 - a$ included in the "every" for any a? $\endgroup$ – dancingvulture Mar 7 at 17:34
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For every $\epsilon > 0$, there exists some $b \in A$ such that $b > \sup A - \epsilon$. In your example, you showed that for $\epsilon = 1 - a$, $b = a$ does not work. But there are other $b \in A$ that will work! For example, $b = \frac{1+a}{2}$ will work in your example.

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  • $\begingroup$ Wouldn't your choice of b fail for $\epsilon = 1 - \frac{1+a}{2}$? That choice of epsilon is still greater than zero, so it counts as part of the "every possible choice" of $\epsilon$ doesn't it? $\endgroup$ – dancingvulture Mar 7 at 17:28
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    $\begingroup$ The point is that for every $\epsilon$, there can be a different $b$. So when you change $\epsilon$, you can pick a new $b$. There does not necessarily exist a $b$ that works for ALL $\epsilon > 0$, but for every $\epsilon > 0$, we can find some suitable $b$. $\endgroup$ – JLinsta Mar 7 at 17:47
  • $\begingroup$ I think I understand now, I just got my dependencies flipped around. Thank you! $\endgroup$ – dancingvulture Mar 7 at 17:49
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    $\begingroup$ No problem! Think about it this way: for $\epsilon > 0$, we can choose $b = \frac{2-\epsilon}{2}$ (the average of $1$ and $1-\epsilon$). This choice of $b$ certainly belongs to $[0,1)$ and satisfies $b > 1 - \epsilon$. $\endgroup$ – JLinsta Mar 7 at 17:52
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for every choice of $\epsilon>0$, there exists an element $a \in A$ ...

means that we choose the number $a$ depending on the value of $\epsilon$. Since $a$ must be allowed to depend on $\epsilon$, you can't have $\epsilon$ depending on $a$.

In other words, when you say $\epsilon = 1-a$, what is this number $a$ you're using? No such variable has been introduced yet.

In formal logic, we call $a$ "bound" by the phrase "there exists an element $a \in A$ satisfying $s - \epsilon < a$" - the variable exists for the purpose of that quantified phrase, and has no meaning outside that phrase.

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  • $\begingroup$ Perhaps my lack of serious background in logic is showing, but when I write $\epsilon = 1 - a$ I'm expressing that, no matter what $a \in A$ we choose to satisfy this inequality, when we run through every possible choice of $\epsilon \gt 0$, that choice of $a$ will fail when $\epsilon = 1 - a$. $\endgroup$ – dancingvulture Mar 7 at 17:40
  • $\begingroup$ Oh wait, you're totally right actually; some a doesn't have to work for all $\epsilon$, it's just that each choice of $\epsilon$ must work for at least one $a \in A$ $\endgroup$ – dancingvulture Mar 7 at 17:47

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