# Real Analysis - Does my counter example disprove this lemma?

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.

• That should say "for every choice of $\epsilon > 0$". – aschepler Mar 7 at 16:56
• Oh, thanks for catching that. – dancingvulture Mar 7 at 16:58
• $\forall \epsilon \exists a$ ... 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. – GEdgar Mar 7 at 17:05
• 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? – dancingvulture Mar 7 at 17:34

## 2 Answers

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.

• 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? – dancingvulture Mar 7 at 17:28
• 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 $b$. – JLinsta Mar 7 at 17:47
• I think I understand now, I just got my dependencies flipped around. Thank you! – dancingvulture Mar 7 at 17:49
• 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 $b > 1 - \epsilon$. – JLinsta Mar 7 at 17:52

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.

• 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 $\epsilon = 1 - a$. – dancingvulture Mar 7 at 17:40
• 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 $a \in A$ – dancingvulture Mar 7 at 17:47