# Is this lemma correct?

The lemma is stated as an alternative definition for a least upper bound $$\mathrm{sup} A - \epsilon < a$$ Let $$A$$ be a set where $$A \subseteq \mathbb{R}$$ and $$\epsilon > 0$$, such that for every choice of $$\epsilon$$, there exists an $$a$$ mathcing the above formula. Now, I know that $$|a - b| < \epsilon$$, for any real $$a,b$$ but this is only the case if $$a=b$$ as $$a=b \iff |a-b| < \epsilon$$ For any choice of $$\epsilon$$. As a side proof, let $$a=b$$ then trivially since $$\epsilon > 0$$ then $$|a - b| < \epsilon$$. For the converse, say that $$a \neq b$$, then $$|a-b|$$ becomes some choice of $$\epsilon$$ itself and absurdly follows that $$|a-b| < |a-b|$$, which is a contradiction, and hence the equivalence holds. Now, all of this is not news so far or what my concern is about...
I noticed that if we apply this result on $$\mathrm{sup} A - \epsilon < a$$ then by doing a bit of manipulation, we would end up with $$\mathrm{sup} A - a < \epsilon$$ (add $$\epsilon$$ on both sides, and subtract $$a$$ on both sides also). Now, this means that $$\mathrm{sup} A - a$$ must be equal, and to prove so, configure the formula as $$|\mathrm{sup} A - a|^2 = (\mathrm{sup} A - a)^2< \epsilon^2$$, then $$\mathrm{sup} A^2 - 2(\mathrm{sup} A)(a) + a^2$$ but assuming them not being equal will imply that they're not less than any choice of $$\epsilon$$ (and the middle term is not going to cancel out since $$2(\mathrm{sup} A)(a) \leq (\mathrm{sup} A)^2 + a^2$$ and they're equal only when $$\mathrm{sup} A$$ and $$a$$ are equal which supports the disproof [I hope]), and thus why I'm concerned with this lemma, and think that it might be false, this would imply that there exists a member in $$A$$ such that $$a = \mathrm{sup} A$$ always... which is not true, as a maximum does not always exist (e.g. let $$A = [1,{1\over n}]$$ for $$n \in \mathbb{N}$$). Did I go wrong somewhere? Am I just flamboyantly stupid and managed to screw up something simple?

As for context, I was re-reading through my analysis textbook, and I saw the first formula given in the display, stated as true iff (if and only if) $$\mathrm{sup} A$$ exists, for some set $$A$$ bounded from above (and it always exists by the axiom of completeness rendering the lemma as true always, in some sense). The first proof for $$|a-b| < \epsilon$$, for every choice of $$\epsilon$$ (for $$\epsilon > 0$$), and real $$a,b$$ I saw in the book, and re-iterated here, as for the second proof, it is completely made by myself, which is why I'm asking here. Thanks in advance.

• I understand I call it a "lemma", while it is not clear what proof it is getting used in... but the book refers to it as a lemma, and it is used in a proof (making it a lemma)... so... Sep 14, 2021 at 11:13

So the definition is that $$\sup A$$ is such that

$$$$\forall_{\epsilon > 0} \exists_{a \in A} : a > \sup A - \epsilon .$$$$

What you're saying is that if $$a,b \in \mathbb{R}$$ are such that if

$$$$\forall_{\epsilon > 0} \lvert a-b \rvert < \epsilon$$$$

then a = b.

That is true, but you can't apply that reasoning to the definition above, because $$a \in A$$ is not a fixed element, i.e. you can imagine that if you change $$\epsilon$$ then the chosen element $$a \in A$$ also changes. For example if $$A = (0,1)$$ then $$\sup A = 1$$. Let's say you choose $$\epsilon = \frac {1}{2}$$, then we can choose $$a = \frac{3}{4}$$, but if we choose $$\epsilon = \frac{1}{8}$$ then we can't choose $$a = \frac{3}{4}$$ anymore, but we can choose for example $$a = \frac{7}{9}$$.

Basically what the definitions states is that there is always some element $$a \in A$$ that is as close as we want to $$\sup A$$. But because this element is not fixed (it can't be if $$\max A$$ doesn't exist) you can't reason the way you did.

What one could do that is similar to what you did would be to find a sequence in A converging to $$\sup A$$ as such:

We know that

$$$$\forall_{n \in \mathbb{N}} \exists_{a_n \in A} : \lvert a_n - \sup (A) \rvert < \epsilon$$$$

Then $$a_n \to \sup A$$.

• A universal quantifier implies an existential quantifier. If $\forall x \psi \to \exists x \psi$, if my proof holds then $\forall x \lnot \psi \to \lnot \exists x \psi$. Hence even if it is not fixed (which is the case anyway since it is arbitrary, the proof must hold, otherwise the proof cannot hold if $a$ is not arbitrary, which is what you're arguing from what I cans see/understand). I don't see a reason to say that the proof assumes $a$ to not be arbitrary. Sep 14, 2021 at 11:30
• Can you prove that, my statement about the proof assumes $a$ to not be arbitrary? If so, I'll mark this as an answer, and thank you. :) Sep 14, 2021 at 11:33
• You proved that $(\forall_{\epsilon > 0} \lvert a - \sup A \rvert < \epsilon )\implies \sup A = a$. But there is no $a \in A$ that has that property, nor does the definition say so. If $a$ wasn't fixed in your proof, then what is the statement of your proof? Supposedly it is that $a = b$, but if $a$ is not fixed what does that mean?But there is no $a \in A$ that has that property, nor does the definition say so. Sep 14, 2021 at 14:43
• It is the purpose of the proof to show that if $\mathrm{sup}A \neq a$ (meaning $a$ is now arbitrary), then $\forall \epsilon \; \exists a \; (\epsilon > 0 \; \to \; (|\mathrm{sup}A - a| < \epsilon))$. If there is no $a$ that has the property of $\mathrm{sup} A = a$ then the inequality doesn't hold for any choice of $\epsilon$. That's my statement, and I invoke it through: $\forall a \; \forall b \; \forall \epsilon \; (\epsilon > 0 \; \to \; (|a - b| < \epsilon))$. Now, what I'm failing to understand is where $a$ stops being arbitrary, can you re-frame perhaps (sorry!)? Sep 14, 2021 at 20:41