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.

  • $\begingroup$ 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... $\endgroup$
    – Math3147
    Sep 14, 2021 at 11:13

1 Answer 1


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

\begin{equation} \forall_{\epsilon > 0} \exists_{a \in A} : a > \sup A - \epsilon . \end{equation}

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

\begin{equation} \forall_{\epsilon > 0} \lvert a-b \rvert < \epsilon \end{equation}

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

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

Then $a_n \to \sup A$.

  • $\begingroup$ 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. $\endgroup$
    – Math3147
    Sep 14, 2021 at 11:30
  • $\begingroup$ 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. :) $\endgroup$
    – Math3147
    Sep 14, 2021 at 11:33
  • $\begingroup$ 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. $\endgroup$ Sep 14, 2021 at 14:43
  • $\begingroup$ 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!)? $\endgroup$
    – Math3147
    Sep 14, 2021 at 20:41

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