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Proof:
Let $S = (1,3)\cup(5,8)$
Clearly $\forall x \in S,\;x<8$. Hence $8$ is an upper bound of $S$.
To prove that 8 is the least upper bound suppose there exists an upper bound less than $8$.
So $\exists m<8 \;s.t \;\forall x \in S,\;x<m$.
But $m<\frac{m+8}2 < 8$ and $\frac{m+8}2 \in S$.
Therefore $m$ is not an upperbound. Hence $8$ is the supremum of $S$.

Is the proof correct ?
I have seen in some proofs, they choose an upper bound $m$ as above and take $x = m+\frac{8-m}2$. Then prove $m < x$ and $x \in S$. Is my proof simpler than that ?

Clemens Bartholdy
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Nimantha
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2 Answers2

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Your proof is very similar to the other proof; note that $$m + \frac{8 - m}{2} = \frac{2m}{2} + \frac{8 - m}{2} = \frac{8 + m}{2},$$ i.e. you're using the same counter-example as the other proof, just written differently.

As for whether this proof works, I would say it doesn't quite work. If I gave you the possible upper bound $m = 0$ (I know it isn't an upper bound, but still) and naively plugged it into your argument, then $\frac{m + 8}{2} = 4 \notin S$, contrary to your claim.

You might want to modify the expression $m + \frac{8 - m}{2}$ to $m + ???$ where $???$ is some expression involving $\frac{8 - m}{2}$ and a max formula...

user764828
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  • It does work. If $m=0$, then it's not an upper bound and the implication $$ m \text{ upperbound of }S \Rightarrow \frac{m+8}{2} \in S $$ is true – AlvinL Apr 06 '20 at 07:22
  • @AlvinLepik It's true when $m = 0$ vacuously (and indeed is either vacuously true, or false), but I don't see where the presented argument deals with this. – user764828 Apr 06 '20 at 07:27
  • I understand what you mean. Why is it that this expression is necessarily in $S$? But you are not challenging this point with your example. – AlvinL Apr 06 '20 at 07:27
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    @AlvinLepik What? Of course I am. That's the point. The OP provides an argument, proving that any $m < 8$ is not an upper bound. Their argument claims that, given any $m < 8$, we have $\frac{m + 8}{2} \in S$ and $\frac{m + 8}{2} > m$. I am providing a counterexample to this claim. – user764828 Apr 06 '20 at 07:29
  • @user764828 Isn't your counter example invalid since m is an upper bound of $S$ ? $m$ cannot be $0$. My argument is if $m < 8$ is an upper bound of $S$, then $\frac{m+8}{2} \in S$. Is it really necessary to state that vacuously true case ? See this answer: https://math.stackexchange.com/a/1234705/351436 . Is that answer incomplete beacuse of he didn't mention the above case. – Nimantha Apr 06 '20 at 07:44
  • @Nimantha My counterexample is valid, because you didn't put any restrictions on $m$. I should therefore be able to choose any $m$ I want, follow your reasoning, and come to the conclusion that the $m$ I chose is not an upper bound. If I try the $m = 0$ case, then your proof does not show me why $0$ is not an upper bound of $S$. If I plug $m = 0$ in your proof, then I get to the claim that $\frac{m + 8}{2} \in S$, then I'd think to myself "Hold on, no it's not!". Nothing else in your proof shows me that $m = 0$ is not an upper bound, so the proof doesn't cover all cases, and thus is not valid. – user764828 Apr 06 '20 at 09:35
  • @Nimantha The linked answer has a similar shortfall. They didn't even bother with claiming that $\frac{2 - M}{2} + M \in A$, which is a necessary step in order to draw the conclusion. If I follow their proof with $M = -6$, then $\frac{2 - M}{2} + M = -2 \notin A$, so the quantity $\frac{2 - M}{2} + M$ may or may not be relevant. – user764828 Apr 06 '20 at 09:40
  • @user764828 I'm really confused. In reality there isn't a $m < 8$ such that $m$ is an upper bound. So considering my implication $A \implies B$, $A$ will be false, then no matter $B$ is true or false, $A \implies B$ will be vacuously true. What should I do to make my proof complete ? – Nimantha Apr 06 '20 at 15:52
  • @Nimantha I agree that it can't both be true that $m < 8$ and that $m$ is an upper bound for $S$. I do agree that $8$ is the supremum. I'm not disputing these facts. The problem is, we haven't proven these facts. How do you know that, if $m < 8$, then $m$ is not an upper bound for $S$? I'm not saying it's false, I'm asking how you prove it. If we had, instead, $S = (-\infty, 8)$, then I'd agree that, given $m < 8$, $m < \frac{m + 8}{2}$ and $\frac{m + 8}{2} \in S$, and therefore $m$ is not an upper bound (a number in $S$ is larger than it). But, this argument doesn't work for the $S$ given. – user764828 Apr 06 '20 at 17:18
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    @Nimantha So, what should you do? You need to show that any $m < 8$ is not an upper bound. Meaning, there is some $k \in S$ such that $k > m$. You have to construct this $k \in S$. You can't simply choose $k = \frac{m + 8}{2}$, because, as we established, this doesn't work for just any $m$. In particular, if we have $m = 0$, we get $k = 4 \notin S$, and it is vitally important that our $k$ be in $S$. Try instead showing that$$k=\max\left{6, \frac{m + 8}{2}\right} \in S$$and $k > m$. If we try $m = 0$ again, we get $k = 6 \in S$, and $m < k$, as required. Try proving it in general. – user764828 Apr 06 '20 at 17:24
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As was noted, the main problem is whether the argument for $\frac{m+8}{2}\in S$ is valid. We suppose $m<8$ is an upper bound of $S$. We have to show $1<\frac{m+8}{2} < 3$ or $5<\frac{m+8}{2}<8$.

Suppose it's not the first one. As $m<8$ we have $\frac{m+8}{2}<\frac{8+8}{2} = 8$. If $\frac{m+8}{2} \leq 1$, then $m\leq -6$, which is impossible, because $m$ is an upper bound of $S$, so we must have $\frac{m+8}{2} \geq 3$. If $\frac{m+8}{2} \leq 5$, then $m\leq 2$, which again would be impossible. Thus $\frac{m+8}{2} >5$.

We also see that $m = \frac{m+m}{2} < \frac{m+8}{2} < 8$, which contradicts $m$ being an upper bound of $S$.

Note that it doesn't matter what $m$ is, exactly. All that matters is the supposition $m<8$ enables us to define a certain element in $S$, which gives us a contradiction.

AlvinL
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