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Let $π‘Ž \in \mathbb{R} $ and $π‘Ž<𝑏$. Make a conjecture about the supremum of (a, b).

My conjecture was this: Since $π‘Ž<𝑏$ then $𝑠𝑒𝑝(π‘Ž,𝑏)=𝑏$.

Proof: Let $c = sup(a,b)$. Let $d$ be the set of all other upper bounds of $(a,b)$. Then $c \leq d$. Further, since $c$ is an upper bound of $(a,b)$ then $c \leq b$ since $a<b$. But, the greatest value of an upper bound that is less than or equal to $b$ is $b$ itself. Therefore, $sup(a,b) = b$.

Is this correct? What improvements can I make?

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Your proof seems correct, but a better idea is to exploit the fact that no element "less than" the least upper bound can be an upper bound for the given set.

So, here if you claim that $\sup \left( a, b \right) = b$, then you should prove that $\forall \epsilon > 0$, $b - \epsilon$ is not an upper bound for $\left( a, b \right)$.

First, we see that $b - \epsilon < b$. Therefore, there is at least one real number between these two, namely $\dfrac{2b - 2\epsilon}{2}$. Now, we observe that depending on the value of $\epsilon$, $b - \epsilon$ can either be inside $\left( a, b \right)$ or outside it. If $b - \epsilon$ is outside $\left( a, b \right)$, it is possible only if $b - \epsilon \leq a$. Hence, in this case the number $\dfrac{a + b}{2} \in \left( a, b \right)$ works so that $b - \epsilon < \dfrac{a + b}{2}$.

If $b - \epsilon \in \left( a, b \right)$, then the number $\dfrac{2b - \epsilon}{2} \in \left( a, b \right)$ works so that $b - \epsilon < \dfrac{2b - \epsilon}{2}$.

Thus, in both the cases we have at least one number $c = \max \left\lbrace \dfrac{a + b}{2}, \dfrac{2b - \epsilon}{2} \right\rbrace \in \left( a, b \right)$ so that $b - \epsilon < c < b$. Hence, $\forall \epsilon > 0$, $b - \epsilon$ is not an upper bound and therefore $b$ must be the "least upper bound" (supremum) for the set.

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  • $\begingroup$ Can you please explain why you got that $b - \epsilon$ is not an upper bound and why this means that $b$ must be the supremum? $\endgroup$ – Masha Apr 30 '19 at 3:11
  • $\begingroup$ The property of upper bound is that every element of the set should not exceed the upper bound. In our case, for every $\epsilon > 0$, we get an element $c$ in the set such that $c > b - \epsilon$. Therefore, by the very definition of "upper bound", $b - \epsilon$ cannot be an upper bound $\endgroup$ – Aniruddha Deshmukh Apr 30 '19 at 3:36
  • $\begingroup$ Also, by "least upper bound" we mean that it is the "least element" in the set of all upper bounds. Therefore, no real number less than this number can be an upper bound. $\endgroup$ – Aniruddha Deshmukh Apr 30 '19 at 3:36
  • $\begingroup$ So, if I understand correct, $b - \epsilon$ cannot be an upper bound because it is less than $c$. But why does this show that $b$ is the supremum? Thank you for your help! $\endgroup$ – Masha Apr 30 '19 at 3:48
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You should say that if $d$ is an upper bound ($d$ is not a set), then $c\leq d$. Hence(instead of further), since $b$(instead of $c$) is an upper bound of $(a,b)$, then $c\leq b$.

Now, for the rest, what you said is true, but it doesn't feel like is formal enough (take in consideration that you are proving something very trivial, so you may want to be very formal in each step). You can just simply write that if $d$ is an upper bound of $(a,b)$, since $b-\epsilon\in (a,b)$ for $\epsilon <b-a$, you get that $d\geq b-\epsilon$ for all $\epsilon <b-a$. Taking limit of $\epsilon\to 0$, you get that $d\geq b$. So, any upper bound is greater or equal than $b$, which implies that $c\geq b$ (because $c$ is the smallest upper bound).

We conclude $c=b$

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