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I would just like it if someone can confirm if my reasoning is correct. Firstly, I proposed that the set had an upper bound of 1 and proved it was true by induction. Then, I supposed that this was not the supremum of the set and hence there must exist another upper bound $1-x < 1$ for this statement to hold. So:

We have $1-x < 1$ and $0 < x < 1$ [clearly $\frac{1}{2}$ is in the set, so, $1-x > 0$ must be true]

$\implies \exists$ $n > \frac{1}{x}$ (such $n \in \mathbb{N}$ exists by the Archimedean Property)

$ \implies -x < - \frac{1}{n}$

$ \implies 1 - x < 1 - \frac{1}{n}$

$ \implies 1 - x < \frac{n}{n} - \frac{1}{n}$

$ \implies 1 - x < \frac{n-1}{n}$

However, this is a contradiction as it has shown that $1-x$ is smaller than some element in the set and so it is not an upper bound. Thus, can I now conclude that 1 is the supremum of the set?

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Helen Byrne
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  • I would explain where the $n$ comes from in your reasoning, i.e. "Take a positive integer $n$ such that $n > \frac1x$." Other than that, your proof looks fine to me. You (claim to) have shown that $1$ is an upper bound, and also that it is the least upper bound of yoour set. – user133281 Jan 05 '15 at 15:59
  • Ok, I added my reasoning for the $n$ to the proof, is it sufficient now? – Helen Byrne Jan 05 '15 at 16:12
  • Supremum is $1$, you was fooled. – Jihad Jan 05 '15 at 16:14

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Your proof is correct. Here I present an easiest way to solve the problem, to know an other technic.

In case where your set is a set where its elements depend on an integer like your set, an easiest way to find the supremum is to choose a sequence wich terms are elements of the set and which limit is an upper bound. $\lim\limits_{n\to+\infty}\dfrac{n-1}{n}=1$ and the terms of this sequence, $\left(\dfrac{n-1}{n}\right)_{n\in\mathbb{N}}$, are elements of the set $X$. Then if 1 is upper bound it's the supremum.

An easiest way is to notice that $\forall n\in\mathbb{N},\,\dfrac{n-1}{n}=1-\dfrac{1}{n}<1$ because $\dfrac{1}{n}>0$.

Using sequences makes things much easier. This exercise is an exemple, where the elements of a set depend on two integers. Finding the supremum without using sequences is very hard.

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Scientifica
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  • So, I should treat it as a sequence and apply the Monotone Convergence Theorem for a simpler method of finding the supremum? Thanks for your help! – Helen Byrne Jan 05 '15 at 16:23
  • You're welcome but in this case I didn't apply this theorem. What I used is the fact that $M=\sup,A$ if and only if $M$ is an upper bound of the set $A$ and it exists a sequence $(a_n){n\in\mathbb{N}}$ such as $\forall n\in\mathbb{N},,a_n\in A$ and $\lim\limits{n\to+\infty}a_n=M$. If we have as an exemple the set $A=\left{\dfrac{4-3n}{2+6n},,n\in\mathbb{N}\right}$ then the limit is $\dfrac{-1}{2}$ which isn't the supremum (in fact it's the infimum because $\dfrac{-1}{2}$ is a lower bound). So pay attention, and sometimes you may have to change the form of the sequence. – Scientifica Jan 05 '15 at 17:53