# Supremum - Proof.

Is this somehow a valid proof ? Thank you in advance.

Let $S=\Big\{(-1)^{n}(1+\frac{1}{n})|n\in\mathbb{N}\Big\} = \{-2,\frac{3}{2},-\frac{4}{3},\frac{5}{4},\ldots,(-1)^{n}(\frac{n+1}{n})\}$. We want to prove that $\sup(S)=\frac{3}{2}$. We see that $\frac{3}{2}\geq (-1)^{n}(\frac{n+1}{n}), \forall n\in\mathbb{N}$. Thus $\frac{3}{2}$ is an upper bound for $S$, and we have $\frac{3}{2}\geq \sup(S)$.

Now, let $M$ be an upper bound of $S$, and suppose that $M<\frac{3}{2}$, then $\frac{3}{2}-M>0$. By the Archimedian property, we can choose an $n$ such that $(\frac{3}{2}-M)n>1$. By simple algebra, we get that $\frac{3}{2}-\frac{1}{n}>M$. Thus $M$ can not be an upper bound, and we have reached a contradiction. Hence $\frac{3}{2}=\sup(S)$.

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I did not read the second paragraph, it is totally unnecessary. You have not quite shown that $\frac{3}{2}$ is $\ge$ any term in the sequence, though you obviously could. Since $\frac{3}{2}$ is in the sequence, there can be no cheaper upper bound. – André Nicolas Sep 5 '13 at 16:47
Well, I've shown that $\frac{3}{2}\geq \sup(S)$. That must be sufficient for showing that $\frac{3}{2}\geq$ any term in the sequence i think ? Correct me if i'm wrong. – New_to_this Sep 5 '13 at 17:06
Sure it is enough. But in my opinion it is better to work more concretely. – André Nicolas Sep 5 '13 at 17:09
Observe that the supremum cannot be less than $\frac{3}{2}$, since $\frac{3}{2}$ is in the sequence. In general, we have to show that nothing cheaper than our candidate will work. That happens to be immediate in this case, since $\frac{3}{2}$ is actually the max. – André Nicolas Sep 5 '13 at 17:23
For the example of these comments, that would do it. One cannot expect a universal recipe, at best there is a general strategy. – André Nicolas Sep 5 '13 at 18:54

Any nonempty set $A$ of real numbers which is bounded above has a smallest upper bound $\sigma=:\sup A$. This number is characterized in the following way: It is arbitrarily well attained from below, maybe even realized, but not surpassed within $A$. When $\sigma\in A$ then $\sigma$ is the maximal element of $A$ and is denoted by $\max A$.
The elements of your set $S$ are the numbers of the form $a_m:=\bigl(1+{1\over 2m}\bigr)$, $\>m\geq1,$ together with the numbers of the form $b_m:=-\bigl(1+{1\over 2m-1}\bigr)$, $\>m\geq 1$. The $b_m$ are all negative, and the $a_m$ ($m\geq1$) form a monotonically decreasing sequence beginning with $a_1={3\over2}>0$. It follows that all elements $x\in S$ are $\leq{3\over2}$ and that one element of $S$ is actually $={3\over2}$. Therefore ${3\over2}\in S$ is the maximal element of $S$ and at at the same time the smallest upper bound of $S$.