Hi everyone I'd like to know if the next proof is correct. I'd appreciate any suggestion mainly in the points marks with (1) and (2).
Theorem: Let $E$ be a nonempty subset of real numbers which has an upper bound. Then it must have exactly one least upper bound.
Proof:
We have to show that if $E$ has at most one lub (least upper bound). Suppose $M$ and $M'$ are lub's for E. Then, $M \le M'$ because $M$ is a lub and $M'$ is an upper bound. Similarly, if we interchanged the roles of $M$ and $M'$, i.e., $M$ is an upper bound and $M'$ is a lub then $M' \le M$. Hence, $M= M'$.
Now we have to show that there exists at least one lub. Let $n$ be a positive natural number, let $M$ be an upper bound of $E$ and let $x_0$ be an element of $E$ (this is possible because $E$ is a nonempty set by hypothesis). Then $x_0 \le M$, thus $M-x_0$ is a positive real number. Furthermore, by the Archimedean property there is a natural number $K$ such that $K/n>M-x_0$, i.e., $x_0+K/n >M$ which means that $x_0+K/n$ is an upper bound of E. Moreover $x_0-1/n$ is not an upper bound for $E$.
We claim that there exists a unique natural number $0\le i \le K$ such that $x_0+i/n$ is an upper bound, but $x_0+(i-1)/n$ is not.
We argue this by contradiction. Suppose there not exists such a $i$. Then, this means that whenever $x_0+(i-1)/n$ is not an upper bound then $x_0+i/n$ it must not be an upper bound. Using induction it is not difficult to see that this would imply $x_0+m/n$ is not an upper bound for every natural number $m$. But $K$ by construction is a natural number, a contradiction. Thus, there exists a $i$ such that $x_0+i/n$ is an upper bound and $x_0+(i-1)/n$ is not an upper bound. To show that this $i$ is unique we argue again by contradiction. Suppose there exists a $i'$ with the desired property and $i' \not= i$. Without loss of generality we may assume that $i' < i$. So, we have $\,i' \le i-1$ and thus $\,x_0+i'/n \le x_0+(i-1)/n$, which implies that $x_0+i'/n$ is not an upper bound, a contradiction.
By the last claim we already know that there is a unique integer such that $x_0+i/n$ is an upper bound for $E$, but $x_0+(i-1)/n$ is not. We now claim that $x_0+(i-1)/n <x_0+i/n$. If were not $\,x_0+(i-1)/n \ge x_0+i/n$ we get a contradiction.
By the denseness of $\mathbb{Q}$ in $\mathbb{R}$ there must be a $q_n\in \mathbb{Q}$ such that $x_0+(i-1)/n < q_n <x_0+i/n$. Thus $q_n-1/n$ is not an upper bound because is less than $x_0+i/n$ which is not an upper bound and $q_n+1/n$ is an upper bound because is bigger than the upper bound $x_0+i/n$.
Let $n,m\in \mathbb{N}- \{0\}$. Since $q_n+1/n$ is an upper bound and $q_m-1/m$ is not an upper bound. Then $q_m-1/m<q_n+1/n$ otherwise we have a contradiction. A similar argument shows that $q_n-1/n<q_m+1/m$. Thus, $-(1/n+1/m)<q_n-q_m<1/n+1/m$, in other words we have $|q_n-q_m|<1/n+1/m$. Let $M$ be a positive integer and $n,m\ge M$ then $|q_n-q_m|<2/M$.
For what we have said above if we define the $\mathbb{Q}$-sequence $(q_n)_{n=1}^{\infty}$ this is $2/M$-steady for every $M$. We claim that $(q_n)$ is $\mathbb{Q}$- Cauchy sequence. So let $\epsilon>0$ be arbitrary it will sufficient to have $2/M< \epsilon$ but this means $M > 2/ \epsilon$ which is possible by the Archimedean principle.
Now since $(q_n)$ is a $\mathbb{Q}$- Cauchy sequence it must be the formal limit of a real number. Let $S$ be such a real number.
To conclude the proof our task is to show that $S$ is the least upper bound of $E$.
(1) We shall show that $S$ is an upper bound. Let $x\in E$ be arbitrary. Since $q_n+1/n$ is an upper bound for every n. So, we have $x \le q_n+1/n$ for every $n$. Thus, $x \le q_n$ and hence $x \le S$ which prove that $S$ is an upper bound
(2) Now we will show that $S$ is the lub. Let $T$ be an upper bound of $E$. Since $q_n-1/n$ is not an upper bound for every n, thus $q_n-1/n \le T$. Then $q_n\le T$ and then $S\le T$ which prove that $S$ is the least upper bound and conclude the proof.
Thanks. 🙂
Best Answer
There’s an oversight in your argument for the existence of a least upper bound for $E$: from $x_0\le M$ you can only conclude that $M-x_0$ is non-negative, not that it’s positive. However, if it’s $0$, then clearly $M$ itself is a least upper bound for $E$, so we can focus on the case in which $M-x_0>0$. Instead of using induction to prove that $i$ exists, you could simply let $$B=\left\{i\in\Bbb N:x_0+\frac{i}n\text{ is an upper bound for }E\right\}$$ and set $i=\min B$: $K\in B$, so $B\ne\varnothing$, and the well-ordering principle applies.
You’ve a typo in the paragraph in which you choose $q_n$: you meant to say that $q_n-\frac1n$ is not an upper bound for $E$ because it’s smaller than $x_0+\frac{i-1}n$, not because it’s smaller than $x_0+\frac{i}n$.
I’ve never seen the term steady in this context, but its meaning is clear from the previous paragraph.
At the end in (1) you cannot go directly from $x\le q_n+\frac1n$ for each $n\in\Bbb Z^+$ to $x\le q_n$ for each $n\in\Bbb Z^+$, because the $q_n$’s need not all be the same. What if, for instance, $q_n=-\frac1{n+1}$, and $x=0$? You can, however, argue as follows. If $S<x$, choose $n\in\Bbb Z^+$ such that $\frac3n<x-S$. Then $S\ge q_n-\frac2n$ (why?), and $x\le q_n+\frac1n$, so $x-S<\frac3n$, which is a contradiction.
You have the same kind of error in (2): the fact that $q_n-\frac1n\le T$ for all $n$ does not imply that $q_n\le T$ for all $n$. For now I’ll leave it to you to see if you can repair this last part.