Prove the least upper bound property using Bolzano Weierstrass theorem.
I know there are quite a fair number of similar questions on the site, but none of them provide satisfactory proofs. Does anyone happen to have a rigorous proof for this?
I will appreciate it very much. Thanks.
1) Bolzano-Weierstrass $\Rightarrow$ Nested Interval Theorem
Let $I_n= [a_n,b_n]$ and $I_1\supset I_2\ldots\supset I_n \supset \ldots$, so for each $n$ we have $a_1\le a_n\le b_1$, thus $(a_n)$ the sequence of leftmost endpoints, it's bounded and so by the BW theorem there is a subsequence $(a_{n_k})$ which converges to some $a\in I_1$, i.e., $a_{n_k}\to a$.
We shall show that $a_{n}\le a$ for all $n$. Suppose for sake of contradiction that there is some $m$ such that $a_m>a$. Let $\varepsilon = a_m-a$. Then there is a $N> m$ such that $|a_{n_k}-a|< a_m -a$, i.e., $a_{n_k}< a_m$ for all $k\ge N$. Choose $k = N$ so $a_{n_N}< a_m$ and then $a_{n_N} \notin I_m$. But $N>m$ and by hypothesis we have $I_m \supset I_{n_N}$, in particular $a_{n_N} \in I_m$, a contradiction.
To conclude we need to show that $a\le b_n$ for all $n$. Let $n$ be arbitrary. Since for any $n$ we have $a_n\le b_n$, in particular $a_{n_k}\le b_{n}$ for all $k \ge n$, so $\lim_k a_{n_k}=a\le b_n$.
Hence $a \in I_n$ for all $n$, i.e., $a \in \bigcap_n I_n$.
2) Nested Interval Theorem $\Rightarrow$ Least Upper bound principle
Let $E$ be a subset of $\mathbb{R}$ which is bounded above and is non-empty. Since $E$ is non-empty there is some $a_1 \in E$ which is not an upper bound. Also because the set is bounded above there is some $b_1$ such that is an upper bound. Let $I_1$ be the closed interval $ [a_1,b_1]$ and let $c$ be the midpoint of $I_1$. So either $c$ is an upper bound or is in $E$. If $c$ is an upper bound consider $I_2= [a_2,b_2]$, where $a_2=a_1$ and $b_2=c$. If were the case in which $c$ is not an upper bound so $a_2=c$ and $b_2=b_1$.
Now suppose we have already chosen the interval $I_n$ such that the leftmost point is in $E$ and the rightmost point is an upper bound. Let $c$ be the midpoint of $I_n$. Thus either $c$ is an upper bound or is in $E$. For the former we let $a_{n+1}=a_n$ and $b_{n+1}=c$ and for the latter $a_{n+1}=c$ and $b_{n+1}=b_{n+1}$. Let $I_{n+1}= [a_{n+1},b_{n+1}]$. Either way $I_{n+1}$ is an interval such that the leftmost-endpoint is in $E$ and the rightmost-endpoint is and upper bound for $E$.
This process yields to a nested interval $I_1\supset I_2\ldots\supset I_n \supset \ldots$ and so we know that their intersection is non-empty. Let $x$ such a point. We claim that $x= \sup E$.
a) We shall show that $x$ is an upper bound. Suppose for sake of contradiction that $x$ is not an upper bound. Then there is some $y\in E$ such that $y>x$. Let $\varepsilon = y-x$ and choose $N$ such that $(b_1-a_1)2^{-N+1}< \varepsilon$. Since $y\le b_n$ because the right-endpoints was chosen to be upper bounds. So for $n\ge N$ we thus have:
$$y-x\le b_n-x\le b_n-a_n=\frac{b_1-a_1}{2^{n-1}}\le \frac{b_1-a_1}{2^{N-1}}< \varepsilon = y-x$$
a contradiction.
b) We shall show that $x$ is the least upper bound. Let $y$ be an upper bound for $E$ and suppose to the contrary that $y<x$. Let $\varepsilon = x-y$. We choose $N$ such that $(b_1-a_1)2^{-N+1}<\varepsilon$. Since $y$ is an upper bound so $a_N\le y$ and $y<x\le b_N$. Then $y\in [a_N,b_N]$. Thus
$$x-y\le b_N-a_N= \frac{b_1-a_1}{2^{N-1}}<\varepsilon = x-y$$
a contradiction. Thus $y\ge x$.
Using (a) and (b) we can conclude that $x$ is the least upper bound of $E$ as desired.
