<Precise Proof> Every non-empty subset of $\mathbb R$ which is bounded above(below) has a least upper bound(greatest lower bound)

Question

Problem_

Every non-empty subset of $\mathbb R$ which is bounded above(below) has a least upper bound(greatest lower bound).

I searched the proofs for the problem, and there were two. However, both proofs had some problems for me to understand.

The first proof I've found was:

Let $S \subseteq\mathbb R$ and let $a_0$ be the largest integer that is a lower bound for $S$. Let $a_1$ be the largest integer between $0$ and $9$ for which $a_0.a_1$ is a lower bound for $S$. Then, generally, let $a_n$ be the greatest integer between $0$ and $9$ for which $a_0.a_1a_2\dots a_n$ is a lower bound for $S$. We claim that $$\lambda=a_0.a_1a_2\dots$$ is the greatest lower bound.

First, we show that $\lambda$ is a lower bound. If not, there $\exists s\in S$ such that $s\lt\lambda$. By Archimedes' condition, there $\exists n\in\mathbb N$ such that $10^{-n}\lt\lambda-s$. Therefore, $a_n$ can be reduced by $1$ in the definition of $\lambda$; or, if $a_n=0$, some earlier $a_m\gt0$ can be reduced by $1$. But this contradicts the definition of $\lambda$.

Then, we show that every lower bound $\mu$ is less than or equal to $\lambda$. If not, $\mu\gt\lambda$, so by Archimedes' condition, there $\exists n\in\mathbb N$ such that $10^{-n}\lt\mu-\lambda$. Therefore, $a_n$ can be increased by $1$ in the definition of $\lambda$; or, if $a_n=9$, some earlier $a_m\lt9$ can be increased by $1$. But this contradicts the definition of $\lambda$.

My question from here is where the definition of $\lambda$ makes the contradiction? I think I'm having confusion due to the lack of explanation about the definition of $\lambda$ and the reason for the contradiction. Could you please explain this?

The other proof I've got was:

Let $A$ and $\mathcal L$ be any non-empty subset of $\mathbb R$ bounded above and a set of all upper bounds of $A$. Suppose that $sup(A)$ does not exist. In other words, $\forall y\in\mathcal L, \exists y'\in\mathcal L$ such that $y'\lt y$. Therefore, $\forall x\in A$ and $\forall y\in\mathcal L, x\lt y$.

If $\exists c$ such that $\forall y\in\mathcal L, c\lt y$ and $\forall\epsilon\in\mathbb R^+, \exists y\in A$ such that $y − c \lt\epsilon$, then $c\notin A$ or $c$ becomes least upper bound. However, this implies that c is an upper bound, contradicting the existance of such c.

Then, if $\exists c_0 \in\mathbb R$ such that $\exists\epsilon\in\mathbb R^+, \forall y\in\mathcal L, c_0\lt y$ and $y − c_0\gt\epsilon$, construct $c_1 = c_0 + \epsilon$. For such $c_n$ satisfying conditions of $c_0$, construct $c_{n+1}$ in the same manner. If no such $\epsilon$ exists, let $c_{n+1} = c_n$. Let $c = \lim_{n\to\infty}c_n$. The constant $c$ clearly exists, because each element in the sequence is bounded above by any elements in $\mathcal L$ and the sequence is nondecreasing. This number acts the same way as the constant $c$ described above. Therefore, $\forall c\in\mathbb R, \exists y\in\mathcal L$ such that $y\lt c$. This concludes that $\forall x\in A$ and $\forall c\in\mathbb R, x\lt c$. Therefore, $A$ is empty, contradicting the assumption that $A$ is a non-empty set.

The second proof was harder to understand than I expected. Even though I read this over 20 times, I'm still confusing about the flow of the proof, such as the roles of $c$, and the reason that the author define $c_0$ again after defining $c$. Besides, the part that the time is mostly devoted to accept is marked as bold: why does $c$ imply "upper bound"? Is there some typo? I deeply focused on this part, but it remains unsolved.

I really appreciate having your ideas. It's fine to answer one of those questions, though it would be perfect when you answer all! Thanks:)

Answer 1

First proof:

In general: The author defines $\lambda$ through its decimal expansion. So, when she writes that $\lambda=a_0.a_1a_2\ldots$, she means that $\lambda=\sum_{n=0}^\infty a_n10^{-n}$. Also, $a_n$ is defined iteratively for $n>0$ through $$a_n=\max\left\{m\in\{0,1\ldots,9\}: \sum_{i=0}^{n-1}a_i\cdot10^{-i}+m\cdot10^{-n}\leq s \text{ for all }s\in S\right\}.$$

First contradiction: Define $\lambda_n= a_0.a_1a_2\ldots a_n$. Now, from the definition of $a_n$ it should be clear that $\lambda_n\leq s$ for any $s\in S$. Also, notice that $$0\leq\lambda-\lambda_n\leq 10^n\qquad\text{for all }n.\qquad(1)$$ Assume $\lambda$ is not a lower bound of $S$. Therefore, there exists $s\in S: s<\lambda$. Now, let $N$ be the smallest natural number such that $\lambda-s>10^{-N}$ (this is well-defined). We have that $$\lambda-10^{-N}>s\geq \lambda_N\Rightarrow \lambda-\lambda_N>10^{-N}$$ which contradicts (1).

Second contradiction: I think you need to think similarly and take into account that $a_n$ is the largest digit that satisfies $\sum_{i=0}^{n-1}a_i\cdot10^{-i}+m\cdot10^{-n}\leq s \text{ for all }s\in S$.

Answer 2

For the first proof, let me try to give more detail about what $\lambda$ is. We are given that $S$ is bounded below, so there exists $x\in \mathbb{R}$ such that $x<s$, for all $s\in S$. There are always integers less than $x$, so there exist integers which are also lower bounds for $S$. However $S\neq\varnothing$, so there exists $z\in\mathbb{Z}$ such that $z$ is not a lower bound for $S$. Then the set $Z_0:=\{z\in\mathbb{Z}:z<s,\forall s\in S\}\subseteq \mathbb{Z}$ is bounded above. Since the elements are integers, we know that $Z_0$ must have a maximal element.

Let $a_0:=\max(Z_0)$. This will be the integer part of our greatest lower bound $\lambda$. Next we move to the $10^{th}$'s place. The set $\{a_0,a_0+\frac{1}{10},a_0+\frac{2}{10},\cdots,a_0+\frac{9}{10}\}$ is finite, and contains a lower bound for $S$, namely $a_0$. So pick $a_1\in\{0,1,2,\ldots,9\}$ maximal such that $a_0+\frac{a_1}{10}$ is a lower bound for $S$.

Similarly, there is a maximal lower bound for $S$ in the set $\{a_0+\frac{a_1}{10},a_0+\frac{a_1}{10}+\frac{1}{100},a_0+\frac{a_1}{10}+\frac{2}{100},\ldots,a_0+\frac{a_1}{10}+\frac{9}{100}\}$. Let $a_2\in\{0,1,2,\ldots,0\}$ be maximal such that $a_0+\frac{a_1}{10}+\frac{a_2}{100}$ is a lower bound for $S$.

We can continue this process: For each $n\in \mathbb{N}$, pick $a_n\in\{0,1,2,\ldots,9\}$ maximal such that $a_0+\frac{a_1}{10}+\frac{a_2}{100}+\cdots+\frac{a_n}{10^n}$ is a lower bound for $S$. Define $\displaystyle{\lambda:=\sum_{n=0}^\infty\frac{a_n}{10^n}}$.

Now we want to show that $\lambda$ is a lower bound, so we assume contrary. If there exists $s\in S$ such that $\lambda>s$, then $\lambda-s>0$. Hence by the Archimedian Principle, there exists $n\in\mathbb{N}$ such that $\lambda-s>\frac{1}{10^n}$. Rearranging, we have $\lambda-\frac{1}{10^n}<s$. But $\lambda-\frac{1}{10^n}=a_0+\frac{a_1}{10}+\frac{a_2}{100}+\cdots+\frac{a_n-1}{10^n}+\cdots <s$. So that means that $a_0+\frac{a_1}{10}+\frac{a_2}{100}+\cdots+\frac{a_n-1}{10^n}$ is a lower bound for $S$. However, we defined $a_n$ to be the maximal digit with this property, which gives us our contradiction.

Note that there is the slight issue of if $a_n=0$ since then $a_{n}-1=-1\notin\{0,1,2,\ldots,9\}$. However this isn't actually a problem since then we carry to the previous digit which itself must have been maximal. If that digit was also $0$, then we keep carrying until we eventually reach our contradiction.

Hope this helps!