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:)
Best Answer
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 $\sum_{i=0}^{n-1}a_i\cdot10^{-i}+m\cdot10^{-n}\leq s \text{ for all }s\in S$.