I am self-learning elementary real analysis from Understanding Analysis by Stephen Abbott. I would like to ask, if some one can help me deduce the correct proof. Below was a first sketch, but I think there are some problems with it.
Let $A \subseteq \mathbf{R}$ be nonempty and bounded above and let $s \in \mathbf{R}$ have the property that for all $n \in \mathbf{N}$, $s + \frac{1}{n}$ is an upper bound for $A$ and $s – \frac{1}{n}$ is not an upper bound for $A$. Show that $s = \sup A$.
Proof.
Let $s = \sup A$.
By the Archimedean property of reals, $\mathbf{N}$ sits inside $\mathbf{R}$. The statement for Archimedean property of real numbers says that:
For every real number $\epsilon > 0$, there exists a natural number $n$, such that $\frac{1}{n} < \epsilon$.
(1) Claim: $s$ is an upper bound for $A$.
Let's pick an arbitarily small, but fixed positive real number $\epsilon > 0$. What we have is:
\begin{align*}
(\exists n \in \mathbf{N}) \text{ such that, }&\frac{1}{n} < \epsilon \\
\implies(\exists n \in \mathbf{N}) \text{ such that, }&-\frac{1}{n} > -\epsilon \\
\implies(\exists n \in \mathbf{N}) \text{ such that, }& s-\frac{1}{\epsilon} < s-\frac{1}{n}
\end{align*}
But,
\begin{align*}
(\forall n \in \mathbf{N}), s – \frac{1}{n} \text{ is not an upper bound}
\end{align*}
Therefore, no matter what that natural number $n$ is, $s – \epsilon$ is not an upper bound.
There must be a $K_\epsilon = s – \frac{\epsilon}{2} \in A$, such that $s – \epsilon < K_\epsilon$.
Since our choice of $\epsilon$ was arbitrary to begin with, $K_\epsilon < s$ for all $K_\epsilon \in A$. Hence, $s$ is an upper bound for $A$.
(2) Claim: $s$ is the least upper bound for $A$.
We are told that $s + \frac{1}{n}$ is an upper bound for $A$ for all natural numbers $n$. Therefore, no matter what the natural number $n$ is $s + \epsilon$ is an upper bound. And, $s < s + \epsilon$ for any $\epsilon > 0$. So, $s$ is the least upper bound.
Best Answer
First of all: s is an upper bound for A. Suppose it is false, i.e. $ \exists a \in A, a-s\gt 0$. Defining $\delta = a-s, \delta \gt 0$. The problem states that for all natural $a\le s+ \frac{1} {n} $. We would have $\forall n \in N, \delta \le \frac{1} {n} $. Which is false. Then, we want to prove s is the least upper bound. $\forall \epsilon \gt 0, \exists n \in N such as \frac{1} {n} \lt \epsilon, so \frac{-1}{n} \gt - \epsilon$. Thus, using the hypothesis of the problem, (since $s- \frac{1} {n} $ is not an upper bound for A, there exists an $a \in A$, (depending on n), such as $s- \frac{1} {n} \lt a$ ), $\forall \epsilon \gt 0, \exists a \in A such as s-\epsilon \lt s -\frac{1} {n} \lt a$.