Prove that a number $u$ is $\sup S$ given certain properties.

Question

Problem

Let $S$ be a nonempty subset of $\mathbb{R}$, and let $u$ be a number with the following properties:

1. for each positive integer $n$, the number $u - \frac{1}{n}$ is not an upper bound of $S$, and
2. for each positive integer $n$; the number $u + \frac{1}{n}$ is an upper bound of $S$.

Prove that $u = \sup S$

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My Attempt at the Proof:

Let $B$ be $\sup S$ and $u \neq B$, where $B, u\in\mathbb{R}$. Also, let $A=u - \frac{1}{n}$ and $C=u + \frac{1}{n}$ for any $n \in\mathbb{N^+}$.

WLOG, consider when $u>B$. As a result of The Least Upper Bound Principle, for any $x\geq C$, $x$ must be an upper bound of $S$, as $x>B$. However, for any $y \le A$, $y$ may be either an upper bound or not. Since $A$ by definition 1 must not be an upper bound in $S$, this is a contradiction. Therefore, $u=B$, such that $u=\sup S$.

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Question

Are there any suggested improvements to my proof? Can anyone verify I'm on the right track? I'm new to real analysis and unfortunately do not have answers to validate my proof.

Answer 1

The conditions $a)$, and $b)$ imply that: $\forall n \geq 1, u-\dfrac{1}{n} <\sup(S)\leq u + \dfrac{1}{n}\Rightarrow |u-\sup(S)| \leq \dfrac{1}{n}\Rightarrow |u-\sup(S)| \leq \displaystyle \lim_{n\to \infty} \dfrac{1}{n} = 0 \Rightarrow u = \sup(S)$

Answer 2

So are you considering a fixed $n$ or all integers $n$ at once (so $A$, $C$ would be sets...). Anyway, the sentence 'However, ... not.' is not valid for a fixed $n$ and if your are trying to say $y \le sup(A)$ where $A$ would a set of $u - 1/n$ for integer $n$, you still would have to prove that the sup is $n$...

Here is one way to approach it.

Suppose $x < u$. Then by the Archimedian Property of $\mathbb{R}$, there exists an integer $n$ such that $n(u-x) >1 $ or $u - \frac{1}n > x$ so all such $x$ cannot be a lower bound. Let $y > u$. Then again by the Archimedian Property, there exists an integer $n$ such that $n(y-u) > 1$ or $y > u+\frac{1}n$ so all such $y$'s are an upper bound. Therefore, it follows that... (what?)

Answer 3

I think this is the closest to your intended proof, just explained more clearly.

Assume $\sup S \not = u$

Let $\varepsilon = |\sup S - u| > 0$, then since $\lim_{n\to \infty}\frac 1 n = 0 \implies \exists n\in \mathbb N$ s.t. $\frac 1 n < \varepsilon$

Case $1$: $\sup S > u \implies u+\frac 1 n < \sup S$ and $u+\frac 1 n$ is an upper bound on $S$ (by assumption). A contradiction with the fact that $\sup S$ is the least upper bound.$\blacksquare$

Case $2$: $\sup S < u \implies u-\frac 1 n > \sup S$ and $u-\frac 1 n$ is not an upper bound on $S$ (by assumption). A contradiction with the fact that $\sup S$ is an upper bound.$\blacksquare$