Question

Establish the following apparently different (but "fancier") characterization of the supremum. Let $A$ be a nonempty subset of $\mathbf{R}$ that is bounded above. Prove that $s=\sup A$ if and only if (i) $s$ is an upper bound for $A$, and (ii) for every $\varepsilon>0$, there is an $a \in A$ such that $a>s-\varepsilon$. State and prove the corresponding result for the infimum of a nonempty subset of $\mathbb{R}$ that is bounded below. Recall that a sequence ( $x_n$ ) of real numbers is said to converge to $x \in \mathbb{R}$ if, for every $\varepsilon>0$, there is a positive integer $N$ such that $\left|x_n-x\right|<\varepsilon$ whenever $n \geq N$. In this case, we call $x$ the limit of the sequence $\left(x_n\right)$ and write $x=\lim _{n \rightarrow \infty} x_n$.

   Establish the following apparently different (but "fancier") characterization of the supremum. Let $A$ be a nonempty subset of $\mathbf{R}$ that is bounded above. Prove that $s=\sup A$ if and only if (i) $s$ is an upper bound for $A$, and (ii) for every $\varepsilon>0$, there is an $a \in A$ such that $a>s-\varepsilon$. State and prove the corresponding result for the infimum of a nonempty subset of $\mathbb{R}$ that is bounded below.
Recall that a sequence ( $x_n$ ) of real numbers is said to converge to $x \in \mathbb{R}$ if, for every $\varepsilon>0$, there is a positive integer $N$ such that $\left|x_n-x\right|<\varepsilon$ whenever $n \geq N$. In this case, we call $x$ the limit of the sequence $\left(x_n\right)$ and write $x=\lim _{n \rightarrow \infty} x_n$.
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Real analysis
Real analysis
N. L. Carothers 1st Edition
Chapter 1, Problem 3 ↓

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** We need to prove that $s = \sup A$ if and only if (i) $s$ is an upper bound for $A$, and (ii) for every $\varepsilon > 0$, there is an $a \in A$ such that $a > s - \varepsilon$.  Show more…

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Establish the following apparently different (but "fancier") characterization of the supremum. Let $A$ be a nonempty subset of $\mathbf{R}$ that is bounded above. Prove that $s=\sup A$ if and only if (i) $s$ is an upper bound for $A$, and (ii) for every $\varepsilon>0$, there is an $a \in A$ such that $a>s-\varepsilon$. State and prove the corresponding result for the infimum of a nonempty subset of $\mathbb{R}$ that is bounded below. Recall that a sequence ( $x_n$ ) of real numbers is said to converge to $x \in \mathbb{R}$ if, for every $\varepsilon>0$, there is a positive integer $N$ such that $\left|x_n-x\right|<\varepsilon$ whenever $n \geq N$. In this case, we call $x$ the limit of the sequence $\left(x_n\right)$ and write $x=\lim _{n \rightarrow \infty} x_n$.
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