Elementary Classical Analysis, Marsden, 2nd ed, Chapter 1, Problem 4

Problem

Show that $d=\inf(S)$ iff $d$ is a lower bound for $S$ and for any $\epsilon > 0$ there is an $x \in S$, such that $d \geq x - \epsilon$.

Answer

($\Rightarrow$) Let $d = \inf(S)$ then obviously $d$ is a lower bound of $S$. For $\epsilon >0$, $d+\epsilon$ is cannot be a lower bound of $S$. Therefore, there exists $x \in S$ such that $x < d + \epsilon$  $\Rightarrow$  $d \geq x - \epsilon$.

($\Leftarrow$) Let $d$ be a lower bound satisfying the given condition. Suppose $c$ is another lower bound of $S$ and $c > d$. Then for $\epsilon:=(c - d)/2$, there exists $x \in S$ such that $d \geq x - \epsilon > x - 2\epsilon = x - (c - d)$  $\Rightarrow$  $c > x$. This contradicts the fact that $c$ is a lower bound.

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