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Problem 19 (Proof By Contrapositive Versus Proof By Contradiction)

Let $a$ and $b$ be real numbers with $a<b$. Consider the set $S=(a,b)=\{x\in \mathbb{R}\mid a<x<b\}$. We know $b$ is an upper bound for this set as if $x\in S$, then by definition of $S$ we have $x<b$, which clearly implies $x\leq b$. To show that $b$ is the supremum of $S$, we must show "If $m$ is an upper bound of $S$, then $b\leq m$." This is an implication of the form "If $P$, then $Q$."

  1. State the contrapositive of this implication.
  2. State the negation of this implication.
  3. Prove that the implication is true by proving that the contrapositive is true.
  4. Now, instead, prove that the implication is true by proving that the negation is false.

As a corollary to this problem, similar reasoning shows that if $S$ is an interval (of the form $(a,b)$, $ [a,b)$, $(a,b] $, or $ [a,b] $), then we have $\sup S=b$ and $\inf S=a$. You may use those facts now without proof.

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