This is an interactive website to help you learn how to turn a proof from the text or from your notes into a proof of a homework problem. Here is a proof I did in class.
Proof: Let \(x\in A\cup B\), and let \(u=\sup\{\sup A, \sup B\}\). These immediately imply that \(\sup A \le u\) and \(\sup B \le u\) because \(u\) is an upper bound for the two-element set \(\{\sup A, \sup B\}\). We want to show that \(u\) satisfies the definition of \(\sup (A\cup B)\). Because \(x\in A\cup B\), either \(x\in A\) or \(x\in B\), so either \(x\le \sup A\le u\) or \(x\le \sup B\le u\). In either case, \(x\le u\) so \(u\) is an upper bound for \(A\cup B\). Now let \(v\) be any upper bound for \(A\cup B\). If \(a\in A\), then \(a\in A\cup B\) so \(a\le v\), making \(v\) an upper bound for \(A\), from which it follows that \(\sup A \le v\). Similarly, \(\sup B \le v\) which implies that \(v\) is an upper bound for the set \(\{\sup A, \sup B\}\). Therefore \(u\le v\) as desired.
How can we adapt this proof to the homework problem I assigned?
Try it yourself first, then click the box below to see the solution.
Show the Proof
Proof: Let \(x\in A\cup B\), and let \(w=\inf\{\inf A, \inf B\}\). These immediately imply that \(\inf A \ge w\) and \(\inf B \ge w\) because \(w\) is a lower bound for the two-element set \(\{\inf A, \inf B\}\). We want to show that \(w\) satisfies the definition of \(\inf (A\cup B)\). Because \(x\in A\cup B\), either \(x\in A\) or \(x\in B\), so either \(x\ge \inf A\ge w\) or \(x\ge \inf B\ge w\). In either case, \(x\ge w\) so \(w\) is an lower bound for \(A\cup B\). Now let \(t\) be any lower bound for \(A\cup B\). If \(a\in A\), then \(a\in A\cup B\) so \(a\ge t\), making \(t\) a lower bound for \(A\), from which it follows that \(\inf A \ge t\). Similarly, \(\inf B \ge t\) which implies that \(t\) is a lower bound for the set \(\{\inf A, \inf B\}\). Therefore, \(w\ge t\) as desired.
The expressions in red needed to be changed, but the rest of the proof stayed the same!