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Knowing That $ A \land B \subset \mathbb{R}$ and they both have a lower bound. Prove that (most likely using the definition of a bound): $$\inf (A \div B)=\min\{ \inf A,\ \inf B \}, $$ where $A \div B$ is the symmetric difference $$ A \div B := (A \cup B) \setminus (A \cap B) $$

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This is false without further specifications. $$A=\{1,\ 2,\ 3\}\implies \inf(A) = 1$$ $$B=\{1,\ 3,\ 5\}\implies \inf(B) = 1$$ Taking these two gives $$A\div B = \{2,\ 5\}\implies \inf(A\div B) = 2\neq \min\{1,\ 1\}$$

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Oh yeah, maybe if I wasn't trying to prove something that is not true but rather think about it in general I would do it myself. Thanks! – Max Oct 29 '12 at 19:08

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