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) $
Prove the lower bound
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analysis
1 Answers
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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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0Oh 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! – 2012-10-29