Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$ A \div B := (A \cup B) \setminus (A \cap B) $$

share|improve this question

1 Answer 1

up vote 0 down vote accepted

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 $$A\div B = \{2,\ 5\}\implies \inf(A\div B) = 2\neq \min\{1,\ 1\}$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.