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I know this might sound silly but, it's easy to convince myself that $|A\cup B|=|A|+|B|-|A\cap B|$ but i'm not sure how i would go about proving it.

Suppose $A'=A\setminus\{a\}$ and $|A|=n+1$ for some positive integer $n$ then $\begin{align} |A\cup B|&=|A'\cup B|+|\{a\}|\\ & =|A'|+|B|-|A'\cap B|+ |\{a\}|\\ & =|A|+|B|-|A\cap B| \end{align}$

So is this correct??

pkjag
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    It might be easier to just write $$A\cup B = A\sqcup (B\setminus(A\cap B)) $$where the RHS is a disjoint union. – Prahlad Vaidyanathan Dec 27 '13 at 14:15
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    Your proof assumes $a \notin B$. So you also have to treat the case $a \in B$ (which is done using the fact that we have $A \cup B = A' \cup B$ and $A \cap B = (A' \cap B) \cup {a}$ in that case). – Joel Cohen Dec 27 '13 at 14:43

3 Answers3

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If A and B are disjoint then it's straight forward that $|A\cup B|=|A|+|B|$. Now, $|A\cup B|=|A\cup(A\cap B)\cup(A^{c}\cap B)|=|A\cup (A^{c}\cap B)|=\\ =|A|+|A^{c}\cap B|=|A|+|B\setminus (A\cap B)|=|A|+|B|-|A\cap B|$

The third equality is due to the fact that $A$ and $A^{c}\cap B$ are disjoint sets.

Kal S.
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I'm not convinced that induction here is the right method..

Hint: Rather prove that $|A\cap B|+|A\cup B|\ =\ |A|+|B|$ by counting the members on both sides of this equation. (Some members will be counted twice, but on both sides, which ones?)

Berci
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Your calculation would be correct if and only if $a\in A$ but $a\not\in B$.

QED
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