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I am learning about Set Theory currently. However, my teacher neglected to explain a concept that has shown up on one of the assigned proofs in the homework. The concept I'm struggling with is doing a proof with representation of the numbers in an element A for example. The proof is:

If $A$ is a set, then let $n(A)$ represent the number of elements in a finite set $A$ Prove that: $$n(A\cup B) = n(A) + n(B) - n(A\cap B).$$

I understand how to prove a regular set theory proof: such as maybe prove that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. You just need to use the definitions of intersection, union, etc. to convert these set statements into their logically equivalent counterparts. I just don't understand what is meant by the proof above. Thank you for your help. It's appreciated.

Austin Mohr
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Conner
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3 Answers3

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Take a point in $A\cup B$ and check that how many times it is counted if we calculate it by the formula $n(A\cup B)=n(A)+n(B)−n(A\cap B)$ see that each point is counted exactly once, hence it gives a proof that the formula is correct!

Arpan1729
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    You may find cases useful: An element of $A \cup B$ belongs either strictly to $A$, strictly to $B$, or to both $A$ and $B$. – Austin Mohr May 10 '17 at 02:08
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Informally, you can "prove" this statement using Venn diagrams. When you add the cardinality of $B$ to the cardinality of $A$, you are adding twice the cardinality of $A\cap B$. So, we need to remove that counting subtrating it.

To give a formal proof to the statement $$n(A\cup B)=n(A)+n(B)−n(A\cap B)$$ you need to know the definition of cardinality and bijection.

See here. Basically, we show that $$n(A\cup B)=n(A\cup (B\setminus A)).$$ Since $A$ and $B\setminus A$ are disjoint, we have $$n(A\cup(B\setminus A))=n(A)+n(B\setminus A).$$ Finally, we show $$n(B\setminus A)=n(B)-n(A\cap B).$$

Cristhian Gz
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The identity can be derived starting from only one basic assumption:  if $X$, $Y$ are disjoint i.e. $X \cap Y = \emptyset$, then $n(X \cup Y)=n(X)+n(Y)\,$.

A direct consequence is that if $Y \subseteq X$ is a subset of $X\,$, then $n(X \setminus Y) = n(X)-n(Y)\,$, which follows from applying the former to the union of disjoint sets $X = (X \setminus Y) \cup Y$.

I understand how to prove a regular set theory proof: such as maybe prove that $\dots$

Then you may find it easier to follow if you start with the identity:

$$ A \cup B \;=\; \left(A \setminus B\right) \;\cup\; (A \cap B) \;\cup\; \left(B \setminus A\right) $$

The three sets on the RHS are mutually disjoint, therefore:

$$ n\left(A \cup B\right) \;=\; n\left(A \setminus B\right) \;+\; n(A \cap B) \;+\; n\left(B\setminus A\right) $$

But $\,A \setminus B = A \setminus (A \cap B)$ where $A \cap B \subseteq A\,$ so $n(A\setminus B) = n(A) - n(A \cap B)\,$, therefore:

$$ \begin{align} n\left(A \cup B\right) \;&=\; \big( n(A) - n(A \cap B)\big) \;+\; n(A \cap B) \;+\; \big( n(B) - n(A \cap B)\big) \\[3px] &=\;n(A)+n(B)-n(A \cap B) \end{align} $$

dxiv
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