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Say we have $3$ sets, $X$, $Y$, and $Z$. I want to prove that the cardinality of the intersection of all three sets is less than or equal to :

the cardinality of $X$ intersect $Y$,
plus the cardinality of $Y$ intersect $Z$,
plus the cardinality of $X$ intersect $Z$.

I am able to reason with examples. Say the sets are all unique and share no common elements. In this case, both sides of the equation evaluate to zero. $0 = 0$

In a case where each of the three sets share a single element, the left side counts this once, and the right side counts this three times. $1$ is less than $3$.

What is a more rigorous way of showing this, perhaps with cardinality and inclusion/exclusion principles without relying on specific examples?

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    Hint: $X\cap Y\cap Z$ is a subset of all of $X\cap Y,Y\cap Z,X\cap Z$. – 2017-02-17

3 Answers 3

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$\textbf{Hint}$: For any $A$ and $B$, $\sharp(A\cap B)\leq \sharp(A)$ and $\sharp(A\cap B)\leq\sharp (B)$, because $A\cap B\subseteq A$ and $A\cap B\subseteq B$. Thus, $$\sharp(A\cap B)\leq \sharp(A)+\sharp(B)$$

In your case... since $X\cap Y\cap Z\subseteq X\cap Y,Y\cap Z,X\cap Z$, $$\sharp(X\cap Y\cap Z)\leq \sharp(X\cap Y)+\sharp(Y\cap Z)+\sharp(X\cap Z).$$

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Hint: Try to find an injective function between the LHS and the RHS

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You even have that the intersection of all three is less than or equal to the intersection of $X$ and $Y$ because any element in $X \cap Y \cap Z$ is also an element of $X \cap Y$.