Cardinality of union of sets whose intersection is the empty set

Question

Let $ n \geq 3 $ an integer such that $\cap_{i=1}^nA_i = \emptyset$.

Prove or disprove: $|\cup_{i=1}^nA_i|=\sum_{i=1}^n|A_i|$

I tried drawing every possible combination of 3 sets to find a counter-proposition, because with what I know about the empty intersection of all 3 sets I can see using the Inclusion-Exclusion principle that :

$|\cup_{i=1}^3A_i|=\sum_{i=1}^3|A_i| - |A_1\cap A_2| - |A_1\cap A_3|- |A_2\cap A_3| $
However I could not manage to find such an example that would falsify this claim.

I could not manage to prove this either - I know this much be correct when the sets are pairwise disjoint, and this is easily provable, which further strengthens my idea that I have to find a counter proposition.

Thanks in advance for any guidance or tips!

Answer 1

This statement is completely false, take the following example for n=3

$A_1 = \{x,y\}, A_2= \{y,z\}, A_3= \{x,z\}$

Then $\cap A_i = \phi$, but $\cup A_i = \{x,y,z\}$, so $|\cup A_i| = 3 , \sum |A_i| = 6$

Clearly not equal.

For general n, let $x_i$ be n distinct elements, and take $A_i$ to be all but the $i^{th}$ element, then again one gets a contradiction.

Answer 2

Let $A_1=\{1\}$ and $A_i=\{2\}$ for $2 \leq i \leq n$. Then $\cap_i A_i$ is empty but $\sum_i |A_i|=n\geq 3 >2=|\cup_i A_i|$.

Answer 3

Let $ X :=\ \{1 ... n\},\ $ where $\ n\ge 3.\ $

Define:

$$ \forall_{k=1}^n\quad A_k :=\ X\setminus\{k\} $$

Then $$\bigcap_{k=1}^n A_k = \emptyset, $$ and

$$ \left|\bigcup_{k=1}^n A_k \right| = n\ < n\cdot(n-1)\ = \sum_{k=1}^n |A_k| $$

Great!

Answer 4

@Sagers gives a good answer in which none of the $A_i$ are empty but the problem does not appear to require that so an even more simple counterexample is $A_1 = \emptyset$ and $A_i = \{1\}$ for $i > 1$.