# Relationship between intersections [closed]

Given $$N$$ sets $$X_1, \dots, X_N$$ and two definitions of intersection $$\cap'$$ and $$\cap''$$, is it possible to show that $$\vert X_i \cap' X_j \vert \le \vert X_i \cap'' X_j \vert, \quad \forall i,j \in \{ 1, \dots, N \}$$ implies $$\vert X_{i_1} \cap' \dots \cap' X_{i_K} \vert \le \vert X_{i_1} \cap'' \dots \cap'' X_{i_K} \vert, \quad \forall X_{i_1} \dots X_{i_K} \in \mathcal{P}(\{ X_1, \dots, X_N \})?$$

• What is a definition of intersection? Mar 20 at 20:06
• There's only one "definition of intersection" I know about - what exactly do you mean, here? Mar 20 at 20:12
• Intersection is essentially related to a definition of identity: which elements of $X_i$ are identical to the elements of $X_j$. Consider two different definitions of identity, and you will correspondingly get two different definition of intersection for the same sets. Mar 20 at 20:15
• What is a definition of identity?
– abx
Mar 20 at 20:50
• You need to actually say that in the question - the language "two definitions of intersection" is totally unclear. Regardless, this seems more appropriate for MSE than MO. Mar 20 at 23:15

As I understand the question, we have two equivalence relations on a set $$X$$ and subsets $$X_1, \ldots, X_N \subseteq X$$, and $$|X_i \cap' X_j|$$ and $$| X_i \cap'' X_j|$$ mean the number of equivalence classes which meet both $$X_i$$ and $$X_j$$, relative to the two equivalence relations, etc.
The answer is no. Let $$X_1 = \{a,b,c\}$$, $$X_2 = \{a,b,d\}$$, $$X_3 = \{a,c,d\}$$. Let the blocks of the first relation be $$\{a\}$$ and $$\{b,c,d\}$$ and let the second relation be the identity relation. Then for any $$i \neq j$$ exactly two blocks meet both $$X_i$$ and $$X_j$$, for either relation. But two blocks of the first relation meet all three sets, while only one block of the second does.
• $X_i\cap'X_j=X_i\cap'X_j\cap'X_k=\{\{a\},\{b,c,d\}\}$ if I understand correctly. Or, $b,c,d$ have the same color, but no two have the same shape. Mar 20 at 22:44
• Actually, my example is unnecessarily complicated. Take three objects, all the same color, all different shapes, and consider the sets $\{a,b\}$, $\{a,c\}$, $\{b,c\}$. Any two of these sets have one color and one shape in common, but all three have one color and no shapes in common. Mar 21 at 15:09