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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 $$ \vert X_{i_1} \cap' \dots \cap' X_{i_K} \vert \le \vert X_{i_1} \cap'' \dots \cap'' X_{i_K} \vert, \quad \forall X_{i_1} \dots X_{i_K} \in \mathcal{P}(\{ X_1, \dots, X_N \})? $$

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    $\begingroup$ What is a definition of intersection? $\endgroup$
    – Ben McKay
    Mar 20 at 20:06
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    $\begingroup$ There's only one "definition of intersection" I know about - what exactly do you mean, here? $\endgroup$ Mar 20 at 20:12
  • $\begingroup$ 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. $\endgroup$
    – Cesare
    Mar 20 at 20:15
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    $\begingroup$ What is a definition of identity? $\endgroup$
    – abx
    Mar 20 at 20:50
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    $\begingroup$ 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. $\endgroup$ Mar 20 at 23:15
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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.

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  • $\begingroup$ I am not sure I understand the example. Could you make the intersections/equivalences explicit? $\endgroup$
    – Cesare
    Mar 20 at 21:49
  • $\begingroup$ $X_i\cap'X_j=X_i\cap'X_j\cap'X_k=\{\{a\},\{b,c,d\}\}$ if I understand correctly. Or, $b,c,d$ have the same color, but no two have the same shape. $\endgroup$ Mar 20 at 22:44
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    $\begingroup$ What Gerry said. Four objects, all different shapes, but the last three have the same color. Every set contains both colors, but only the first shape belongs to all three sets. $\endgroup$
    – Nik Weaver
    Mar 20 at 23:18
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    $\begingroup$ 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. $\endgroup$
    – Nik Weaver
    Mar 21 at 15:09

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