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It is not so clear to me whether this relation is transitive. I have two ideas, one - relation transitive, other - non-transitive.

  1. I have so proved that you are transitive:

Let $M_1RM_2 \iff M_1 \cap M_2 = S$, $M_2RM_3 \iff M_2 \cap M_3 = S$, for a fixed $S \in P(N)$, with $S \in N$, $S \neq \varnothing$.

$M_2$ occurs in both equations $\implies$ $M_1$, $M_2$, $M_3$ have the same elements in common, since they are linked to $\cap$.

I.e., $M_1 \cap M_2 = S \iff M_1RM_3$, i.e., transitively.

  1. But I can also think with the counterexample (maybe I don’t quite understand what “a fixed $S$” is):

Let $M_1 = \{1, 2\}$, $M_2 = \{2, 3\}$, $M_3 = \{3, 4\}$.

Then $M_1 \cap M_2 = \{2\} = S_1$, $M_2 \cap M_3 = \{3\} = S_2$.

But then $M_1 \cap M_3 = \varnothing = S_3$. Not transitive, since $S_3$ should not be empty.

I cannot understand where I make the mistakes.

Rócherz
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Daniel Perfilev
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  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Nov 02 '23 at 23:01
  • Both 1 and 2 are wrong. Please note: $S$ is FIXED. There is no "$S_1, S_2, S_3$". There is a single set $S$ out there such that any pair of $M_1, M_2$ are related only if their intersection is this unchanging set $S$. In (2), If $M_1$ is related to $M_2$, that would mean $S = {2}$. Otherwise they would not be related. But that means $M_2$ is not related to $M_3$ since their intersection is ${3}$, not $S$. So the fact that $M_1$ is also not related to $M_3$ says nothing about whether $R$ is transitive. – Paul Sinclair Nov 03 '23 at 18:35
  • (1) is false because your explanation of it is nonsense. It is literally you waving your hands and babbling because you didn't bother to look closely at what is going on. $R$ is not transitive, but you need to consider the counter-example more carefully. – Paul Sinclair Nov 03 '23 at 18:42

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