How to write the set of intersections of sets in a collection of partitions

Question

Let $A$ be a non empty set and $X$ be the set of all partitions of $A$. If $T \subseteq X$, I want to write the set of intersections of sets in the partitions of $T$. Let me explain myself:

1. If $T = \{ S_1, S_2 \}$, I want the set $S = \{C \cap D: C \in S_1, D \in S_2\}$
2. If $T = \{ S_1, S_2, S_3 \}$, I want the set $S = \{C \cap D \cap E: C \in S_1, D \in S_2, E \in S_3\}$

So my question is: if $T \subseteq X$, how can I write $S$?.

My take is this: we can write $T = \{S_i: i \in I\}$ for fome index $I$. Now, every $T_i$ is a colection of sets itself, so we can write, for every $i \in I$, $T_i = \{C_{i,j}: j \in J_i\}$, for some index $J_i$. So in my head $S = \{N: N = \bigcap_{i \in I} (\bigcup_{j \in J_i} C_{i,j})\}$. Is this correct, or there is something to change? Thank you.

Answer 1

This answer assumes that $T$ is already defined as $T = \left\{T_{i} \mid i \in I\right\}$ for some index set $I$.

Take some $B \in S$. By definition of $S$, we have

$$B = \bigcap_{i \in I} B_{i}$$

for some collection of sets $\left\{B_{i} \mid i \in I\right\}$ such that for each $i \in I$, we have $B_{i} \in T_{i}$. Hence, we can define $S$ simply as

$$S = \left\{\bigcap_{i \in I} B_{i} \mid B_{i} \in T_{i}\right\}$$

Update

Assume that we must continue from the definitions already provided by the OP.

Take some $N \in S$. By definition of $S$, we have

$$N = \bigcap_{i \in I} C_{i, j}$$

for some collection of sets $\left\{C_{i,j} \mid i \in I\right\}$ such that for each $i \in I$, we have $j \in J_{i}$. Hence, $S$ is simply defined as

$$S = \left\{\bigcap_{\substack{i \in I \\ j \in J_{i}}} C_{i, j}\right\}$$

without any further characterization in the definition.

Answer 2

This is slightly easier to write if you think about the actual cells in the partitions. Ultimately, I suppose, the goal is to find the coarsest partition that refines all the partitions in your collection.

So given $a\in A$, what will be the cell containing $a$? Well, exactly $\{b\in A\mid\forall i\in I\,\forall S\in S_i:a\in S\to b\in S\}$. So call this $T(a)$, and now the refinement is simply $\{T(a)\mid a\in A\}$. This also ensures that the result is a partition (i.e., it will not have any empty sets).