given a group X. S, T are partitions of X.
U = { A ∩ B : A $\in$ S ∧ B $\in$ T} \ {∅}
prove that U is a partition of x.
prove that U is a partition of x.
i understand that this is a patiotion of a partitions, so therefor it exisit the the condistions needed to ba a called a partition.
but i don't know how to prove that in a formal way.
For proving that $U$ is a partition of $X$ first we have to prove that
$U_{11}\cup U_{12}\cup \ ... \ \cup U_{1m} \cup U_{21} \ ... \ \cup U_{nm} = X$
Where $U_{ij}=A_i \cap B_j$
Being $S=A_1 \cup \ ... \ \cup A_n$
And $T=B_1 \cup \ ... \ \cup B_m$
So let be $x \in X$
For being $S$ a partition fof $X \Rightarrow \exists A_i \in S $ $\ $with $x \in A_i$
For being $T$ a partition fof $X \Rightarrow \exists B_j \in T $ $\ $with $x \in B_j$
Then $x \in A_i \cap B_j$
As this happens to all $x \in X \ $ we conclude that $U_{11}\cup U_{12}\cup \ ... \ \cup U_{1m} \cup U_{21} \ ... \ \cup U_{nm} = X$
Now we just have to see that $U_{ij} \cap U_{kl} = \emptyset \ $ if $\{ i,j \} \not = \{ k,l \} $
Suppose that $U_{ij} \cap U_{kl} \not= \emptyset \ $
Then $\exists y\in X \ $ with $y \in U_{ij} \ $ and $y \in U_{kl}$ $\Rightarrow y \in A_i, \ y\in B_j,\ y\in A_k \; $and $\ y \in B_l$
In particular $A_i \cap A_k \not= \emptyset \Rightarrow S \ $ is not a partition of $X$ $\ \ \ $which is a contradiction.