Partition of a Nonempty Set $X$ Let $X$ be a nonempty set, and $\{A_\alpha : \alpha\in I\}$ be a partition of $X$. If $B\subseteq X$ such that $A_\alpha\cap B\neq\emptyset$ for every $\alpha\in I$, is $\{A_\alpha\cap B : \alpha\in I\}$ a partition of $B$?
I just need a starting point on how to think about this. Is there any logic that says taking an intersection between a partition of $X$ and a subset of $X$ gives you a partition of that subset?
Thanks!
 A: Ask yourself some questions:


*

*Given some $x\in B$ is there some $\alpha\in I$ such that $x \in A_\alpha \cap B$?

*Could there be more than one such $\alpha$?

*If there is more than one---call them $\alpha$ and $\beta$---is $A_\alpha\cap B$ the same set as $A_\beta \cap B$?

A: In a previous question of yours I have written the three properties of a partition.
So we need to verify three things:


*

*$A_\alpha\cap B\neq\varnothing$ for all $\alpha\in I$, which is the working assumption.

*$A_\alpha\cap A_\beta=\varnothing$ whenever $\alpha\neq\beta$, which you can deduce from the fact the $\{A_\alpha\mid\alpha\in I\}$ is a partition of $X$.

*For every $x\in B$ you have some $\alpha\in I$ such that $x\in A_\alpha\cap B$. You can show that this holds because $x\in X$, therefore it has to be in some $A_\alpha$, thus in $A_\alpha\cap B$.


In mathematics if you get lost it is often a good idea to work step by step from the definitions. If you want to show that something is a partition just take the definition of a partition and check that the conditions hold, or try and produce a counterexample.
