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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?


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up vote 2 down vote accepted

In a previous question of yours I have written the three properties of a partition.

So we need to verify three things:

  1. $A_\alpha\cap B\neq\varnothing$ for all $\alpha\in I$, which is the working assumption.
  2. $A_\alpha\cap A_\beta=\varnothing$ whenever $\alpha\neq\beta$, which you can deduce from the fact the $\{A_\alpha\mid\alpha\in I\}$ is a partition of $X$.
  3. 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.

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Thank you! You're far better than any professor I've had – roboguy12 Feb 16 '12 at 18:20
@roboguy12: I'm glad that you find my answers useful! :-) – Asaf Karagila Feb 16 '12 at 18:33

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 $A_\beta \cap B$?
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So, since Aa is a partition of X and B is a subset of X, an x in B must also be in Aa, right? So then I can say since x is in Aa and B then x is in the intersection. – roboguy12 Feb 16 '12 at 18:11
If alpha does not equal beta then x is not in both, because since Aa is a partition its subsets are pairwise disjoint. Ok, this is kinda making sense – roboguy12 Feb 16 '12 at 18:17
If $\alpha\neq\beta$, it might still be the case that $A_\alpha=A_\beta$. It depends on how you do your indexing. You could partition a set into two disjoint exhaustive subsets and list one of them twice and the other once as your index runs through three possible values. – Michael Hardy Feb 16 '12 at 19:09
For example, suppose $A_1=\{1\}$, $A_2=\{1\}$, $A_3=\{2\}$. Then $\{A_1,A_2,A_3\}$ is a partition of the set $\{1,2\}$ into two disjoint subsets. You open up that possibility by using an indexed family of sets, rather than just some set of sets. In a way, your notation is more complicated than what the situation requires. – Michael Hardy Feb 16 '12 at 19:10

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