How can we prove $P(A \cap B) = P(A) \cap P(B)$?

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SUMMARY

The discussion focuses on proving the equality of two sets, specifically that $P(A \cap B) = P(A) \cap P(B)$. To establish this, participants emphasize the necessity of demonstrating that every element of $P(A \cap B)$ is contained within $P(A) \cap P(B)$ and vice versa. The proof involves utilizing the definitions of power sets and intersections, highlighting the importance of showing that both sets contain the same elements. The discussion provides a structured approach to proving set equality through subset relationships.

PREREQUISITES
  • Understanding of set theory concepts, particularly power sets and intersections.
  • Familiarity with the notation and properties of probability, specifically $P(A)$ and $P(B)$.
  • Basic knowledge of logical proofs and subset relationships.
  • Experience with mathematical reasoning and formal definitions.
NEXT STEPS
  • Study the definitions and properties of power sets in detail.
  • Learn about set operations and their implications in probability theory.
  • Explore formal proof techniques, particularly those involving subset proofs.
  • Investigate additional examples of set equality proofs in probability contexts.
USEFUL FOR

Mathematicians, students of probability theory, and anyone interested in formal proofs and set theory concepts will benefit from this discussion.

tmt1
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$P(A \cap B) = P(A) \cap P(B)$

How can we prove this to be true?
 
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Hi tmt,

To show two sets $X$ and $Y$ are equal, you must prove that every element of $X$ belongs to $Y$ and vice versa. In your case, you must prove that every element of $P(A\cap B)$ belongs to $P(A)\cap P(B)$, and every element of $P(A) \cap P(B)$ belongs to $P(A\cap B)$. Use the definitions of the power set and intersection $\cap$ to do it.
 
Normally, the way one proves that two sets are equal is to show they contain exactly the same elements-which is equivalent to showing they are subsets of each other.

Here's how one such proof might begin:

Suppose $X \in P(A \cap B)$. Then $X \subseteq A \cap B$.

Now $A \cap B \subseteq A$, and $A \cap B \subseteq B$, by the definition of "$\cap$".

So, $X \subseteq A$, and...
 

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