Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $ Starting from the left side, $ (A \cup B) \cap C = $ By distributive law, ( distributing the $\cap C$), we have $ (A \cap C ) \cup (B \cap C) = $ Therefore, $ (A \cap C ) \cup (B \cap C) = (A \cap C) \cup (B \cap C)$ If I start from the right, I have = $(A \cap C) \cup (B \cap C) $
Prove $ (A \\cup B) \\cap C$ = $(A \\cap C) \\cup (B \\cap C)
Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $ Starting from the left side, $ (A \cup B) \cap C = $ By distributive law, ( distributing the $\cap C$), we have $ (A \cap C ) \cup (B \cap C) = $ Therefore, $ (A \cap C ) \cup (B \cap C) = (A \cap C) \cup (B \cap C)$ If I start from the right, I have = $(A \cap C) \cup (B \cap C) $
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Proof for $A \\cup B = B$ if and only if $A \\subset B$
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
The equality you mention results in: $$\lambda(A\cup B)=\lambda(A\cap B^c)+\lambda(A\cap B)+\lambda(A^c\cap B)$$ because we are dealing with a union of disjoint sets.
Prove that if $A$ and $B$ are measurable, then $\\lambda(A)+\\lambda(B ...
The equality you mention results in: $$\lambda(A\cup B)=\lambda(A\cap B^c)+\lambda(A\cap B)+\lambda(A^c\cap B)$$ because we are dealing with a union of disjoint sets.
That conclusion would be as wrong as it is to claim $\operatorname{int}(A\cup B) = \operatorname{int}(A)\cup\operatorname{int}(B)$. For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false).
Is $\\operatorname{int}(A\\cup B)=\\operatorname{int}(A)\\cup ...
That conclusion would be as wrong as it is to claim $\operatorname{int}(A\cup B) = \operatorname{int}(A)\cup\operatorname{int}(B)$. For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false).
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