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    1. Prove: If X⊆Y then P(X)⊆P(Y).
    2. Prove: P(A∩B) = P(A) ∩ P(B) with detail please
    3A. Prove that If A⊆B or B⊆A then P(A∪B) = P(A) ∪ P(B)
    3B. Prove #3A from the opposite direction, meaning: if P(A∪B) = P(A) ∪ P(B) then A⊆B or B⊆A
    4. N is a set of natural numbers N={0,1,2....,}
    For every n⊆ N => An = {x∈ N | 0≤ x ≤ n}
    Prove or Disprove the following:
    a) A0 = Ø
    b) ∀nN An ⊆ An+1
    c) n∈ N An = N
    d) ∀n∈Nk∈Nm∈N |Am - An| = k
    e) ∀n∈Nm∈N ((Am = {x2 | x∈An}) ↔ (m=n ^ n<2))
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    Hello, I've moved this to maths study help for you. Please post what you have attempted so far for the questions.
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    (Original post by bright_sunshine)
    1. Prove: If X⊆Y then P(X)⊆P(Y)
    To start you off..

    By drawing a Venn diagram, we have

    X \subset Y \Rightarrow Y = X \cup (X' \cap Y)

    and

    X \cap (X' \cap Y) = \varnothing

    So what does that say about the required probabilities?
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    (Original post by bright_sunshine)
    2. Prove: P(A∩B) = P(A) ∩ P(B) with detail please
    This is meaningless - have you mistranscribed it? If so, it would make more sense to put up the original question.
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    (Original post by atsruser)
    This is meaningless - have you mistranscribed it? If so, it would make more sense to put up the original question.
    He/she uses the notation P(X) to stand for the power set of X, not probabilities. In either case, massive apology for you having wasted your time and efforts; the OP has posted a duplicate thread here with quite a bit of answers.

    @OP: Refrain from posting duplicate threads; it wastes the time and effort of everybody trying to help you and is just impolite. Closed.
 
 
 
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