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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) ∀n∈N 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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    Zacken
    TeeEm
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    (Original post by bright_sunshine)
    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) ∀n∈N 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))
    What's up? D'you want us to do all your homework for you? Provide some thought at the very least.

    Start off with: Let x \in X then \{x\} \in \matcal{P}(x) - so...?
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    (Original post by Zacken)
    What's up? D'you want us to do all your homework for you? Provide some thought at the very least.

    Start off with: Let x \in X then \{x\} \in \matcal{P}(x) - so...?
    Yes the first one i think i pretty much got it ..
    If P(X) is a subset of P(Y), what this means is that every subset of X is also a subset of Y. Therefore the union of all subsets of X is itself a subset of Y. But the union of all subsets of X is X itself, and therefore X⊆Y
    is that correct?

    this is not my homework, these are revision question for my test on sunday, and i'm getting stuck with them. I asked for detail so i could see step by step to understand better..
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    2 is only true if A & B are independent ?
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    (Original post by bright_sunshine)
    Yes the first one i think i pretty much got it ..
    If P(X) is a subset of P(Y), what this means is that every subset of X is also a subset of Y. Therefore the union of all subsets of X is itself a subset of Y. But the union of all subsets of X is X itself, and therefore X⊆Y
    is that correct?

    this is not my homework, these are revision question for my test on sunday, and i'm getting stuck with them. I asked for detail so i could see step by step to understand better..
    Your implication is the wrong way round, you need to show that X \subseteq Y \implies P(X) \subseteq P(Y).
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    (Original post by joostan)
    Your implication is the wrong way round, you need to show that X \subseteq Y \implies P(X) \subseteq P(Y).
    So since x is a subset of y, therefore the the sets of x are also in the sets of y (if not, he cannot be a subset), and the subset of x = p(x) must contain x, same with y,
    therefore p(x) is a subset of p(y)?
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    (Original post by bright_sunshine)
    2. Prove: P(A∩B) = P(A) ∩ P(B) with detail please
    (Original post by the bear)
    2 is only true if A & B are independent ?
    She's talking about the powerset, not probabilities. You need to do this in two stages. First show that \mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B) and then the reverse. To do so:

    Take any subset X \subseteq A \cap B then X \in \mathcal{P}(A) \cap \mathcal{P}(B), etc... (remember that X is in both A and B, so...)
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    (Original post by bright_sunshine)
    So since x is a subset of y, therefore the the sets of x are also in the sets of y (if not, he cannot be a subset), and the subset of x = p(x) must contain x, same with y,
    therefore p(x) is a subset of p(y)?
    You're getting there, though your argument isn't set out very clearly. I'm also not sure what you mean by the bolded part.
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    (Original post by Zacken)
    She's talking about the powerset, not probabilities. You need to do this in two stages. First show that \mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B) and then the reverse. To do so:

    Take any subset X \subseteq A \cap B then X \in \mathcal{P}(A) \cap \mathcal{P}(B), etc... (remember that X is in both A and B, so...)
    can you please give me a little push with #4? my teacher hasn't gone through indexes so roughly and it's a part of my revision...
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    (Original post by joostan)
    You're getting there, though your argument isn't set out very clearly. I'm also not sure what you mean by the bolded part.
    bolded was by mistake, what else i missing?
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    (Original post by bright_sunshine)
    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
    This is easy: A_0 = \{x \in \mathbb{N} : 0 \leq x \leq 0\} = \{x \in \mathbb{N} : x = 0\} = \{\cdots \} is the set the empty set or not?
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    (Original post by bright_sunshine)
    bolded was by mistake, what else i missing?
    Nothing really, you just need to express the argument more precisely - you have if X \subseteq Y then \forall A \subseteq X we must have A \subseteq Y.
    You can tidy up the conclusion from here.
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    (Original post by Zacken)
    This is easy: A_0 = \{x \in \mathbb{N} : 0 \leq x \leq 0\} = \{x \in \mathbb{N} : x = 0\} = \{\cdots \} is the set the empty set or not?
    it is
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    (Original post by bright_sunshine)
    it is
    Why do you think so?
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    (Original post by Zacken)
    This is easy: A_0 = \{x \in \mathbb{N} : 0 \leq x \leq 0\} = \{x \in \mathbb{N} : x = 0\} = \{\cdots \} is the set the empty set or not?
    what i did was: not sure it's right..

    A0 = {Ø}
    = A-B <=> (A∈B) ^ (B∈A)
    A0 = {Ø} = Ø <=> {Ø} ∈ Ø (doesn't take place) ^ Ø ⊆ {Ø} (takes place)


    therefore, its not true
 
 
 
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