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Math Help - Power set of B

  1. #1
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    Power set of B

    Is there such an expression as \forall B \mathcal,{P}(B)?

    It's very hard for me imagine there would be more than one B in \mathcal{P}(B).
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  2. #2
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    Quote Originally Posted by novice View Post
    Is there such an expression as \forall B \mathcal,{P}(B)?
    It's very hard for me imagine there would be more than one B in \mathcal{P}(B).
    Can you possibly fill out that question with more details?

    Is it just \mathcal{P}(B)=\{Y:Y\subseteq B\}~?
    If B has n elements the \mathcal{P}(B) has 2^n elements.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Can you possibly fill out that question with more details?

    Is it just \mathcal{P}(B)=\{Y:Y\subseteq B\}~?
    If B has n elements the \mathcal{P}(B) has 2^n elements.
    \forall B \in \mathcal{P}(B)(A \subseteq B)

    Could it be a typo?

    Does \{Y:Y\subseteq B\} mean \forall Y \in \mathcal{P}(B)(Y \subseteq B)
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    Quote Originally Posted by novice View Post
    Is there such an expression as \forall B \mathcal,{P}(B)?
    I don't recognize that as a meaningful formula.

    What is it that you're trying to say there?
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  5. #5
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    Quote Originally Posted by novice View Post
    \forall B \in \mathcal{P}(B)(A \subseteq B)
    That's a formula, though it's an odd one. It seems not what you intend.

    What you probably mean is

    \forall A \in \mathcal{P}(B)(A \subseteq B)

    Quote Originally Posted by novice View Post
    Does \{Y:Y\subseteq B\} mean \forall Y \in \mathcal{P}(B)(Y \subseteq B)
    No.

    \{Y:Y\subseteq B\}

    is a term; it's a "name"; it names a particular set; it doesn't make an assertion.

    \forall Y \in \mathcal{P}(B)(Y \subseteq B)

    is not a term, but rather it's a formula; it makes an assertion (once we agree on what 'B' stands for, or, alternatively, once we decide it should be taken in the sense of a universal generalization on 'B').

    You very much need to get that logic book I recommended, as soon as possible; it will get you on the right track to writing correct symbolizations.
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  6. #6
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    Quote Originally Posted by MoeBlee View Post
    I don't recognize that as a meaningful formula.

    What is it that you're trying to say there?
    I am supposed to provide an example of a set A that satisfies  \forall B P(B)(A\subseteq B).

    A= \varnothing seems to fit the expression, but I know there is only one B in P(B)

    If B={1,2}, P(B)={ \varnothing {1},{2},B}. I wondering why \forall B is used. It doesn't make sense.
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  7. #7
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    Quote Originally Posted by MoeBlee View Post
    You very much need to get that logic book I recommended, as soon as possible; it will get you on the right track to writing correct symbolizations.
    It's coming my way.
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  8. #8
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    Quote Originally Posted by novice View Post
    A that satisfies  \forall B P(B)(A\subseteq B).
    I don't recognize that as a formula. Yes, it doesn't make sense. I can't find an A that satisfies it, because it's not even a formula.

    The earlier formula you gave was

    \forall B \in \mathcal{P}(B)(A \subseteq B)

    That is equivalent to:

    For all B (If B is a subset of B, then A is subset of B).

    An A that satisfies that is B itself, or any other A that is a subset of B, depending on what B is.

    Though maybe you've not rendered the formula correctly (it is rather odd)?
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  9. #9
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    It surely is a typo of some sort.
    I know of several books that use this notation, P(B:A).
    It stands for A\in P(B) and P(B:A)=\{X\in P(B):A\subseteq X\} .
    All subsets of B that have A as a subset.

    Is that close to what you think it may mean?
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  10. #10
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    Thank you both. I am going scrape off everything I got so far and study logic first to rebuild it from ground up.
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  11. #11
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    Quote Originally Posted by novice View Post
    It's coming my way.
    You'll enjoy it. You're doing the right thing.
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