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Math Help - problem involving families of sets

  1. #16
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    Re: problem involving families of sets

    Quote Originally Posted by emakarov View Post

    The important thing is that your second goal means \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z=\emptyset, i.e., \mathcal{F} \cap \mathcal{G}\subseteq\{\emptyset\}; it does not guarantee the converse inclusion.
    makarov , I was looking at this and I am not convinced why \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z=\emptyset] is same
    as \mathcal{F} \cap \mathcal{G}\subseteq\{\emptyset\}.

    The first part just means that whenever ,  z \in (\mathcal{F}\cap \mathcal{G}) , that z has to be an empty set. But the second part means

    \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z \in \{\emptyset\}]

    can you comment ?
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  2. #17
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    Re: problem involving families of sets

    Both \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z=\emptyset] and \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z \in \{\emptyset\}] start in the same way. As far as the end goes, z=\emptyset\Leftrightarrow z\in\{\emptyset\}.
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  3. #18
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    Re: problem involving families of sets

    Hi

    I think I realised my mistake. What I actually proved was

    \left[(\cup \mathcal{F})\cap(\cup \mathcal{G})=\emptyset \right]\Rightarrow (\mathcal{F}\cap \mathcal{G}) \subseteq \{\emptyset\}

    as was pointed out by makarov. I was confused by the consequent
    part. Lets put it separately.

    \forall x\left [x \in (\mathcal{F}\cap \mathcal{G})\Rightarrow x \in \{\emptyset\}\right ]

    lets call  \math{A}=(\mathcal{F}\cap \mathcal{G})
    for the purpose of brevity. so above implication is

    \forall x[x \in \mathcal{A}\Rightarrow x \in \{\emptyset\}]

    so when \mathcal{A}=\{\emptyset\} , and if we let
    x=\emptyset then both antecedent and consequent are
    TRUE and all is well. But in other case when \mathcal{A}=\emptyset and if we let x=\emptyset then antecedent is FALSE but consequent is TRUE . Though
    in both case the implication itself is TRUE. I think that was the
    source of confusion.

    Thanks everybody
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