Page 2 of 2 FirstFirst 12
Results 16 to 18 of 18

Thread: problem involving families of sets

  1. #16
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    208

    Re: problem involving families of sets

    Quote Originally Posted by emakarov View Post

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

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

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

    can you comment ?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790

    Re: problem involving families of sets

    Both $\displaystyle \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z=\emptyset]$ and $\displaystyle \forall z[z \in (\mathcal{F}\cap \mathcal{G})\Rightarrow z \in \{\emptyset\}]$ start in the same way. As far as the end goes, $\displaystyle z=\emptyset\Leftrightarrow z\in\{\emptyset\}$.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    208

    Re: problem involving families of sets

    Hi

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

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

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

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

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

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

    so when $\displaystyle \mathcal{A}=\{\emptyset\}$ , and if we let
    $\displaystyle x=\emptyset $ then both antecedent and consequent are
    TRUE and all is well. But in other case when $\displaystyle \mathcal{A}=\emptyset$ and if we let $\displaystyle 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
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Inquiry about proofs involving families of sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 11th 2012, 06:14 AM
  2. Solving a problem involving the closure of 2 sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 22nd 2010, 11:14 AM
  3. proof involving differences of sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Oct 13th 2009, 03:36 PM
  4. Try these proofs involving sets
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Apr 15th 2009, 02:29 AM
  5. Proof involving sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Oct 19th 2008, 12:00 PM

/mathhelpforum @mathhelpforum