problem involving families of sets

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• Jul 23rd 2011, 08:26 AM
issacnewton
Re: problem involving families of sets
Quote:

Originally Posted by emakarov

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 ?
• Jul 23rd 2011, 12:40 PM
emakarov
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, $\displaystyle z=\emptyset\Leftrightarrow z\in\{\emptyset\}$.
• Jul 25th 2011, 06:32 AM
issacnewton
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
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