# Math Help - problem involving families of sets

1. ## Re: problem involving families of sets

Originally Posted by emakarov

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 ?

2. ## 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, $z=\emptyset\Leftrightarrow z\in\{\emptyset\}$.

3. ## 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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