Mathematics Asked by Andrea Burgio on November 28, 2020

I think it’s not so difficult but I struggling a little to figure it out, I want to make sure I’m correct, is $F$ a set of the form:

$F={ {(a, A), (b, A), …}, {(α, B), (β, B), …}, …}$ for all $a,b,…∈A$ and $α,β,…∈B$, where $A,B,…⊆M$?

and $⋃F$ a set of the form:

$⋃F={(a, A), (b, A), …,(α, B), (β, B), …}$?

Thank you!

$y$ is an element of $F$ iff it has the shape $y=Atimes{A}$ where $Asubseteq M$ and $Aneqvarnothing$.

The equality in your title already states this and cannot essentially be improved, so you do not have to bother about $F$ itself.

$x$ is an element of $bigcup F$ iff $xin y$ for some $yin F$.

Referring to the first line we get:

$x$ is an element of $bigcup F$ iff $xin Atimes{A}$ for some $A$ with $Asubseteq M$ and $Aneqvarnothing$.

That means that $x$ must have the shape $x=(a,A)$ where $ain A$.

Proved is now that:$$bigcup F={(a,A)mid ain Asubseteq M}$$
The condition $Aneqvarnothing$ does not need to be mentioned here because $ain A$ *implies* that $Aneqvarnothing$.

Correct answer by drhab on November 28, 2020

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