Set theory question

Aion

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May 8, 2018
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Suppose that B is a set and F is a family of sets. Prove that \(\displaystyle \bigcup

\left\{ A - B \mid A \in F \right\}\subseteq \bigcup (F-\mathcal{P}(B))\)

Where \(\displaystyle A - B\) is equivalent to \(\displaystyle A \cap B'\)

My question the following:

Is it possible to rewrite the set \(\displaystyle \left\{ A - B \mid A \in F \right\}\) into the form \(\displaystyle \left\{x \mid ... \right\}\) and if so, how/why?
 
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Can I just write \(\displaystyle \left\{ A - B \mid A \in F \right\}\) = \(\displaystyle \left\{ x \mid \exists A \in F (x = A - B)\right\}\) ?

Yea I know this isnt formal mathematical logic. This book i'm reading is quite vague sometimes lol.
 
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Suppose that B is a set and F is a family of sets. Prove that \(\displaystyle \bigcup
\left\{ A - B \mid A \in F \right\}\subseteq \bigcup (F-\mathcal{P}(B))\)
Where \(\displaystyle A - B\) is equivalent to \(\displaystyle A \cap B'\)
My question the following: Is it possible to rewrite the set \(\displaystyle \left\{ A - B \mid A \in F \right\}\) into the form \(\displaystyle \left\{x \mid ... \right\}\) and if so, how/why?
\(\displaystyle A\setminus B=A\cap B^c\) in LaTeX [ tex]A\setminus B=A\cap B^c[ /tex]
So if \(\displaystyle x\in\bigcup \{A\setminus B\mid A\in\mathscr{F}\}\) then \(\displaystyle \exists H\in\mathscr{F}\) such that \(\displaystyle x\in H\setminus B\)
Suppose that \(\displaystyle x\notin\bigcup (F\setminus\mathcal{P}(B))\) then \(\displaystyle (\forall G\in\mathscr{F}\setminus \mathcal{P}(B)][x\notin G]\).
But we know that \(\displaystyle x\in H\setminus B \) hence \(\displaystyle x\in\bigcup (F\setminus\mathcal{P}(B))\).
 
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