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Verify one of DeMorgan’s Laws for sets:

$$\bigcap \{S\setminus U:U \in \mathcal U\} = S \setminus \bigcup \{U :U \in \mathcal U\}.$$

Can anyonw show me how to do this? a little confused, thanks

Michael Hardy
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Pasie15
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  • Begin with: Let $$x \in \bigcap_{ U \in \mathcal{U}}S\backslash U $$ Then show this implies $$x \in S\backslash\bigcup_{ U \in \mathcal{U}}U$$ which establishes that $$\bigcap_{ U \in \mathcal{U}}S\backslash U \subseteq S\backslash\bigcup_{ U \in \mathcal{U}}U$$ Then do the opposite to complete the proof. – graydad Nov 28 '14 at 23:00
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    @graydad : \backslash does not provide proper spacing in things like "$A\setminus B$"; instead you see "$A\backslash B$". \setminus is standard. ${}\qquad{}$ – Michael Hardy Nov 28 '14 at 23:13
  • That does look better. Thank you for mentioning that – graydad Nov 28 '14 at 23:13

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\begin{align} & x\in\bigcap \{S\smallsetminus U:U \in \mathcal U\} \\[8pt] \iff & \forall U \in\mathcal U \quad x\in S\smallsetminus U \\[8pt] \iff & \forall U\in\mathcal U\quad ( x\in S\ \&\ x\not\in U) \\[8pt] \iff & x\in S\ \&\ \forall U\in\mathcal U\quad x\not\in U \\[8pt] \iff & x\in S\ \&\ \lnot\exists U\in \mathcal U\quad x\in U \\[8pt] \iff & x\in S\ \&\ x\not\in\bigcup \{ U: U\in\mathcal U \} \\[8pt] \iff & x\in S\smallsetminus \bigcup \{ U: U\in\mathcal U \}. \end{align}

Michael Hardy
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