(New page: Category:Set Theory Category:Math == Theorem == Let <math>\{E_{\alpha}\}</math> be a (finite or infinite) collection of sets <math>E_{\alpha}</math>. Then, <br/> <center><math>(...)
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Revision as of 11:45, 6 October 2013


Theorem

Let $ \{E_{\alpha}\} $ be a (finite or infinite) collection of sets $ E_{\alpha} $. Then,

$ (\bigcup_{\alpha}E_{\alpha})^C = \bigcap_{\alpha}(E_{\alpha}^C) $


Proof

$ \begin{align} x\in (\bigcup_{\alpha}E_{\alpha})^C &\Rightarrow x \notin \bigcup_{\alpha}E_{\alpha} \\ &\Rightarrow x \notin E_{\alpha} \;\forall \alpha \\ &\Rightarrow x \in E_{\alpha}^C \;\forall\alpha \\ &\Rightarrow x \in\bigcap E_{\alpha}^C \\ &\Rightarrow (\bigcup_{\alpha}E_{\alpha})^C \subset \bigcap_{\alpha}(E_{\alpha}^C) \end{align} $


Conversely,

$ \begin{align} x \in \bigcap_{\alpha}(E_{\alpha}^C) &\Rightarrow x \in E_{\alpha}^C\;\forall\alpha \\ &\Rightarrow x \notin E_{\alpha}\; \forall \alpha \\ &\Rightarrow x \notin\bigcup_{\alpha}E_{\alpha} \\ &\Rightarrow x \in \bigcup_{\alpha}E_{\alpha})^C \\ &\Rightarrow \bigcap_{\alpha}(E_{\alpha}^C)\subset (\bigcup_{\alpha}E_{\alpha})^C \end{align} $

Therefore,

$ (\bigcup_{\alpha}E_{\alpha})^C = \bigcap_{\alpha}(E_{\alpha}^C) $

$ \blacksquare $


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