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 $


References

  • W. Rudin, "Basic Topology" in "Principles of Mathematical Analysis", 3rd Edition, McGraw-Hill Inc. ch 2, pp 33.


Back to list of all proofs

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=Complement_of_unions_mh&oldid=59165"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009