Question

De Morgan's laws proof

Answer 1

\[\left(\bigcup_nS_n\right)^c=\bigcap_nS_n^c\\\text{Proof:}\\\text{let }x\in\left(\bigcup_nS_n\right)^c\implies x\cancel\in\left(\bigcup_nS_n\right)\implies\cancel\exists n: x\in S_n\\\text{therefore, for each }n\text{, we have that}\\ x\in S_n^c\implies x\in\bigcap_nS_n^c\\\text{hence }\left(\bigcup_nS_n\right)^c=\bigcap_nS_n^c\]

I want to do a similar proof for De Morgan's other law, but I only get as \[\left(\bigcap_nS_n\right)^c=\bigcup_nS_n^c\\\text{Proof:}\\\text{let }x\in\left(\bigcap_nS_n\right)^c\implies x\cancel\in\left(\bigcap_nS_n\right)\implies\cdots ?\]then I can't say whether or not \(x\in S_n\) for any individual \(S_n\) or not, so how should I reason it?

Answer 2

Yes I have followed the wiki proof, but I want to do this one in the same, less rigorous fashion, as the previous, as my book suggests is fairly easy.

Answer 3

why not try to go reverse way ,
\(x \in \cap S_n^c\)

Answer 4

great idea, let me toy with that

Answer 5

which implies \(x \in S^c_n \)
and so on, till you reach ....yep, try it.

Answer 6

oh wow, that took two seconds XD
thanks!

Answer 7

lol, most welcome ^_^