Hi ya. can anyone give me a hint on how to start showing the following

 

:For any artitary subsets of R

 

1)a complement of a complement is the original set

2)complement of a union is the intersection of the complements.

 

cheers.

Just show that the two sets are equal.

 

1) Let A be any subset of R, then [math]A^c = \{x \in \mathbb{R} \; : \; x \nin A\}[/math].

[math](A^c)^c = \{x \in \mathbb{R} \; : \; x \nin A^c\}[/math]. Now every x in A is in this set and every x in this set is in [math]A^c[/math]

 

2) Show that if [math]x \in (\bigcup_{i \in I} A_i)^c[/math] then it is in [math]\bigcap_{i \in I} A_i^c[/math] and vice versa.

 

The first part would be an argument of the type, that if x is in the complement of your union, then it is not in the union, hence in none of the A_i, therefore it would be in all of the complement of A_i and thus also in the intersection. The other way around goes along the same lines.

 

Mandrake

The second part can be done using induction as well, for its easier to prove for 2 sets and then do inductively.