# Another Set Theory Proof(s)

1. Nov 18, 2009

### congtongsat

Problem:

(i)A$$\subseteq$$B $$\Leftrightarrow$$ A$$\cup$$B = B
(ii) A$$\subseteq$$B $$\Leftrightarrow$$ A$$\cap$$B = A

and

For subsets of a universal set U prove that B$$\subseteq$$A$$^{c}$$ $$\Leftrightarrow$$ A$$\cap$$B = empty set. By taking complements deduce that A$$^{c}$$$$\subseteq$$B $$\Leftrightarrow$$ A$$\cup$$B = U. Deduce that B = A$$^{c}$$ $$\Leftrightarrow$$ A$$\cap$$B = empty set and A$$\cup$$B = U.

Can't wrap my head around the last question at all. The i and ii seem simple but I'm just not getting it to work.

2. Nov 18, 2009

### clamtrox

(i) If A is empty, claim is true trivially. If it's not, then take an element of A, $$x \in A$$.

Suppose A is a subset of B. What does this mean for x? Use the definition of the cup operation :) Then suppose $$A \cup B = B$$ and do the same.

For (ii) you might want to assume that A is not empty because the thing you're trying to prove does not generally hold if both A and B are empty (mathematicians are weird...)

For the last bit, use again some element of b, $$y \in B$$. Show that if $$y \in A^c$$ then it can't be in A.