TITLE: Let $A,B$ and $U$ be sets so that $A\subseteq U$ and $B\subseteq U$. Prove that $A\subseteq B$ iff $(U\setminus B)\subseteq(U\setminus A)$.
QUESTION [1 upvotes]: Let $A,B$ and $U$ be sets so that $A\subseteq U$ and $B\subseteq U$. Prove that $A\subseteq B$ iff $(U\setminus B)\subseteq(U\setminus A)$.
The forward implication is easy to prove but i got stuck at the backward implication, i.e. $(U\setminus B)\subseteq(U\setminus A)\Rightarrow A\subseteq B$. Can someone help me out with it? Thank you.

REPLY [1 votes]: well, as others have pointed out:
You've proven $A \subset B \implies U/B \subset U/A$ so 
let $C = U/B$ and $D = U/A$ and you have
$C \subset D \implies U/D = U/(U/A) = A \cap U = A \subset U/C = U/(U/B) = B\cap U = B$.
But if you want to do it directly:
$U/B \subset U/A$.
Let $a \in A$ (so $a \in U$).  Then $a \not \in U/A$.  If $a \in U/B$ then $a \in U/A$ which would be a contradiction.  So $a \not \in U/B$.  So $a \in B$ (as $a \in U$).
So $A \subset B$.
Or here is a third way:
Let $a \in A$ but $a \not \in B$.  So $a \in A \implies a \in U$ and $a \not \in B \implies a \in U/B \implies a \in U/A$.  But $a \in A$ so that's a contradiction.  So $a \in A \implies a \in B$.  
So $A \subset B$.
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