# Proof: $A\cup B = \Omega \Longleftrightarrow{A^c \subseteq B}$

I need to prove $$A\cup B = \Omega \Longleftrightarrow{A^c \subseteq B}$$

How can I do it? This is what I've got so far, but I don't know if it is a valid demonstration:

$A\cup B = \Omega \Longleftrightarrow{x\in{A}\vee x\in{B}}\Longleftrightarrow{x\notin{A}\rightarrow{x\in{B}}}\Longleftrightarrow{A^c\subseteq B}$

Is this correct? If not, how can I prove that? Thanks in advanced.

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There is the problem of the missing quantifiers, identified by Arturo Magidin. Also, if you have any doubts, don't do things in one line. You need to prove two things, $A \cup B\rightarrow A^c\subseteq B$, and the implication that runs the other way. Prove them separately, you will retain better control of the logic. Now for the first direction. Suppose that $A\cup B=\Omega$. Let $x\in A^c$. So $x$ is not in $A$. Since $x\in A\cup B$, $x$ must be in $B$. I leave the other direction to you. – André Nicolas Nov 15 '11 at 18:17

## 1 Answer

The second clause should really be "For all $x$, $x\in A\lor x\in B$" (or $\forall x(x\in A\lor x\in B)$), and the quantifier should be repeated in the next one; otherwise, it looks correct.

(Except that you are proving, not proofing).

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