Proof of union of a set and its complement is equivalent to a universe THEOREM
Let U be the universe, and let A be a subset of U. Then
$$ A \cup A^c = U $$
PROOF
Let x be any object. We must show that
$$ x \in ( A \cup A^c ) \iff x \in U $$
From the definition of complement we know that, $ A^c = U - A $. Thus 
$$  [x \in A \lor x\in(U-A)] \iff x \in U $$
$$ [x \in A \lor ( x\in U \land x\notin A)]=\iff x \in U  $$
$$  [(x \in A \lor x \in U) \land ( x \in A \lor x \notin A)] \iff x \in U $$
$$  [(x \in A \lor x \in U) \land (x \in A \implies x \in A)] \iff x \in U $$
Now I use earlier proved statement $ A \cup B = B$ iff $A \subset B$ and other basic definitions, so I deduce the next thing
$$ U \cap A = U $$
But A is subset and I know that $ B \cap A = A$ iff $ A\subset B$, that was proved earlier as well. So at last I've got that $ U \cap A = A $, and it's not what I wanted. Where is the mistake?
 A: Let $A\subseteq U$. 
Then for every element $x$ of $U$ exactly one of the following statements is true:


*

*$x$ is an element of $A$.

*$x$ is not an element of $A$.


By definition $A^{\complement}$ is the subset of $U$ that contains exactly all elements of $U$ that are not elements of $A$.
So we could also state that for every $x\in U$ exactly one of the following statements is true:


*

*$x\in A$

*$x\in A^{\complement}$
This together comes to the same as: $A\cup A^{\complement}=U$ and $A\cap A^{\complement}=\varnothing$.
A: I'm sure your proof is fine but this follows almost by definition.
Definition $A^c = \{x\in U| x\not \in A\}$ and $A$ tautologically is  $\{x \in U| x \in A\}$.
So $A\cup A^c = \{x\in U| x\in  A$ or $x \not \in A\}= \{x \in U\} = U$.
A: Recall that $A\subset U$ and $A^c=U\setminus A=\{x\in U:x\notin A\}\subset U$ and so $A\cup A^c\subset U$. Again for $x\in U$ if $x\in A$ then $x\in A\cup A^c$ else if $x\in A^c$ then $x\in A\cup A^c$. So as a whole $U\subset A\cup A^c$.
A: You are overcomplicating things. We can begin by proving $U \subseteq A \cup A^C$:
Let $x \in U$. If $x \in A$, we are done! Otherwise, $x \in A^c$, so we are also done.
The other inclusion is trivial
