Show that $A \subseteq B \Leftrightarrow \overline{A} \cup B = U$ I'd like to show that if $A \subseteq B \Leftrightarrow \overline{A} \cup B = U$
I'd like to understand the way I am suppose to approach set theory challenges. Since this is not something intuitive I don't know where to start when approaching such challenges.
Apart from not being sure how to start, since there is a $\Leftrightarrow$ I suppose that I need to provide in both directions, am I correct? Please help with the right reasoning for this proof.
Updated:
U here is the universe of discourse
 A: Yes, you have to prove both implications. For the forward implication, suppose that $A\subseteq B$; we want to show that $\overline A\cup B=U$. Clearly $\overline A\cup B\subseteq U$, so we really only have to show that $U\subseteq\overline A\cup B$. To do that, let $x\in U$. If $x\in B$, then certainly $x\in\overline A\cup B$, so suppose that $x\notin B$. Then $x\notin A$, since $A\subseteq B$, so $x\in\overline A$, and again we have $x\in\overline A\cup B$. Thus, $U\subseteq\overline A\cup B$, and we conclude that $\overline A\cup B=U$.
For the other direction I suggest proving the contrapositive instead: show that if $A\nsubseteq B$, then $\overline A\cup B\ne U$. Use the fact that if $A\nsubseteq B$, then there is some $a\in A\setminus B$.
A: Here it is an alternative proof for the sake of curiosity.
Let us prove the implication $(\Rightarrow)$ first.
If $A\subseteq B$, then $A - B = A\cap B^{c} = \varnothing$. Consequently, one has that
\begin{align*}
U & = (A - B)\cup(A - B)^{c}\\\\
& = (A\cap B^{c})\cup(A\cap B^{c})^{c}\\\\
& = \varnothing\cup(A^{c}\cup B)\\\\
& = A^{c}\cup B
\end{align*}
and we are done.
Now we shall prove the converse implication $(\Leftarrow)$.
Indeed, one has that
\begin{align*}
U = A^{c}\cup B & \Rightarrow A\cap U = A\cap(A^{c}\cup B)\\\\
& \Rightarrow A = (A\cap A^{c})\cup(A\cap B)\\\\
& \Rightarrow A = \varnothing\cup(A\cap B)\\\\
& \Rightarrow A = A\cap B\\\\
& \Rightarrow A\subseteq B
\end{align*}
and we are done.
Hopefully this helps!
