Prove that $A' \subseteq B$ if and only if $A \cup B = U$

Question

Consider subsets $A$, $B$ and $C$ of the universe $U$

Prove that $A^c \subseteq B$ if and only if $A \cup B = U$

I know that there are two directions for the problem:

1. If $A^c \subseteq B$ then $A \cup B = U$

   1.1 Suppose $A^c \subseteq B$ then prove that $A \cup B \subseteq U$

   1.2 Suppose $A^c \subseteq B$ then prove that $U \subseteq A \cup B$

2. If $A \cup B = U$ then $A^c \subseteq B$

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Attempt:

1.1 Suppose $A^c \subseteq B$. Let $x \in A^c$ then we know $x\in B$ so $x\in A\cup B$. Since universal set $U$ contains all set, then $A \cup B \subseteq U.$

1.2 Suppose $A^c \subseteq B$. Let $x \in A^c$ then we know $x\in B$. As well $x\in U$ --- we know that $U = A \cup A^c$.
If $x \in A$, then $x \in A \cup B$.
If $x \in A^c$, then $x \in B$ since $A^c \subseteq B$. Then, $x \in A\cup B$.
So in either cases, $x \in A \cup B$. Therefore $U \subseteq A\cup B$.

2. I'm not sure how to approach this direction. Could I do something similar to part 1.2 and split into cases of $A$ and $A^c$ from $U$ and argue?

Answer 1

Assume $A\cup B = U$. By De morgan's law, one has $A^c \cap B^c = U^c = \varnothing$. This means $A^c \subset B$.