How to prove that $A\cap B\subseteq C$ and $A^c\cap B\subseteq C$ imply that $B\subseteq C$? How do you solve this problem??
Suppose that $A\cap B\subseteq C$, and $A^c\cap B\subseteq C$. Prove that $B$ is a subset of $C$.
I don't know where even to begin
Can anyone help?
Thank you
 A: Hint:
Pick an element $b\in B$. If it's in $A$, conclude that it's in $C$, and if it isn't in $A$, it's in the complement of $A$ -- conclude that it is in $C$.
Those are the two options, so we covered all of them.
Work closely with the definitions, and you'll be fine.
A: Note from the venn diagram $B=(A\cap B)\cup(A^c\cap B)$. You are given $A\cap B\subseteq C$$A^c\cap B\subseteq C$. If you take the union both sides (you can do that), we have $(A\cap B)\cup(A^c\cap B)\subseteq C \cup C$ which implies $B\subseteq C$
A: Venn says:

The upper case is for $A \subseteq C$ while the lower case is for $A \nsubseteq C $.
It is evident that the region $B$ is enclosed in $C$ $(B\subseteq C)$ if and only if both the yellow and the green regions are enclosed in $C$.
We have
\begin{align}
 (A \cap B \subseteq C) \land (A^c\cap B\subseteq C) &\implies ((A \cap B) \cup (A^c\cap B))\subseteq C \\&\implies ((A \cup A^c)\cap B)\subseteq C \\&\implies (U \cap B) \subseteq C \\&\implies B \subseteq C.
\end{align}
Where $U$ is the universal set.
A: You could also think of it another way: What if $B \nsubseteq C$? Then you have several possibilities:
$(A \subseteq C) \land (A^C \nsubseteq C) \implies A^C\cap B\nsubseteq C \\
(A \nsubseteq C) \land (A^C \subseteq C) \implies A\cap B\nsubseteq C \\
(A \nsubseteq C) \land (A^C \nsubseteq C) \implies (A\cap B\nsubseteq C) \land (A^C\cap B\nsubseteq C) \\
(A \subseteq C) \land (A^C \subseteq C) \implies C = U \implies B \subseteq C
$
Regardless, all these cases would contradict the original (given) statements, or, as in the last case, the assumption that $B \nsubseteq C$.
EDIT: Last two cases added, and previous cases edited, after @Asaf Karagila pointed out the incompleteness/incorrectness of my answer.
