Prove that $B^{c} \subseteq A^{c} \Rightarrow A \subseteq B$ How do I prove that $B^{c} \subseteq A^{c} \Rightarrow A \subseteq B$?
I tried:
Take $x \in B^{c}$. We want to prove $\forall x \in A, x \in B$
By definition of the subset, $\forall x \in B^{c}, x \in A^{c}$
Then I got stuck not knowing where to go...
 A: You want to show that $A\subseteq B$ under the assumption that $B^c\subseteq A^c$.
To do so we have to show that for $x\in A$ we have $x\in B$.
So let $x\in A$. Then $x\notin A^c$. Since $B^c\subseteq A^c$ this means $x\notin B^c$. So $x\in B$, and we are done.
Alternative:
Let $x\in A$. Suppose $x\notin B$. Then $x\in B^c$, and therefore $x\in A^c$, which is a contradiction, as $x\in A$ and $x\in A^c$ can not hold.
A: I think the contrapositive can be used, modus tollens.
Modus tollens tells us that if $p \implies q$, then $\neg q \implies \neg p$, where $\neg$ indicates logical negation.
$x\in B^c \implies x \in A^c$ by hypothesis.
Here, $p=x \in B^c$, so $\neg p= x \in B$
We also have $q=x\in A^c$, so $\neg q=x \in A$.
Modus tollens gives us $\neg q \implies \neg p$, thus we conclude $x \in A \implies x\in B$. The conclusion follows by definition of a subset.
Alternative approach:
$p \implies q$ is logically equivalent to the disjuction $q \vee  \neg p$.
So we have as our hypothesis  $x \in A^c \vee x \in B$. The veracity of the disjunction implies at least one of the two statements is true. If we have that $x\in A$, then the first half of the disjunction is false, so the second half must be true, thus $x \in B$.
