How to proof for all sets $A$ and $B$: if $A^{\complement} \subseteq B$ then $ B^{\complement} \subseteq A$

Question

I think that there is a proof for this because we have an implication which we can translate to an or statement:

$\left[\text{If } A^{\complement} \subseteq B\text{ then }B^{\complement} \subseteq A\right]\equiv \left[\neg(A^{\complement} \subseteq B)\text{ or } B^{\complement} \subseteq A\right]$

so we would only have to proof the last part, or the first part, but i dont know how to do that without a Venn diagram, because only using a Venn diagram really wouldn't be a proof.

Answer 1

Let it be that $A^{\complement}\subseteq B$ and $x\in B^{\complement}$.

Then $x\in A$ since if not then $x\in A^{\complement}\subseteq B$ contradicting that $x\in B^{\complement}$.

This proves that $B^{\complement}\subseteq A$.

Answer 2

I am quite new to maths, so take my answer with care.

$A,B\subset E$, we have $A^c:=\lbrace x\in E \mid x\not\in A\rbrace$ so $A^c\subset B$ means $$\forall x\in E,\quad x\not\in A \implies x\in B\quad (1)$$ Assume that now. Let $x\in E$ such that $x\not\in B$, from contrapositive of $(1)$ it follows that $x\in A$. Hence $$\forall x \in E,\quad x\not\in B \implies x\in A$$ Wich means exactly $B^c\subset A$.