# $A \subseteq B$ if and only if $B' \subseteq A'$?

I have already tried to prove it in my own way and would like to share my results in hope that a mathematician somewhere can tell if I am correct. I believe that the proof consists of 2 parts: First, prove left to right that $$A \subseteq B \implies B' \subseteq A'$$. Second, prove right to left that $$B' \subseteq A' \implies A \subseteq B$$.

In the first part, we assume $$A \subseteq B$$. We have $$x \in A' \implies x \not \in A$$ and since $$A \subseteq B$$, then $$x \not \in B$$. On the other hand, $$x \in B' \implies x \not \in B$$ and $$x \in A'$$ therefore $$A' \cap B' = B'$$ which means $$B' \subseteq A'$$.

The second part of the proof is more or less the same. Is my approach correct? Thanks.

• $(A\subseteq B)\iff (\overline{B}\subseteq \overline{A})$ is equivalent to saying $(x\in A\,\Rightarrow\, x\in B)\iff (x\not\in B\,\Rightarrow\, x\not\in A)$, which is a fact just because the statements are contrapositives of each other. Contraposition is a law that states $(P\,\Rightarrow\, Q)\iff (\lnot Q\,\Rightarrow\, \lnot P)$, which you can use here. – user26486 May 3 '15 at 22:11
• Please follow this guide next time. It will make your questions much easier to read. – Arthur May 3 '15 at 23:29
• For the love of God, please punctuate. Write complete sentences. And write in LaTeX. It makes the problem so much easier to read. – AJY Dec 8 '15 at 21:51
• Very close. But $x \notin A$ does not mean $x \notin B$. A = {1,2} B= {1,2,3}, 3 $\notin$ A. On the other hand $x \notin B$ does mean $x \notin A$. – fleablood Dec 8 '15 at 22:18

Your argument is rather unclear and not quite correct. What you want to do is first assume $A \subset B$ and let $x \in B'$. Then, since $x \not\in B$ and since $A \subset B$, we have that $x \in A'$. So $B' \subset A'$. The other direction follows from the previous implication since $A'' = A$ and $B'' = B$ so $B' \subset A' \implies A \subset B$
Suppose $A\subset B$. Then $x\in A \implies x\in B$. Suppose that $x\in B^C$, but $x\not\in A^C$, then $x\in A, x\not\in B$, this implies $A\not\subset B$, a contradiction. So $B^C \subset A^C.$
Suppose conversely that $B^C\subset A^C$, and that $\exists x: x\in A, x\not\in B$. Then $x\in B^C, x\not \in A^C$ and so, $B^C\not\subset A^C$ a contradiction. So $A\subset B$.
Let "$p$" be the proposition "$x\in A$", and let "$q$" be the proposition "$x\in B$".
$(A\subseteq B)$ is equal to "$\forall x, (x\in A \implies x\in B)$". Similarly $(B'\subseteq A')$ is equal to "$\forall x, (x\notin B \implies x\notin A)$". Now use Truth Table below: \begin{array}{c|c|c|c|c} p & q & \lnot p &\lnot q & p \Rightarrow q & \lnot q \Rightarrow \lnot p\\ \hline T & T & F &F & T & T\\ \hline T & F & F & T & F & F\\ \hline F & T &T & F& T &T\\ \hline F & F &T & T& T &T\\ \end{array}