2
$\begingroup$

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.

$\endgroup$
4
  • 7
    $\begingroup$ $(A\subseteq B)\iff (\overline{B}\subseteq \overline{A})$ 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. $\endgroup$ – user26486 May 3 '15 at 22:11
  • 1
    $\begingroup$ Please follow this guide next time. It will make your questions much easier to read. $\endgroup$ – Arthur May 3 '15 at 23:29
  • 2
    $\begingroup$ For the love of God, please punctuate. Write complete sentences. And write in LaTeX. It makes the problem so much easier to read. $\endgroup$ – AJY Dec 8 '15 at 21:51
  • $\begingroup$ 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 $x \notin A$. $\endgroup$ – fleablood Dec 8 '15 at 22:18
1
$\begingroup$

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 $B' \subset A' \implies A \subset B$

$\endgroup$
1
$\begingroup$

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 $A\subset B$.

This shows both directions of the equivalence and completes the proof.

$\endgroup$
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.