0

Started my proof by assuming that $A \subset B$ and $C \subset D$. Then I proceeded by supposing that $x \in A \setminus D$. So $x \in A$ and $x \notin D$. Is it valid to conclude that $x \notin C$ since $C \subset D$? I'm assuming it is always true but I'm not sure if it is a valid argument I can use for formal proofs.

Aside from that, I'm also lost as to how I should continue my proof. For now I continued with my proof mentioned above and added "since $A \subset B$, then $x \in B$. Also, $x \notin C$ so it follows that $x \in B \setminus C$. Therefore, $A \setminus D \subset B \setminus C$. " Although, I'm not really sure if this is a valid argument. For instance, I'm having doubts with $x \in B$ and $x \notin C$ since $x$ is also not an element of $D$. Therefore, there's a possibility that $x \in B \setminus D$ (??). Any help with regards as to how I should properly and formally prove this will be a great help.

Edit: Thanks a lot to everyone who verified my approach!

  • You can write $A\setminus D$ as $A \setminus D$ – jjagmath Apr 22 '21 at 16:44
  • There is a possibility that $x \in B \setminus D$, because it's definitely true: you've proven both $x \in B$ (since $x \in A$) and $x \notin D$ (since $x \in A \setminus D$). That isn't a problem; it doesn't contradict what you want to show, which is $x \in A \setminus C$. – Theo Bendit Apr 22 '21 at 16:50

2 Answers2

1

You're correct stating that if $x\not\in D$, since $C\subset D$ then $x\not\in C$. If it isn't clear to you, look at the contraposition (if $x\in C$, since $C\subset D$ then $x\in D$).

In the second part, you proved that for every element $x\in A\setminus D$, you have $x\in B\setminus C$, therefore $A\setminus D\subset B\setminus C$. If $x\in B$ and $x$ is not an element of $D$ then automatically $x\in B\setminus D$.

Maxence1402
  • 590
-1

Your argument is right. In symbols, you're essentially doing $$A\subset B\quad\text{and}\quad D^c\subset C^c\implies A\cap D^c\subset B\cap C^c$$ using the fact that $M\subset N\iff N^c\subset M^c$, which is right.

Martund
  • 14,706
  • 2
  • 13
  • 30