1
$\begingroup$

How can I prove that for any $A, B$ if $A\subseteq B$ and $B\subseteq C$, then $(C-A)\cup (B-A)\subseteq C$?

I've been working on this question and I haven't really made much progress with it. I know that I can rewrite it as $(C \cap A^c) \cup (B\cap A^c)$. I'm pretty sure that if $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$. If this is the case then wouldn't $(C \cap A^c) = \emptyset$ and $(B \cap A^c) = \emptyset$ or am not understanding something with set theory? Thank you for the help.

$\endgroup$
  • 1
    $\begingroup$ Rewrite each statement as an implication in memberships. $\endgroup$ – J.G. Dec 9 '18 at 22:38
  • $\begingroup$ Consider $A=\{1\}, B=\{1,2\}, C=\{1,2,3\}$. We have $A\subseteq B\subseteq C$. But the complement $A^c$ of $A$ with respect to $C$ (as it's important to stress what the complement is in respect to) is $\{2,3\}$, so $C\cap A^c=\{2,3\}$ and $B\cap A^c=\{2\}$. $\endgroup$ – Shaun Dec 9 '18 at 22:39
  • $\begingroup$ $A \subset C$ means everything in $A$ is in $C$. $C\setminus A$ means all the stuff in $C$ that isn't in $A$. If $A$ is in $C$ then $C\setminus A$ is all the stuff in $C$ that isn't in $A$. That needn't be empty. Maybe you are confusing $C \setminus A$ with $A \setminus C$? Setminus is not commutative. $A \setminus C$ is all the stuff in $A$ that is not in $C$, but everything in $A$ is in $C$ so $A\setminus C = \emptyset$. But $C\setminus A$ is just... $C \setminus $A$. $\endgroup$ – fleablood Dec 9 '18 at 23:09
1
$\begingroup$

Quite simply: by definition, $C-A\subseteq Cˆ$ and $B-A\subseteq B$. Further, $B\subseteq C$...

$\endgroup$
1
$\begingroup$

It's not the case that $X \subseteq Y$ implies $Y-X = \emptyset$. Indeed if $X = \{0\}$ and $Y = \{0,1\}$ then $Y - X = \{1\}$.

We do have, however, that $X \subseteq Y$ implies $Y - X \subseteq Y$. Applying this to what you have so far, conclude that $C- A \subseteq C$ and $B - A \subseteq B \subseteq C$. Hence $(C-A) \cup (B-A) \subseteq C$ too.

$\endgroup$
1
$\begingroup$

For every $x\in (B-A)\cup (C-A)$, either $x\in B-A$ or $x\in C-A$ (or both). For every $x\in B-A$, $x\in B$ hence $x\in C$. For every $x\in C-A$, obviously $x\in C$. Hence if $x\in (B-A)\cup (C-A)$, surely $x\in C$. By definition, this means that it is a subset of $C$.

$\endgroup$
0
$\begingroup$

Draw Venn Diagrams. This should be obvious.

enter image description here

Now just prove it with Element chasing or definitions or whatever you like.

$\endgroup$

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.