Hi,
I am new here ,
I have stuck to Proved or disproved This sentence :
If for any A,B,C groups exist A\C=B\C And A ∩ C =B ∩ C , then A⊆ B.
Thanks very much for help.
Please, please carefully reread the above. I think that is not what you really mean.
Do you really mean groups? Or is it sets or collections?
Are you aware that $A\setminus C=A\cap C^c$ where $C^c$ is the complement of C. That is not usual group theory.
You write prove: $A\subseteq B$ really? $A$ is a subset of $B~?$
Let $x \in A$. If $x \in C$, then $x \in A \cap C = B\cap C$, so $x \in B$.
If $x \notin C$, then $x \in A \setminus C = B\setminus C$, so $x \in B$.
Since either $x \in C$ or $x \notin C$, this covers all possibilities, and $x \in B$ for any $x \in A$. This is the definition of a subset.
We know that $A\setminus C=B\setminus C$ or $A\cap C^c=B\cap C^c$
$ \begin{align*}A\cap C^c&=B\cap C^c \\\text{and given }A\cap C&=B\cap C\\\text{we have }(A\cap C)\cup(A\cap C^c)&=(B\cap C)\cup(B\cap C^c)\\A\cap(C\cup C^c)&=B\cap(C\cup C^c)\\A\cap~ \mathcal{U})&=B\cap~ \mathcal{U}\\A&=B \end{align*}$
I have another questions of this : "If for any A,B,C sets exist C\A=C\B And A ∩ C =B ∩ C , then A⊆ B."
I have tried to solve like that :
Let x∈A. If x∈C, then x∈A∩C=B∩C, so x∈B.
If x∉C, then x!∈C∖A=C∖B.
Is that Okay ? ( At begging I want to take x∈C but I need element from set A to show that exist : A⊆ B )
Thanks
No, that does not follow.
Let $x \in A$ is the first step. Now, you have an element of $A$.
Now, you have two cases:
Case $x \in C$: Then $x \in A\cap C = B\cap C \subseteq B$ shows $x \in B$.
Case $x \notin C$: Then $x \in A \setminus C = B\setminus C \subseteq B$ shows that $x \in B$.
Yes you right , becuase on the page exercises they have written that , that sentence not true and I give them contradict example.
I have another difficult question that I don't know how to solve :
"For any A,B,C Sets if exist (A∪B)\(A∩B) =(A∪C)\(A∩C) ⇐⇒ B=C "
Thanks for help.
$\Rightarrow$: Let $x \in B$. Show that $x \in C$.
Case 1: $x \in A$. Then, $x \notin (A\cup B) \setminus (A\cap B) = (A\cup C) \setminus (A \cap C)$. Since $x \in A$, this implies $x \in A\cup C$. Since $x \notin (A\cup C) \setminus (A \cap C)$, it must be that $x \in A\cap C \subseteq C$, so $x \in C$.
Case 2: $x \notin A$. Then $x \in (A\cup B)$, but $x \notin A\cap B$. So, $x \in (A\cup B) \setminus (A\cap B) = (A\cup C) \setminus (A\cap C)$. Since $x \notin A$, you know $x \notin A\cap C \subseteq A$. Therefore, it must be that $x \in C$.
$\Leftarrow$: Let $x \in C$. Show that $x \in B$.
Similarly as above.
Thanks very much ,
in the original question I have A "Triangle symbol" B = A "Traingle Symbol" C , I can write this like I have written as : "(A∪B)\(A∩B) =(A∪C)\(A∩C)" ?
(I havn't write the traingle symbol that is symetric difference because I don't know how to write this in this box if you could explain me I will be gratefull )
I have to rewrite the question with this : "(A∪B)\(A∩B) =(A∪C)\(A∩C)" to solve it like you do ?