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Can you conclude that $A = B$ if $A$, $B$, and $C$ are sets such that

(a) $A \cup C = B \cup C$

No, the sets $A=\{1,2\}, B=\{3,4\}, C=\{1,2,3,4,5\}$ disprove this, because $A \cup C = B \cup C$ but $A\neq B$

(b) $A \cap C = B \cap C$

No, the sets $A=\{1,2\}, B=\{1,4\}, C=\{1\}$ disprove this because $A \cap C = B \cap C=\{1\}$, however $A\neq B$

(c) $A \cup C = B \cup C$ and $A \cap C = B \cap C$

Having Trouble with (c)

This is what I have based on your response

L1 A ∪ C = B ∪ C
L2 A ∩ C = B ∩ C
L3 ∀x(x ∈ A ∪ x ∈ C = x ∈ B ∪ x ∈ C    )    
L4 ∀x(x ∈ A ∪ x ∈ C = x ∈ B ∪ x ∈ C    )    
L5 ∀x(x ∈ A ∩ x ∈ C = x ∈ B ∩ x ∈ C)
L6 ∀x(x ∈ A ∩ x ∉ C = x ∈ B ∩ x ∉ C)
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We are given sets $A,B$ and $C$ such that $A \cup C = B \cup C$ and $A \cap C = B \cap C$. We wish to conclude $A = B$.

To show $A = B$ it suffices to show $A \subseteq B$ and $B \subseteq A$. I will go through one of these arguments leaving you to show the other.

So, let's show that $A \subseteq B$, i.e. for every $a$ in $A$ we have $a$ in $B$:

Since $a \in A$ and $A \subseteq A \cup C$ it follows that $a \in A \cup C$. Now since $A \cup C = B \cup C$ if follows that $a \in B \cup C$.

Ok, now we ask a question, is $a$ in $C$? Well, it might be or it might not be. But either it is in $C$ or it is not. So there are two cases:

1.) Suppose $a$ is in $C$. Then $a$ is in $A$ and $C$, i.e. $a \in A \cap C$. Since $A \cap C = B \cap C$ (by our assumption) it follows that $a \in B \cap C$, in particular $a$ is in $B$.

2.) Suppose $a$ is not in $C$. Ok, well, we know that $a$ is in $B \cup C$, so if $a$ is not in $C$ then it must be in $B$.

Thus, in both cases $a$ is in $B$, which is exactly what we wanted. Thus we have shown $A \subseteq B$.

Now you must use a similar argument to show that $B \subseteq A$. When you have this, it follows (by definition of set-equality) that $A = B$.

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  • $\begingroup$ Hmm i appreciate the quick response... but idk, i think the equals signs are throwing me off? this is what I have L1 A ∪ C = B ∪ C _______ L2 A ∩ C = B ∩ C _______ L3 ∀x(x ∈ A ∪ C = B ∪ C) ______ L4 ∀x(x ∈ A ∪ C = x ∈ B ∪ C)______ $\endgroup$ – user155456 Jun 5 '14 at 19:43
  • $\begingroup$ ok, I'm not sure I understand your difficulty, but I edited a bit and gave the beginning of an argument. Has this helped? $\endgroup$ – user50948 Jun 5 '14 at 19:47
  • $\begingroup$ should I be trying to get rid of the C's somehow? I guess that's what I'm confused about, im not sure where providing that 'a' is an element of A and B is going to get me? If i say a a∉C, then wouldn't I just be running in circles changing all of the other elements to ∉? Sorry dude I'm very much a noob to this stuff. Edited to show my 'progress' $\endgroup$ – user155456 Jun 5 '14 at 19:56
  • $\begingroup$ Ok, no problem, try and read through my edited version. $\endgroup$ – user50948 Jun 5 '14 at 20:10
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Let it be that $A\cup C=B\cup C$ and $A\cap C=B\cap C$ and assume that $a\in A\wedge a\notin B$.

Then $a\in A\cup C=B\cup C$ and $a\notin B$ tells us that $a\in C$.

However, then $a\in A\cap C=B\cap C\subset B$ and a contradiction is found.

We conclude that $\forall x\;\left[x\notin A\vee x\in B\right]$ or equivalently $A\subseteq B$.

Likewise we find $B\subseteq A$ hence $A=B$.

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We know $A = B$ iff both the following: $$x \in A \implies x \in B\\ x \in B \implies x \in A $$

We also know $$ x\in A \implies x \in A\cup C $$ $$ x \in A\cap C \implies x \in A, x\in C $$ $$ \begin{align} x\in A \cup C \implies &x \in A, \text{or}\\ &x\in C \end{align} $$ which transposes to: $$ x\not\in C \implies (x\in A\cup C \implies x \in A) $$

We assume: $$ A\cap C = B \cap C,\\ A\cup C = B\cup C $$

Select $x \in A$. Then either $x\in C$ or $x\not\in C$.

  1. $x\in C$: $$ \begin{align} x\in A,x\in C &\implies x\in A\cap C\\ &\implies x\in B\cap C\\ &\implies x\in B \end{align} $$

  2. $x\not\in C$ $$ \begin{align} x\in A &\implies x\in A\cup C\\ &\implies x\in B\cup C\\ x\not\in C,x\in B\cup C &\implies x\in B \end{align} $$

In either case, $x\in A \implies x\in B$. The arguments work the same with $A$ and $B$ switched, so $x\in B \implies x\in A$. Thus, we conclude $A=B$.

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