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I am stuck. I know that I need to prove $A$ is a subset of $B$ and that $B$ is a subset of $A$, and that in each case all elements in A are elements in $B$, but I don't know where to go from there. Any help is appreciated!

4 Answers 4

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$$A=(A-C)\cup(A\cap C)=(B-C)\cup(B\cap C)=B$$

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Note that we have that $$A \cap C^{c}=B \cap C^{c}$$$$A \cap C=B \cap C$$ So we have that $$(A \cap C^{c}) \cup (A \cap C)=(B \cap C^{c}) \cup (B \cap C) \tag{1}$$ However, we know that $$(A \cap C^{c}) \cup (A \cap C)=A \cup (C^{c} \cap C)=A \tag{2}$$From the Distributive Property. So $(1)$ and $(2)$ gives us that $$A=B$$ as desired.

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    if $C=\emptyset$, $C^C=?$ (https://books.google.de/books?id=sxr4LrgJGeAC&lpg=PA14&hl=de&pg=PA29#v=onepage&q&f=false) – 2017-02-15
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    @mle If $C$ is empty the result follows trivially, so then just consider the case where $C$ is nonempty. – 2017-02-15
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    @S.C.B, thank you! If $C=\emptyset$, Do you define $A \setminus C:=A \cap C^C$? Do you use two definitions, one for empty set and another for not emptyset? – 2017-02-15
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    @mle See [here](https://en.wikipedia.org/wiki/Complement_(set_theory)#Properties_2). It's not a defenition, it is a property which follows from the definition of $A \setminus C$. (By the way, the notations are confusing for me, because the US and Korea use different notations). – 2017-02-15
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    @mle $A \setminus \emptyset := \{x \in A\}\cap \{x \notin \emptyset\},$ but $\forall x\in A$ we have that $x \notin \emptyset$; hence, $A \setminus \emptyset = A.$ – 2017-02-15
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    @S.C.B, I read: "Let A and B be two sets in a universe U", here ist U=C? – 2017-02-15
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    @David, ok.. $F\setminus G:=\{x|x \in F \wedge x \notin G\}$, but in ZFC I can not prove that $F\setminus G=F\cap G^C$ without an universe $U$ – 2017-02-15
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    @mle Why is $U=C$? – 2017-02-15
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    @mle, why do you think that $U = C$. We of course have that $C \subseteq U$, but $U$ is simply a set that contains all elements of interest. Also, by context, it would seem unlikely that this question was asked regarding any axiomatic set theories (it's even tagged with elementary-set-theory)... – 2017-02-15
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    ok, it is so naive.. than we have in $U$ and here that $$\begin{align} \emptyset \cup (A- C)=(A -C)&=A- (A\cap C)\\&=A-(B\cap C)\\&=(A-B) \cup (A-C)\\&\to \emptyset=(A-B) \\&\to A=B\end{align}$$ – 2017-02-15
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In full detail, let us define $U$ to be the "universe" which contains all elements of interest in our context. We then naturally have a few definitions and identities, namely

  1. For any set $S \subseteq U$, $S^c := \{x \in U\} \cap \{x \notin S\}$
  2. If $S \subseteq U$ then $S \cup S^c = U$
  3. If $S \subseteq U$ then $S = S \cap U$
  4. If $S_1,S_2,S_3 \subseteq U$ then $S_1 \cap (S_2 \cup S_3) = (S_1\cap S_2) \cup (S_1 \cap S_3)$
  5. If $S_1,S_2 \subseteq U$ then $S_1 \setminus S_2 = S_1 \cap S_2^c$

Then we have $$\begin{align} A = A\cap U &= A \cap (C \cup C^c) \\&= (A \cap C) \cup (A \cap C^c) \\ &= (A\cap C) \cup (A \setminus C) \\ &= (B\cap C) \cup (B \setminus C)\\&= (B \cap C) \cup (B \cap C^c) \\ &= B \cap (B\cup B^c) = B \cap U = B.\end{align}$$ In order the equalities hold by: 3, 2, 4, 5, problem statement, 5, 4, 2, 3.

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As you said, we can start by supposing $x\in A$. We want to prove that $x\in B$. Then, consider whether or not $x\in C$. If $x\in C$, then $x\in A\cap C$. But this means that $x\in B\cap C$, so $x\in B$. If instead $x\notin C$, then $x\in A\setminus C$. But this means that $x\in B\setminus C$, so, again $x\in B$. Thus, $A\subset B$.

The reverse inclusion is almost the same. Just replace $A$ and $B$ above.