2
$\begingroup$
  • Given: $A \cap B= \emptyset $
  • Prove: $A\setminus B=A$.
  • My thoughts:

$x \in \emptyset \Rightarrow x \in A \cap \emptyset $ and $ x \in A \cap B \Rightarrow x \in A \cap B \cap \lnot B $ and $ x \in A \cap B \Rightarrow x \in (A \cap \lnot B) \cap B$ and $ x \in A \cap B \Rightarrow x \in (A\setminus B) \cap B $ and $ x \in A \cap B \Rightarrow A\setminus B=A $

  • 1
    Why $x \in A \cap B =>x \in A \cap B \cap \lnot B $.? – 2017-02-19
  • 1
    Show that $\{x \in A : x \notin B \} \subseteq A$ and then that $A \subseteq \{x \in A : x \notin B \}$. One direction is trivial by definition. – 2017-02-19
  • 1
    The empty set axiom states: $(\exists y)(\forall x)\neg(x \in y)$. Thus, there is no element $x \in \emptyset$... – 2017-02-19
  • 1
    My thoughts: you need to use words in your proofs more often. – 2017-02-19

5 Answers 5

1

Def. let be $A,B$ sets: $$A\setminus B:=\{x|x \in A \wedge x \notin B\} $$

We prove following:

The$_0$. let be $A,B$ sets: $$ A \setminus (A \cap B)=A \setminus B$$ Proof: $$ \begin{align} x \in A \setminus (A \cap B) &\leftrightarrow x \in A\wedge (x \notin (A\cap B)) \\ &\leftrightarrow x \in A \wedge (x \notin A \vee x \notin B) \\ &\leftrightarrow (x \in A \wedge x \notin A) \vee (x \in A \wedge x \notin B) \\ &\leftrightarrow x \in A \setminus A \vee x \in A \setminus B \\ &\leftrightarrow x \in \emptyset \vee x \in A \setminus B \\ &\leftrightarrow x \in \emptyset \cup A \setminus B \\ &\leftrightarrow x \in A\setminus B \end{align}$$

Now we prove your theorem:

The$_1$. let be $A,B$ sets: $$A \cap B=\emptyset \to A \setminus B=A$$ Proof: we use The$_0$. therefore: $$ A \setminus B= A \setminus (A \cap B)= A \setminus \emptyset= A$$

0

It is hard to follow what is going on here. You should adopt a systematic approach.

To prove that $S_1 = S_2$, where $S_1$ and $S_2$ are sets, you must prove that $S_1 \subseteq S_2$ and $S_2 \subseteq S_1$.

This then means that you prove that whenever $x \in S_1$ then also $x \in S_2$ and whenever $x \in S_2$ then also $x \in S_1$.

0

If $x \in A \setminus B$ then certainly $x \in A$ , so $A \setminus B \subseteq A$.

If $x \in A$, then $x \notin B$, otherwise $x \in A \cap B = \emptyset$, contradiction. So $x \in A \setminus B$ by definition, hence $A \subseteq A \setminus B$, so we have equality of sets.

If you like "equation style" proofs ($X$ denotes the "universe" with respect to which we take complements):

$$A =A \cap X = A \cap (B \cup B^c) = (A\cap B) \cup (A \cap B^c) = \emptyset \cup (A \cap B^c) = A \cap B^c= A \setminus B$$

0

Without words, arguments tend to be pretty hard to follow. Here's how I'd do it:

It's clear that $A \setminus B \subseteq A$. So let's show the converse. Assume $x \in A.$ We need to that show $x \in A \setminus B$, in other words $x \in A$ and $x \notin B$. Well, we already know $x \in A$. So all we need to do is show $x \notin B$. Assume toward a contradiction that $x \in B$. Then $x \in A \cap B$. So $x \in \emptyset$, a contradiction. We conclude that $x \notin B$, as required. This completes the proof.

0

If you don't want to use subset relationships, you can use co-implications: $$ x \in A\setminus B \iff x \in A \ \wedge x \not\in B \iff x \in A$$ where the last co-implication holds since $A \cap B = \emptyset$ and so, if $x \in A$ it cannot be in $B$.