The definition of a symmetric difference of two sets, that my book provides, is: Set containing those elements in either $A$or $B$, but not in both $A$ and $B$.
So, in set builder notation, I figured that it would be $A⊕B = \{x| (x \in A \vee x \in B) \wedge (x \notin A \wedge x \notin B)\}$
Is this correct? I have some properties of symmetric differences of sets to prove, and I wanted to prove them by subsets, and by using definitions and laws of logic.
$\endgroup$6 Answers
$\begingroup$As written, you say that $x$ is in at least one of them, and $x$ is in neither of them. In other words, you've defined the empty set. To fix it, you could change $$x\notin A\wedge x\notin B$$ to $$x\notin A\vee x\notin B.$$
$\endgroup$ 1 $\begingroup$Taking your first sentence and putting it in symbols we have $A\oplus B=A\cup B- A\cap B$. Now for any sets $X$ and $Y$, the set difference $X-Y$ put in set-builder notation is $\{x\in X: x\not\in Y\}$. So $$A\oplus B=\{x\in A\cup B:x\not\in A\cap B\}=\{x\in A\cup B:x\not\in A\vee x\not\in B\}.$$
$\endgroup$ $\begingroup$Notation $\oplus$ is not apropriate for symmetric difference of sets we must use $\Delta$ or $\ominus$ $$A\Delta B=(A\cup B)\setminus(A\cap B)=(A\setminus B)\cup(B\setminus A $$
$\endgroup$ 2 $\begingroup$no. $A \oplus B = \{ x \in A \cup B \mid x \in A \setminus B \text{ or } x \in B \setminus A \}$.
The definition you gave could be fixed by changing $x \notin A \wedge x \notin B$ to $\lnot (x \in A \wedge x \in B)$
$\endgroup$ 3 $\begingroup$The most simple definition of set difference I know is $$ x \in A ⊕ B \;\;\equiv\;\; x \in A \;\;\not\equiv\;\; x \in B $$ for all $x$, $A$, and $B$, where $\equiv$ is logical equivalence (often written as $\Leftrightarrow$) and $\not\equiv$ is inequivalence ('exclusive or', occasionally written as $\not\Leftrightarrow$).
See another answer of mine for a practical application of this definition (but using the different symbol $\Delta$).
(Note, by the way, that $\equiv$ and $\not\equiv$ are both and mutually associative: $(P \equiv Q) \not\equiv R$ is equivalent to $P \equiv (Q \not\equiv R)$, so it is perfectly safe to leave out the parentheses.)
$\endgroup$ $\begingroup$The definition your book provides is almost correct. As others have pointed out, apparently you mistyped $x\notin A\wedge x\notin B$. I just want to point out that this definition is clearly equivalent to $A\bigtriangleup B=(A\cup B)\setminus(A\cap B)$. $x\in A\vee x\in B$ means the same as $x\in A\cup B$ and $x\notin A\vee x\notin B$ means the same as $x\notin A\cap B$.
$\endgroup$
