Definition Of Symmetric Difference

Question

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.

Answer 1

As written, you say that $x$ is an 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 $$x\notin A\vee x\notin B.$$

Answer 2

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 $$A\oplus B=\{x\in A\cup B:x\not\in A\cap B\}=\{x\in A\cup B:x\not\in A\vee x\not\in B\}.$$

Answer 3

Notation $\oplus$ is not apropriate for symmetric difference of sets we must use $\Delta$ or $\ominus$ $$A\Delta B=(A\cup B)\setminus(A\cap B)=(A\setminus B)\cup(B\setminus A $$

Answer 4

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 $\lnot (x \in A \wedge x \in B)$

Answer 5

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.)