$A=\emptyset $ if and only if $B = A \bigtriangleup B$

Question

Is this true?

If $A$ and $B$ are sets, then $A=\emptyset $ if and only if $B = A \bigtriangleup B$.

If $A=\emptyset$ then $B=\emptyset$ too?

Could someone help me please?

Answer 1

Definition: $$A \bigtriangleup B := (A \setminus B) \cup (B\setminus A)$$

Proof: $A=\emptyset \leftrightarrow A \bigtriangleup B=B$ $$"\to"$$$$\begin{align} A \bigtriangleup B &:= (A \setminus B) \cup (B\setminus A) \\ &= (A \cup B) \setminus (A\cap B) \text{ (by Proof} _0) \\ &=(\emptyset \cup B) \setminus (\emptyset \cap B) \text{ (because } A=\emptyset)\\ &= B\setminus \emptyset \\ &=B\end{align} $$ $$"\leftarrow"$$$$A \bigtriangleup B=B=\emptyset \bigtriangleup B \to A=\emptyset \text{ (by Proof} _3)$$

In addition, I used following:

Proof$_0$: $(A \cup B) \setminus (A \cap B)=(A \setminus B) \cup (B\setminus A) $ \begin{align*} (A \cup B)\setminus (A\cap B) &= ((A \cup B) \setminus A) \cup ((A \cup B) \setminus B) \text{ (by Proof} _1)\\ &=(B \cup A) \setminus A) \cup ((A \cup B) \setminus B) \\ &= (B \setminus A) \cup (A \setminus B) \text{ (by Proof} _2)\\ &=(A \setminus B) \cup (B\setminus A) \end{align*}

Proof$_1$: $A\setminus(B\cap C)=(A\setminus B) \cup (A\setminus C)$ \begin{align*} x\in A\setminus(B\cap C) &\leftrightarrow x \in A \wedge x \notin (B \cap C)\\ &\leftrightarrow x \in A \wedge (x \notin B \vee x \notin C)\\ &\leftrightarrow (x \in A \wedge x \notin B) \vee (x \in A \wedge x \notin C) \\ &\leftrightarrow x \in (A \setminus B) \vee x \in (A\setminus C) \\ &\leftrightarrow x \in (A\setminus B) \cup (B\setminus C) \end{align*}

Proof$_2$: $(A \cup B)\setminus B=A \setminus B$ \begin{align*} x \in(A \cup B)\setminus B &\leftrightarrow x \in (A\cup B) \wedge x \notin B \\ &\leftrightarrow (x \in A \vee x \in B)\wedge x \notin B\\ &\leftrightarrow (x \in A \wedge x \notin B) \vee (x \in B \wedge x \notin B) \\ &\leftrightarrow x \in (A\setminus B) \vee x \in (B\setminus B) \\ &\leftrightarrow x \in ((A\setminus B) \cup \emptyset )\\ &\leftrightarrow x \in (A\setminus B) \end{align*}

Proof$_3$: $E\bigtriangleup F=E \bigtriangleup G \to F=G$ $$\begin{align} F &=\emptyset \bigtriangleup F \\ &= (E\bigtriangleup E) \bigtriangleup F \\ &= E\bigtriangleup(E\bigtriangleup F) \\ &= E \bigtriangleup (E \bigtriangleup G) \\&= (E \bigtriangleup E) \bigtriangleup G \\ &= \emptyset \bigtriangleup G \\ &= G\end{align}$$

Answer 2

Noone mentioned characteric functions yet: let $\chi_A$ be the function such that $\chi_A(x)=1$ if $x$ belongs to $A$ and $0$ otherwise.

As $\chi_{A\cap B}=\chi_A\chi_B$ and $\chi_{A\triangle B}=\chi_A+\chi_B-2\chi_{A\cap B}=\chi_A+\chi_B-2\chi_{A}\chi_{B}$, we have

$$ A\triangle B=B \Leftrightarrow \chi_{A\triangle B}=\chi_B \Leftrightarrow \chi_A+\chi_B-2\chi_{A}\chi_{B} =\chi_B \Leftrightarrow \chi_A(1-2\chi_{B}) \Leftrightarrow \chi_A=0 \Leftrightarrow A=\emptyset $$

The second-to-last equivalence follows from the fact that $1-2\chi_B=\pm 1$.

Answer 3

$A \Delta B$ is the set of things that belong to exactly one of $A$ and $B$.

So, if $A = \emptyset$, then $A\Delta B = (A\backslash B) \cup (B\backslash A) = B$.

If $A\Delta B = B$, then $A\backslash B = \emptyset$, or $A\subset B$, but $B\backslash A = B$ then $A = \emptyset$.

Answer 4

$A\triangle B$ is the set of elements in either $A$ or $B$ but not both.

If $A=\emptyset$ then $A\triangle B$ contains all the elements of $B$ and none that are not.

If $B=A\triangle B$ then every element of $B$ is in $A\triangle B$ and no element not in $B$ is in $A\triangle B$, which is only permissible if $A$ is empty.

So $A=\emptyset \iff A\triangle B=B$

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However, a fact of $B=A\triangle B$ places no restraint on the contents of $B$, and $B$ being not empty is no prohibition against $A\triangle B=B$.

Answer 5

I hope you know that $A\bigtriangleup B=(A\setminus B)\cup(B\setminus A)$.

Then, if $A=\emptyset$, obviously $A\bigtriangleup B=(\emptyset\setminus B)\cup(B\setminus\emptyset)=B$.

For the other direction, note that $A\setminus B$ never is a subset of $B$, unless it is empty. So $A\subseteq B$. But $B\setminus A=B$, so if $A\neq\emptyset$, then $B\setminus A\subsetneq B$. Thus $A=\emptyset$.