Prove that if $A\Delta B\subseteq A$ then $B\subseteq A$.

Question

Not a duplicate of

Prove that if $A \bigtriangleup B\subseteq A$ then $B \subseteq A.$

Prove that if $A \mathop \triangle B \subseteq A$ then $B\subseteq A$

This is exercise $3.5.5$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Prove that if $A\Delta B\subseteq A$ then $B\subseteq A$.

Here is my proof:

Suppose $A\Delta B\subseteq A$. Let $x$ be an arbitrary element of $B$. Suppose $x\notin A$. From $x\in B$ and $x\notin A$, $x\in B\setminus A$. Ergo $x\in(B\setminus A)\cup(A\setminus B)$. From $A\Delta B\subseteq A$ and $x\in(B\setminus A)\cup(A\setminus B)$, $x\in A$ which contradicts the assumption that $x\notin A$. Therefore $x\in A$. Thus if $x\in B$ then $x\in A$. Since $x$ is arbitrary, $\forall x(x\in B\rightarrow x\in A)$ and so $B\subseteq A$. Therefore if $A\Delta B\subseteq A$ then $B\subseteq A$. $Q.E.D.$

Is my proof valid$?$ Is there a way to prove the above statement directly $($not using proof by contradiction$)?$ If there is, then please provide me with hints and not complete answers.

Thanks for your attention.

Answer 1

Hint: If $B\Delta A= (B\setminus A) \cup (A\setminus B) \subset A$, what can you say about $B\setminus A$?

Answer 2

Your argument looks good. For a direct proof:

$$A\triangle B = (B\backslash A) \cup (A\backslash B)$$

Thus, $$B\backslash A \subset A.$$ On the other hand

$$ B\backslash A = B \cap A^c \subset A^c.$$

So $$ B\backslash A \subset A \cap A^c = \emptyset.$$

Therefore $$ B \backslash A = \emptyset$$ and $B\subset A$ follows.

Answer 3

An identity is $$B = ((B \Delta A) \cap B) \cup (A \cap B)$$

(an $x \in B$ is either not in $A$ and then it's in the left hand side, or it is in $A$ and then it's in the right hand side)

Now we apply that $A \Delta B \subseteq A$ so that

$$B \subseteq (A \cap B) \cup (A \cap B) = A \cap B \subseteq A$$

and we're done.