Prove that if [imath]A\cap B = \emptyset[/imath] then [imath]A\cap B^c=A[/imath]

[Ozma]

I have to show that for any pair of sets [imath]A,B \subseteq X[/imath], if [imath]A\cap B=\emptyset[/imath] then [imath]A\cap B^c=A[/imath]; with [imath]B^c[/imath] I mean the complement of [imath]B[/imath] in [imath]X[/imath].

Since the intersection of two sets is always a subset of both sets, it is immediate that [imath]A \cap B^c \subseteq A[/imath].

Let now [imath]x \in A[/imath]. By hypothesis it is [imath]A \cap B=\emptyset[/imath], hence for any [imath]x \in X[/imath] it is [imath]x \notin A \cap B[/imath]. That is, for any [imath]x \in X[/imath] it is [imath]x \in (A \cap B)^c[/imath] and so [imath]x \in A^c \cup B^c[/imath]. But we assumed that [imath]x \in A[/imath], so from [imath]x \in A^c \cup B^c[/imath] it follows that [imath]x \in B^c[/imath]. Hence we have both [imath]x \in A[/imath] and [imath]x \in B^c[/imath], that is [imath]x \in A \cap B^c[/imath]. Since [imath]x \in A[/imath] is arbitrary, it follows that [imath]A \subseteq A \cap B^c[/imath]. It this proof valid?

 

[blamocur]

Looks good to me.

 

[Steven G]

I do not think that this proof is valid! You failed to say at the end that combining both results we can conclude that A∩B^c=A.
The point I am making is that nowhere in your proof did you state that you got the results that you wanted to prove!

 

[pka]

I have to show that for any pair of sets [imath]A,B \subseteq X[/imath], if [imath]A\cap B=\emptyset[/imath] then [imath]A\cap B^c=A[/imath] Hence we have both [imath]x \in A[/imath] and [imath]x \in B^c[/imath], that is [imath]x \in A \cap B^c[/imath]. Since [imath]x \in A[/imath] is arbitrary, it follows that [imath]A \subseteq A \cap B^c[/imath]. It this proof valid?

If [imath]x\in A[/imath] then [imath]x\notin B[/imath] because [imath]A\cap B=\emptyset[/imath].
That means [imath]x\in B^c[/imath] so that [imath]A\subseteq A\cap B^c[/imath]
Suppose that [imath]x\in A\cap B^c[/imath] then [imath]x\in A.[/imath]
That means [imath](A\subseteq A\cap B^c)\wedge (A\cap B^c\subseteq A)[/imath]
What can we now conclude?