Prove that $A - B=A\cap B^c$

The problems is exactly what is stated in the title.

Have been stuck on this for a while now. Thanks. 

Set Theory Real Analysis Algebra
Staceypc Staceypc
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Answer

Let $x \in A \setminus B$. By definition $x \in A$ and $x \not \in B$, or equivalently $x \in B^c$; it follows that $x \in A \cap B^c$, so $A \setminus B \subseteq A \cap B^c $.

Let $x \in A \cap B^c$. By definition $x \in A$ and $x \in B^c $, or equivalently $x \not \in B$; it follows that $x \in A \setminus B$, so $A \cap B^c \subseteq A \setminus B$.

The double inclusion holds, so $A \setminus B = A \cap B ^c$.

Alessandro Iraci Alessandro Iraci
1.7K
The answer is accepted.
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