Prove that $A - B=A\cap B^c$
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 $A \setminus B = A \cap B ^c$.
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
1.7K
-
Thank you so much :)
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 3111 views
- $3.00
Related Questions
- Fields and Galois theory
- How to find specific heat of a turpentine question?
- Algebra Word Problem 1
- Analyzing the Domain and Range of the Function $f(x) = \frac{1}{1 - \sin x}$
- Evaluate $\int \ln(\sqrt{x+1}+\sqrt{x}) dx$
- Rouche’s Theorem applied to the complex valued function $f(z) = z^6 + \cos z$
- Length of a matrix module
- A Problem on Affine Algebraic Groups and Hopf Algebra Structures
