# Numerical sets

#### Leandrowski

Sort on true or false this sentence:

(A-B)∪(A∩B) = A

Ok, we know that: (A-B) = (x|x ∈ A ∧ x ∉ B)
we also know that: (A∩B) = (x|x ∈ A ∧ x ∈ B)
resulting in the sentence = (x|x ∈ A ∧ x ∉ B) ∪ (x|x ∈ A ∧ x ∈ B)

True or false? Why? I need help.

OBS: I'm Brazilian, sorry about my English.

#### skeeter

Math Team
$A-B = A \cap B^c$

$(A \cap B^c) \cup (A \cap B) = A \cap (B^c \cup B) = A \cap \emptyset = A$

#### skipjack

Forum Staff
True. As {x|x ∈ A ∧ x ∉ B} ∪ {x|x ∈ A ∧ x ∈ B} combines the cases where x ∉ B and where x ∈ B, one of which must be true, it simplifies to {x|x ∈ A}, which is A.

If you want a formal proof from your axioms, let us know what those axioms are.

#### Maschke

Sort on true or false this sentence:

(A-B)∪(A∩B) = A
Here's another proof. To prove two sets equal we need to prove two things:

1) The left side is a subset of the right side; and

2) The right side is a subset of the left side.

Let's do (2) first. Suppose $a \in A$. Then $a \in A - B$; and since $a$ is an element of one of the disjuncts of a union, it's in the union. Therefore the right side is a subset of the left side.

Now (1). Say $x \in (A - B ) \cup (A \cap B)$. Either $x \in A - B$ or $x \in A \cap B$.

If $x \in A - B$ then $a \in A$ by definition of set difference and we're done.

If $x \in A \cap B$ then $x \in A$ by def of intersection. Either way $x \in A$ and we're done.