Real Analysis Proofs

1 ♦ 0 = 0.a

0.a = 0.a

0.a = (0 + 0).a [Since 0 is the additive identity]
0.a = 0.a + 0.a [Distributive Axiom]

-0.a + 0.a = -0.a + 0.a + 0.a [Adding (-0.a) to both sides, the additive inverse of 0.a]
0 = (-0.a + 0.a) + 0.a [Associative Axiom]

0 = 0 + 0.a

0 = 0.a [Since 0 is the additive identity] ♠

2 ♦ -a = -1(a)

(-a) = (-a)

(-a) = (-a) + 0 [Additive Identity]

(-a) = (-a) + 0.a [Since 0 = 0.a (By above theorem (1))]
(-a) = (-a) + (1 + (-1))a [Additive Inverse of 1]

(-a) = (-a) + (1.a + (-1)a) [Distributive Axiom]
(-a) = (-a) + (a + (-1)a) [Multiplicative Identity]
(-a) = ((-a) + a) + (-1)a [Associative Axiom]
(-a) = 0 + (-1)a [Additive Identity]

(-a) = (-1)a [Additive Identity] ♠

3 ♦ -(a + b) = (-a) + (-b)

-(a + b) = -(a + b)

–(a + b) = (-1)(a + b) [By above theorem (2)]
(-1)(a + b) = (-1)(a) + (-1)(b) [Distributive Axiom]

-(a + b) = (-a) + (-b) [By above theorem (2)]

♦ Sup(A + B) = Sup(A) + Sup(B)

∀ a ∈ A and ∀ b ∈ B;
a ≤ Sup(A) and b ≤ Sup(B)

∴ a + b ≤ Sup(A) + Sup(B)
∴ Sup(A) + Sup(B) is an upper bound of [A + B]

By definition of Sup(A + B), Sup(A + B) is the least upper bound of [A + B]. ∴ Sup(A + B) ≤ Sup(A) + Sup(B) -- (1)

Let ε > 0;

∃a ∈ A and ∃b ∈ B such that a + ε/2 ≥ Sup(A) and b + ε/2 ≥ Sup(B)
a + b + ε ≥ Sup(A) + Sup(B)
a + b ≥ Sup(A) + Sup(B) - ε

Since a + b ≤ Sup(A + B);
Sup(A + B) ≥ a + b ≥ Sup(A) + Sup(B) - ε ∴ Sup(A + B) ≥ Sup(A) + Sup(B) - ε ⇒ Sup(A + B) ≥ Sup(A) + Sup(B) -- (2)

(1) and (2) ⇒ Sup(A + B) = Sup(A) + Sup(B) ♠