Intervals

Defenition

Let A ≤ R; A is said to be an interval if following condition is satisfied.

If a, b ∈ A with a < b, then
a < x < b ⇒ x ∈ A

Eg: A = {1, 2, 3, 4}
2, 3 ∈ A & 2 < 2.5 < 3 & 2.5 ∉ A

Therefore A is not an interval

Defenition

Let a < b

1. (a, b) = {x | x ∈ ℝ & a < x < b} - Open Interval
2. [a, b] = {x | x ∈ ℝ & a ≤ x ≤ b} - Closed Interval
3. (a, b] = {x | x ∈ ℝ & a < x ≤ b} - Open & Closed Interval
4. [a, b) = {x | x ∈ ℝ & a ≤ x < b} - Closed & Open Interval

For all the above 4 cases, the length of the interval is (b - a). There are intervals of infinite length.

Defenition

Let a ∈ ℝ

1. (a, ∞) = {x | x ∈ ℝ & a < x}
2. [a, ∞) = {x | x ∈ ℝ & a ≤ x}
3. (-∞, a) = {x | x ∈ ℝ & a > x}
4. (-∞, a] = {x | x ∈ ℝ & a ≥ x}

Defenition

Let x ∈ ℝ

The absolute value of x is the size of x. It's defined,

		x if x ≥ 0
|x|	{
		-x if x < 0

Theorem

Let a > 0

1. |x| < a ⇔ -a < x < a
2. |x| ≤ a ⇔ -a ≤ x ≤ a

Now we can say;
{x | |x| < 5} = {x | -5 < x < 5} = (-5, 5)

Lemma

∀x ∈ ℝ, -|x| ≤ x ≤ |x|

Proof

If x = 0; this is trivial
So let x ≠ 0. Now |x| > 0
x ≤ |x| with |x| > 0

∴ -|x| ≤ x ≤ |x| from part ii. of the Theorem.

Triangle Inequality

∀x, y ∈ ℝ, |x + y| ≤ |x| + |y|

Proof

Let x, y ∈ ℝ

Case 1: |x| + |y| = 0
Then x = 0 = y, then the result is trivial

Case 2: |x| + |y| > 0
From the lemma;
-|x| ≤ x ≤ |x|
-|y| ≤ y ≤ |y|

∴ -(|x| + |y|) ≤ x + y ≤ |x| + |y|

from part ii. of the Theorem,
|x + y| ≤ |x| + |y|

Triangle Inequality - Alternative Version

∀x, y ∈ ℝ, |x| - |y| ≤ |x - y|

Proof

Take a = x - y; b = y

∀a, b ∈ ℝ |a + b| ≤ |a| + |b|

|(x - y) + y| ≤ |x - y| + |y|
|x + (-y + y)| ≤ |x - y| + |y|
|x| ≤ |x - y| + |y|

|x| - |y| ≤ |x - y|

Definition

1. If ∃u ∈ ℝ s.t. x ≤ u ∀x ∈ A, We say that u is an upper bound of A. When this happens, A is said to be bounded above. A = (-∞, 5) → A is bounded above but not bounded below.
2. If ∃v ∈ ℝ s.t. v ≤ x ∀x ∈ A, We say that v is a lower bound of A. When this happens, A is said to be bounded below. A = (3, ∞) → A is bounded below but not bounded above.
3. If A is both bounded below & above, it is said to be bounded. A = (3, 5) → A is bounded.

Definition

Let A ≤ ℝ which is bounded above. Suppose ∃u₀ ∈ ℝ with the following properties,

1. u₀ is an upper bound of A.
2. If u is any upper bound of A; u₀ ≤ u

Then we say that u₀ is the Supremum of A & write u₀ = sup(A)

Let A ≤ ℝ which is bounded below. Suppose ∃v₀ ∈ ℝ with the following properties,

1. v₀ is the lower bound of A.
2. If v is any lower bound of A, v ≤ v₀

Then we say that v₀ is the Infimum of A & write v₀ = Inf(A)

Theorem

If a < b, then ∃r ∈ ℝ s.t. a < r < b

We can prove it using r = ^{(a + b)} / ₂

Eg: Let A = [1, 2]
Prove that Sup(A) = 2

∀x ∈ A, x ≤ 2
∴ 2 is an upper bound of A. Let u be any upper bound of A.

Assume u < 2,
by theorem ∃r ∈ ℝ s.t. u < r < 2
1.5 ∈ A ⇒ 1.5 ≤ u
1.5 ≤ u < r < 2

∴ r ∈ A with u < r [Contradiction]
∴ 2 ≤ u
∴ 2 = Sup(A)

Completeness Property - I

Let A ≤ R which is bounded above. Then Sup(A) exists. Proof is omitted. In fact this can be never rigorously proven.

Completeness Property - II

Let A ≤ R which is bounded below. Then Inf(A) exists. Proof is omitted. But this can be rigorously proven by assuming the first one.