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.

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