### 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

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

2, 3 ∈ A & 2 < 2.5 < 3 & 2.5 ∉ A

*Therefore A is not an interval*### Defenition

Let a < b

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

- (a, ∞) = {x | x ∈ ℝ & a < x}
- [a, ∞) = {x | x ∈ ℝ & a ≤ x}
- (-∞, a) = {x | x ∈ ℝ & a > x}
- (-∞, 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

- |x| < a ⇔ -a < x < a
- |x| ≤ a ⇔ -a ≤ x ≤ a

Now we can say;

{x | |x| < 5} = {x | -5 < x < 5} = (-5, 5)

{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.

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|

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|

∀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

- 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. - 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. - 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 ∃u

_{0}∈ ℝ with the following properties,- u
_{0}is an upper bound of A. - If u is any upper bound of A; u
_{0}≤ u

Then we say that u

_{0}is the**Supremum of A**& write u_{0}= sup(A)
Let A ≤ ℝ which is bounded below. Suppose ∃v

_{0}∈ ℝ with the following properties,- v
_{0}is the lower bound of A. - If v is any lower bound of A, v ≤ v
_{0}

Then we say that v

_{0}is the**Infimum of A**& write v_{0}= Inf(A)### Theorem

If a < b, then ∃r ∈ ℝ s.t. a < r < b

We can prove it using r =

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)

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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