Download DEFINITIONS Math 355 1. For a real number x, the absolute value of

DEFINITIONS
Math 355
1. For a real number x, the absolute value of x, denoted by |x|, is defined by
½
|x| =
x,
if x > o
−x, if x ≤ 0
2. The absolute value function has the following properties:
(a) |x| ≥ 0 ∀x ∈ R.
(b) |x| = 0 if and only if x = 0.
(c) For every real number x, | − x| = |x|.
(d) For all real numbers x and y, |xy| = |x||y|.
(e) Let a > 0. Then |x| ≤ a ⇐⇒ −a ≤ x ≤ a and
|x| < a ⇐⇒ −a < x < a.
(f) For every real number x, −|x| ≤ x ≤ |x|.
(g) Triangle Inequality: For all real numbers x and y, |x + y| ≤ |x| + |y|
(h) Reverse Triangle Inequality: For all real numbers x and y, ||x| − |y|| ≤ |x − y|.
3. Let X be a nonempty set and d be a metric on X. Then the ordered pair (X, d) is called a metric
space. The function d is also called distance function or simply distance. Often d is omitted and
one just writes X for a metric space if it is clear from the context what metric is used.
4. For the set R define d : R × R −→ R by d(x, y) = |x − y|. This is called the usual metric on R
and the phrase ”the metric space R” will always refer to this metric space.
p
5. For the set R2 define d : R2 × R2 −→ R by d((x1 , y1 ), (x2 , y2 )) = (x1 − x2 )2 + (y1 − y2 )2 . This
is called the usual metric on R2 and the phrase ”the metric space R2 ” will always refer to this
metric space.
p
6. For the set R3 define d : R3 ×R3 −→ R by d((x1 , y1 , z1 ), (x2 , y2 , z2 )) = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 .
This is called the usual metric on R3 and the phrase ”the metric space R3 ” will always refer to
this metric space.
7. Let X be a nonempty set. Define d : X × X −→ R by
½
d(x, y) =
0
1
if x = y
if x =
6 y
Then d is a metric on X and is called the discrete metric on X and (X,d) is called the discrete metric space.
8. CAUCHY-SCHWARZ INEQUALITY: ∀x, y ∈ Rn |x.y| ≤ |x||y|
9. The Triangle Inequality in Rn : ∀x, y ∈ Rn |x + y| ≤ |x| + |y|.
Let x, y ∈ Rn . Then |x + y|2 = (x + y) · (x + y) = x · x + y · y + 2x · y ≤ |x|2 + |y|2 + 2|x.y| ≤
|x|2 + |y|2 + 2|x||y| = (|x| + |y|)2 . Take the square root of both side to have |x + y| ≤ |x| + |y|.
Verification of (M4 ): If P1 = (x1 ,p
y1 ), P2 = (x2 , y2 ), and P3 = (x3 , y3 ). p
Then |P1 − P2 | =
p
(x1 − x2 )2 + (y1 − y2 )2 , |P1 −P3 | = (x1 − x3 )2 + (y1 − p
y3 )2 and |P2 −P3 | = (x2 − x3 )2 + (y2 − y3 )2 .
2 + (y − y )2 = |P −P | =
Now using the triangle inequality, |x+y| ≤ |x|+|y|, we have (x1 − x3 )p
1
3
1
3
2 + (y − y )2 +
|P
−
P
−
P
|
=
|(P
−
P
)
+
(P
−
P
)|
≤
|P
−
P
|
+
|P
−
P
|
=
(x
−
x
)
2
3
1
2
2
3
1
2
2
3
1
2
1
2
p1
(x2 − x3 )2 + (y2 − y3 )2
NOTE: For this class, we will assume the following theorem:
The sum of the lengths of two sides of a triangle is greater than or equal to the length of the third
side. (Equality holds when the three vertices lie on a line; that is, the triangle is degenerate).