1) Metric Spaces Flashcards
Metric space defn
Definition 1.1 (Metric Space). A metric space (X, d) is a set X together with a function d : X × X → R that satisfies the following properties
(i) d(x, y) ⩾ 0; and d(x, y) = 0 ⇐⇒ x = y (positive definite); (ii) d(x, y) = d(y, x) (symmetric);
(iii) d(x,z)⩽d(x,y)+d(y,z)(triangleinequality).
The function d is called the metric.
Standard distance between two reals
d(x, y) = |x − y|.
Distance between two points on a plane
d(x, y) = d((x_1, x_2), (y_1, y_2)) = √[(x_1 − y_1)^2 + (x_2 − y_2)^2].
x,y vectors
E.g 1.2 standard metric
X = R. The standard metric is given by d1(x, y) = |x − y|.
E.g of metrics
Exponential one -proof
Generic of discrete -proof
d(x,y)=|e^x −e^y|;
d(x, y) =
|x−y| if|x−y|⩽1,
1 if |x − y| ⩾ 1.
Discrete metric
d(x,y) =
{1 ifx̸=y,
{0 ifx=y.
The Euclidean metric in R^n
If X= R^m. The standard metric
x
= (x_1, …,x_m)
y
= (y_1, …,y_m)
Then
d_2(x,y)=sqrt[(x_1 −y_1)^2 +(x_2 −y_2)^2 +…+(x_m −y_m)^2]
Euclidean metric links
linked to the inner-product (scalar product),
x · y = x_1y_1 + x_2y_2 + . . . + x_my_m, since it is just sqrt[(x − y).(x − y)]
d_∞(x,y) in R^m
d∞(x,y)=
max{|x1 −y1|,|x2 −y2|,…,|xm −ym|}.
metric on R^m
d_1
d_1(x,y)=|x1 −y1|+|x2 −y2|+…+|xm −ym|.
d1 , d2 , d∞ are all ____ and ____
d1 , d2 , d∞ are all translation-invariant
d(x + z, y + z) = d(x, y)),
and positively homogeneous (i.e., d(kx, ky) = |k|d(x, y)
Metrics
For X = C[a,b]
d1 , d2 , d∞
d_2 (f, g)
= √ [∫_[a,b] |f(x) − g(x)|^2 dx]].
a
Again, this is linked to the idea of an inner product,
d_1(f, g) =
[∫_[a,b] |f(x) − g(x)|dx]].
the area between two graphs
d∞(f, g) = max{|f(x) − g(x)| : a ⩽ x ⩽ b}, the maximum vertical separation between two graphs.
Example 1.3. On C[0, 1] take
f(x) = x and g(x) = x^2 and calculate
d_1 (f,g)
d1(f, g) = the area between two graphs ∫_[b, a]
|f(x) − g(x)| dx
= [0.5x^2 - (1/3)x^3] on interval [0,1]
= 1/6
Example 1.3. On C[0, 1] take
f(x) = x and g(x) = x^2 and calculate
d_2 (f,g)
d_2 (f, g)
= √ [∫_[a,b] |f(x) − g(x)|^2 dx]].
(x-x^2)^2 = x^2 +x^4 -2x^3
= (1/3)x^3 + 0.2x^5 -0.5x^4
= √(1/30)
Example 1.3. On C[0, 1] take
f(x) = x and g(x) = x^2 and calculate
d_ ∞(f,g)
d∞(f, g) = max{|f(x) − g(x)| : a ⩽ x ⩽ b}, the maximum vertical separation between two graphs.
d∞(f,g) =
max _{x∈ [0,1]} |x−x^2|= 1/4
(* y=x-x^2 , dy /dx= 1-2x has turning point x=0.5, y= 0.25 maximum second deriv =-2 *
1/4.