Define metric spaces with examples.
Answers
Answered by
15
here is your answer.
A metric space is a set M with a metric on it. A metric is a function d that satisfies the following properties for all pairs (j,k) of elements in M:
1 d(j,k)=0⟺j=k for all j,k∈M (Distance between the same element is zero)
d(j,k)>0 for all distinct j,k∈M (Distance is always positive)
2 d(j,l)≤d(j,k)+d(k,l) for all j,k,l∈M (Triangle Inequality)
d(j,k)=d(k,j) for all j,k∈M (The distance between two elements is the same)
Technically, the second point follows from the other three like so:
Assume that j≠k
d(j,k)+d(k,j)≥d(j,j) by Property 3
d(j,k)+d(j,k)≥d(j,j) by Property 4
2d(j,k)≥0 by Property 1 and addition
d(j,k)≥0 by Multiplication by 12
Since we assumed at the start that j≠k, then d(j,k)≠0 by Property 1, so d(j,k)>0
A metric space is a set M with a metric on it. A metric is a function d that satisfies the following properties for all pairs (j,k) of elements in M:
1 d(j,k)=0⟺j=k for all j,k∈M (Distance between the same element is zero)
d(j,k)>0 for all distinct j,k∈M (Distance is always positive)
2 d(j,l)≤d(j,k)+d(k,l) for all j,k,l∈M (Triangle Inequality)
d(j,k)=d(k,j) for all j,k∈M (The distance between two elements is the same)
Technically, the second point follows from the other three like so:
Assume that j≠k
d(j,k)+d(k,j)≥d(j,j) by Property 3
d(j,k)+d(j,k)≥d(j,j) by Property 4
2d(j,k)≥0 by Property 1 and addition
d(j,k)≥0 by Multiplication by 12
Since we assumed at the start that j≠k, then d(j,k)≠0 by Property 1, so d(j,k)>0
PrincessNumera:
gr8
Answered by
5
Answer:
Two vectors are orthogonal if the angle between them is 90 degrees. If two vectors are orthogonal, they form a right triangle whose hypotenuse is the sum of the vectors. Thus, we can use the Pythagorean theorem to prove that the dot product xTy = yT x is zero exactly when x and y are orthogonal.
Similar questions