Problems 1.1 | 𝑑(𝑥,𝑦)=√(𝑥1 −𝑦1)^2+(𝑥2−𝑦2)^2 Metric Space Chapter 01 | Functional Analysis Kreyszig

Описание: Problems 1.1 | 𝑑(𝑥,𝑦)=√(𝑥1 −𝑦1)^2+(𝑥2−𝑦2)^2 Metric Space Chapter 01 | Functional Analysis Kreyszig

𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) | Question No 6 | Problems 1.1 | Metric Space | Chapter 01 | Problems 1.1 | Introductory Functional Analysis with Applications | Erwin Kreyszig

𝑑(𝑥,𝑦)=√(|𝑥_1−𝑦_1 |+|𝑥_2−𝑦_2 |

Book Name : Introductory Functional Analysis with Applications

By : Erwin Kreyszig


Chapter Number : 01
Chapter Name : Metric Space

Lecture Number : 8
By (Name) : Awais Rasool

Exercise Number : 1.1
Problems Number: 1.1

Question Number : 08
Part Number : 0
Example Number : 03

Awais Rasool Shah

Topics Name :

Chapter 1. Metric Spaces
1.1 Metric Space
1.2 Further Examples of Metric Spaces
1.3 Open Set, Closed Set, Neighborhood
1.4 Convergence, Cauchy Sequence, Completeness
1.5 Examples. Completeness Proofs
1.6 Completion of Metric Spaces

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Definition (Metric space , Metric).
Consider a non-empty set 𝑥 and a function 𝑑 : 𝑥∗𝑥  𝑅^+ "∪{0}".
This function 𝑑 is called metric on 𝑥 if following conditions are holds:
1) 𝑑(𝑥,𝑦)≥0
2) 𝑑(𝑥,𝑦)=0 𝑖𝑓𝑓 𝑥=𝑦
3) 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) (Symmetry)
4) 𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)
The set X with d is called metric Space and written as (X, 𝑑).



Show that 𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) is a metric on 𝑅^2.
Where 𝑥=(𝑥_1,𝑥_2) and y=(𝑦_1,𝑦_2) ∀𝑥_1,𝑥_2,𝑦_1,𝑦_2∈𝑅.
Sol:
𝑑(𝑥,𝑦)≥0
𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) ≥0
The square of any real number is always non-negative and the sum of a
non-negative number is also non-negative , and the square root of a
non-negative number is also non-negative. so this property is satisfied.
𝑑(𝑥,𝑦)=0 𝑖𝑓𝑓 𝑥=𝑦
𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) =0
This implies that
"⟺" 〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 =0
Since both terms are non-negative, each term must individually be zero:
⟺ 〖〖(𝑥〗_1−𝑦_1)〗^2=0 ⟺ 〖(𝑥_2−𝑦_2)〗^2=0
⟺ 𝑥_1−𝑦_1=0 ⟺ 𝑥_2−𝑦_2=0
⟺ 𝑥_1=𝑦_1 ⟺ 𝑥_2=𝑦_2

𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) (Symmetry)
𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 )
=√(〖〖((−1)(𝑦_1−𝑥〗_1))〗^2+〖〖((−1)(𝑦_2−𝑥〗_2))〗^2 )
=√(〖〖〖(−1)〗^2 (𝑦_1−𝑥〗_1)〗^2+〖〖〖(−1)〗^2 (𝑦_2−𝑥〗_2)〗^2 )
=√(〖〖 (𝑦_1−𝑥〗_1)〗^2+〖〖 (𝑦_2−𝑥〗_2)〗^2 )
=𝑑(𝑦,𝑥)
𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥)
𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)
𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 )
Let 𝑥=(𝑥_1,𝑥_2) , y=(𝑦_1,𝑦_2 ) and 𝑧=(𝑧_1,𝑧_2)

CASE: 1
When three points 𝑥 , y and 𝑧 are collinear.
√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) =√(〖〖(𝑥〗_1−𝑧_1)〗^2+〖(𝑥_2−𝑧_2)〗^2 ) +√(〖〖(𝑧〗_1−𝑦_1)〗^2+〖(𝑧_2−𝑦_2)〗^2 )
𝑑(𝑥,𝑦)=𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)  (i)

CASE: II
When three points 𝑥 , y and 𝑧 are non-collinear.
√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) √(〖〖(𝑥〗_1−𝑧_1)〗^2+〖(𝑥_2−𝑧_2)〗^2 ) +√(〖〖(𝑧〗_1−𝑦_1)〗^2+〖(𝑧_2−𝑦_2)〗^2 )
𝑑(𝑥,𝑦) 𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)  (ii)
For equation (i) and (ii).
𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦)
Thus, 𝑑(𝑥,𝑦)=√(〖〖(𝑥〗_1−𝑦_1)〗^2+〖(𝑥_2−𝑦_2)〗^2 ) satisfies all the properties of a metric, proving that it is indeed a metric on 𝑅^2.