Question

please can someone explain what open balls, closed balls and spares are in a metric space

Answer 1

The best way to think about it, so that it makes sense with the name, is with the standard metric on \(\mathbb{R}^2\). Give me a radius \(\delta\) and some point \(a\) and the open ball about \(a\) is all the points within \(\delta\) of \(a\). So its like you surround \(a\) with an open ball.

Answer 2

or circle...

Answer 3

They are two important conditions in metric I think we should mention them right ?

Answer 4

3, he didnt ask about metrics...

Answer 5

The concept stays the same through different metrics but they no longer match the name. The taxi cab metric will gave an open ball that looks like a diamond, and the infinity metric a square....

Also note that the actual ball is not the border, its all the stuff inside(for an open ball). So if you take a basket ball and fill it with air, then the open ball is the air(in R^3 with euclid metric). I hope that makes sense.

Answer 6

I do not know what a spare is

Answer 7

repost that

Answer 8

let E =R endowed with the metric Do defined by
Do(x,y)= 1, if x is not y and 0 if x =y for arbitrary x,y element of R . compute the ball B(1;1/2)

Answer 9

well that says that everything but 1, has a distance of 1 from 1. And we want to know about the points within 1/2 of 1. So there is only one. what is it?

Answer 10

Do(1,1.4)=1
Do(1,1.2)=1
Do(1, 0.8)=1
Do(1,1)=0
Do(1,0.6)=1

Answer 11

lol it is a weird metric, everything is at 1 unit away from everything

Answer 12

how many points have a distance of less than 1?

Answer 13

not everything :)

Answer 14

Ah except the self

Answer 15

Right, so here is a metric that gives an open ball that is a singleton. Does this make sense? @GIL.ojei ?

Answer 16

So Sir , what is the question asking us to find and how did u get all those point s and equate them to 1 and how was tour conclusion made??

Answer 17

\(B_{Do}(1;1/2)=\{x\in R \mid Do(1,x)<\frac{1}{2}\}=\{1\}\)

Answer 18

The question wants the set of all points that are within distance 1/2 of 1. But with this metric, everything, except 1, is distance 1 from 1. So the only point in the set is 1 itself.

Answer 19

because the distance from 1 to 1 is 0.

Answer 20

if this was the euclidean metric we would have the interval (0.5, 1.5)

Answer 21

i think "n" points require "n-1" dimensions for this metric to be valid/used

Answer 22

I don't know what you mean.

Answer 23

It passes all the rules of a metric on a set.

Answer 24

at least in euclidean metric in \(\mathbb{R}^n\)...
if we have 3 points, then they can be at 1 unit away from each other only if they are at corners of an equilateral triangle - two dimensions

Answer 25

similarly if we have 4 points, we must go to 3-space where the points can be at vertices of a tetrahedron or something .. its hard to visualize for more points idk lol

Answer 26

A metric is a binary operation on the set. It takes only two elements as an argument .

Answer 27

err not a binary operation but from XxX to R.

Answer 28

XxX to R is a binary operation which takes two operands as input and spits out one real number as output right

Answer 29

I think a binary operation on X has to have X itself as the codomain

Answer 30

But yes.

Answer 31

\(\circ : X \times X \rightarrow X\)

Answer 32

That's a binary operation... but anyway.

Answer 33

But I think I see what you are trying to do and that is think about shapes with this metric. I am not willing to take that jump tonight :)

Answer 34

Exactly! I am trying to visualize, which is forbidden sometimes in real analysis haha!

Answer 35

How would we define a square with a normal metric?

Answer 36

I see your point, taxicab metric works well i think ?

Answer 37

I want to think about a square in this metric with that definition.

Answer 38

Well not even that. I am just saying how ever we define a square with the normal distance function on R^2, lets use that definition on this metric and try and think of what a square looks like.

Answer 39

there are only two possible values for distances here : {0, 1}

Answer 40

So a square with side length 1, lets say the unit square and look at the point (0,1/2)

normally we would have all the points that are 1 unit away in one direction and we get only one point (1,1/2)

But with this metric we get EVERYTHING... lol

Answer 41

So most shapes will give everything.

Answer 42

I think the only purpose of this metric is to ask this question :) ok 5am good night

Answer 43

Haha that is really weird to visualize! xD

Answer 44

the metric is just the one that induces the discrete topology, so balls of radius \(r<1\) only contain one point: \(B_{r\,<\,1}(p)=\{p\}\)

Answer 45

https://en.wikipedia.org/wiki/Discrete_space#Definitions

this is because all the points are isolated

Answer 46

please guys, you have been arguing and i do not understand one bit please, what are the steps in solving the quation i gave and what would be the final answer

Answer 47

@ikram002p

Answer 48

@oldrin.bataku

Answer 49

i need to see the definition in your book of metric space in your book + which class is this
to help you more ^_^

Answer 50

i am in my finals in national open university. here is a link to the book

Answer 51

where is it ? :D

Answer 52

http://www.nou.edu.ng/uploads/NOUN_OCL/pdf/SST/MTH%20401.pdf

Answer 53

@zzr0ck3r

Answer 54

@triciaal

Answer 55

@dan815

Answer 56

sorry can't help on this I don't know

Answer 57

let E =R endowed with the metric Do defined by Do(x,y)= 1, if x is not y and 0 if x =y for arbitrary x,y element of R . compute the ball B(1;1/2)

Answer 58

what is the definition of a ball

Answer 59

the open ball of radius \(r\) about \(p\) in a metric space \((X,d)\) is defined as $$B(p;r)=\{x\in X:d(p,x)<r\}$$

Answer 60

yes
that was the definition in my book, they gave just open balls, closed balls and sphare

Answer 61

but did not define a ball

Answer 62

in this case, it doesn't matter whether they ask for open or closed balls, because there are no points other than \(p\) that are up to or within distance \(1/2\) of \(p\), so the ball in our discrete metric is a singleton: $$B(p;1/2)=\{p\}$$

Answer 63

ok, so what nxt

Answer 64

remember the definition of the metric here: $$d(x,y)=\left\{\begin{matrix}0&\text{if }x=y\\1&\text{if }x\ne y \end{matrix}\right.$$

Answer 65

@GIL.ojei Explain what you understand.

Answer 66

yes i remember

Answer 67

for example, suppose our space consisted of the following points \(p,q,r\). we know: $$d(p,p)=0\\d(p,q)=1\\d(p,r)=1$$ so the only thing within a distance of \(1/2\) is \(p\), since \(d(p,q)=d(p,r)=1>1/2\) and \(d(p,p)=0<1/2\)

Answer 68

yes, i know that

Answer 69

okay, and that's it

Answer 70

that's the problem you asked about, @ganeshie8 answered it hours ago

Answer 71

and @zzr0ck3r

Answer 72

so, how did he get does points like d(1;0.8)=1 I MEAN THE 0.8

Answer 73

OK, WHAT ABOUT THE COMPUTATION OF THIS AND SOLVE COMPLETELY WITH STEPS, PLEASE

Answer 74

OK, WHAT ABOUT THE COMPUTATION OF THIS AND SOLVE COMPLETELY WITH STEPS, PLEASE

Answer 75

let E =R endowed with the metric Do defined by Do(x,y)= 1, if x is not y and 0 if x =y for arbitrary x,y element of R . compute the ball B(1;5)

Answer 76

HELLO

Answer 77

can some one please answer

Answer 78

hello

Answer 79

You need to tell us when you don't understand us. We were not arguing we were discussing math. What do you not understand?

Answer 80

does it mean that hat they told us to do is to find points from 1 to <5?

Answer 81

Do you understand that \(B(1, \frac{1}{2})\) is the set of all points that are within one half of 1?

Answer 82

yes that is 1<x<1/2 right?

Answer 83

So with the normal metric we would get stuff like 0.6,0.7,0.8,0.9,1.1,1.2,1.3

Answer 84

no with the standard metric it would be

0.5<x<1.5

Answer 85

Do you see that? \(1\pm0.5\)

Answer 86

yes, e - neighborhood of 1

Answer 87

So this is with the standard metric, but we are not in that metric. In this metric distance works differently than you are used to.

The distance between any two different points is 1

Answer 88

So now there are no points within 1/2 of 1 (except 1 itself)

Answer 89

because everything has distance 1

Answer 90

So now there are no points within 1/2 of 1 (except 1 itself) ,, please give mare example on it

Answer 91

yes exactly

Answer 92

the distance between 1 and 3 is 1
the distance between 1 and 7 is 1
the distance between 1 and 900000000000 is 1

Answer 93

waw

Answer 94

So when you ask me what are all the points within 1/2 of 1, I tell you there is only 1 and that is 1 itself

Answer 95

ok

Answer 96

So now you tell me the answer to this question and I will know you understand

\(B(56, 0.7)=?\)

Answer 97

1 following my definition of d(x,y)