Question

let x=R and p be the discrete metric on X defined by p(x,y) = 8 if x is not equal to y and 0, if x=y. compute Bo(-1)

Answer 1

@zzr0ck3r

Answer 2

@Loser66

Answer 3

@dan815

Answer 4

@misty1212

Answer 5

PLEASE HELP

Answer 6

@Kainui

Answer 7

@oldrin.bataku

Answer 8

@Hero

Answer 9

@ganeshie8

Answer 10

sorry, how have you defined Bo(-1) ?

Answer 11

don't know.that is how the quation came

Answer 12

Bo(-1)? What is this?

Answer 13

Usually with a ball we have a center and a radius.

Answer 14

sometimes you will see notation \(B_{\delta}(a)\) which means a ball of radius \(\delta\) about \(a\).

Answer 15

You sure it says \(Bo(-1)\)?

Answer 16

yes.

Answer 17

why @Michele_Laino ?

Answer 18

n that case the answer is the subsequent set:
{-1}

Answer 19

sorry I have made a typo

Answer 20

since we have the discrete metrics

Answer 21

So radius \(0\)?

Answer 22

What is \(Bo(-1)\)?

Answer 23

well i have note seen that king of quation before, i only know the one b(x,r) so explain sir

Answer 24

what you wrote it is the only possible interpretation of Bo(-1), if we want an open ball centerd atx=-1 with radius 0

Answer 25

Did it say \(B_0(-1)\) or \(Bo(-1)\)???

Answer 26

i think B0(-1)

Answer 27

B zero(-1)

Answer 28

I think radius must be positive in every definition I have seen.

Answer 29

so, it is an open ball whose center is located at x=-1 and whose radius is zero

Answer 30

ok but can radius be= 0?

Answer 31

Yeah, that does not make sense in any book I have used. Even wiki has that \(r>0\)/

Answer 32

Well if the book asked you the question, then that is what they want. Normally subscript is the radius, but all books I use define an open ball to have a positive radius.

Answer 33

the point x=-1 belongs to B0(-1)

Answer 34

It makes no sense.

Answer 35

An open ball is all the points <r, we cant have <0.

Answer 36

Because a metric is never negative.

Answer 37

In my textbook I see that an open ball centerd at x_0 contains x_0

Answer 38

yes, but not if the radius is 0.

Answer 39

because an open ball is defined for all points STRICTLY less than the radius. Keep looking in your book and I bet it defines an open ball with a positive radius, else we broke all of math.

Answer 40

radius is given by the metric, and we have the discrete metric

Answer 41

lol ok, you agree that given a metric \(\rho\) we have \(B_{\delta}(a):=\{x|\rho(a, x) < \delta\}\)?

Answer 42

Then what you are saying is \(\rho(-1,-1) =0<0\) YIKES!

Answer 43

I am saying that for sure any ball contains its center, I am also saying the question asked here with radius 0 does not make sense.

Answer 44

sorry, we have this metric:
\[\begin{gathered}
d\left( {x,y} \right) = 8,\quad x \ne y \hfill \\
d\left( {x,x} \right) = 0 \hfill \\
\end{gathered} \]

Answer 45

yes

Answer 46

A ball with \(r=0\) makes no sense. I have said exactly nothing about the definition of this metric.

Answer 47

so radius=0 means that B0(-1)={-1}

Answer 48

NOOOOOOOOOOOOOOOOOOOOO

Answer 49

that implies \(d(-1,-1)<0\) and it is not!

Answer 50

please, my battery is low. i willl look at the solving and solution in the next 30mins.. thank you sirs

Answer 51

http://math.stackexchange.com/questions/787892/open-ball-of-radius-r-0-is-empty

Answer 52

if you are going to accept a zero radius you will ALWAYS get the empty set for ANY ball.

Answer 53

let me know, for you what is the discrete topology? @zzr0ck3r

Answer 54

every set it open

Answer 55

that is the definition

Answer 56

since discrete topology exists as you know

Answer 57

sure, infinite of them, but that has nothing to do with what we are talking about.

Answer 58

we have a discrete topology when we use the discrete metric

Answer 59

Do you agree that this is the definition of a ball???"

\(B_r(a)=\{x|d(x,y)<r\}\)

Yes or no?

Answer 60

of an open ball on metric \(d\)

Answer 61

no, since we have to use
\[\Large ...\leqslant r\]

Answer 62

that is a CLOSED ball....

Answer 63

else [3,4] is open in R.

Answer 64

Here is our problem. :) I have said about 3 times we have strict inequality.

Answer 65

that is the definition of a neighborhood

Answer 66

there are closed and open neighborhoods man

Answer 67

just google what I am saying, google open ball....

Answer 68

or think of an open ball in R, i.e. an open interval (a,b). By your definition [2,5] is an open ball...of course it is not.

Answer 69

with the standard metric...

Answer 70

or go read the thing I posted. you pick :)

Answer 71

There is no point to allow 0 as a radius on an open or closed ball. if it is open you will ALWAYS get the empty set. If it is a closed ball you will ALLWAYS get the center as a singleton set.

Answer 72

[2,5] is a closed set, of course, nevertheless I beleve that using the discrete metric we get a discrete topology

Answer 73

what does any of this have to do with the descrete topology. yes this generates a topology in which every set is open, and thus the topology is descrete. But that has nothing to do with what you are saying...

Answer 74

A constant metric will always give the descrete topology.

Answer 75

it is not true, since a discrete topology is made by isolated points

Answer 76

what is not true?

Answer 77

a costant metric doesn't gives a discrete topology, we have to see how is defined

Answer 78

Not constant out side of d(x,x)=0

Answer 79

\(d(x,y)=a\) for \(x\ne y\) where \(a>0\) and \(d(x,y) = 0\) when \(x=y\) will always generate the descrete topology.

But again, I have no idea why we are talking about this. There is no reason to even bring in topology to this. The only reason it was said in the question is to let you know the answer...

Answer 80

I still don't know your point and the answer \(\{-1\}\) is not right for an open ball. I have proved why.

Answer 81

my reasoning is very simple, since an open ball centerd at x_0 is a neighborhood of x_0, then x_0 has to belong to that open ball, and using the fact that we have the discrete metric then our openball is made of the point-set {-1}

Answer 82

namely the discrete topology

Answer 83

it is an open neighborhood. you can easily google these definitions man, I am not making them up. That word open makes all of your reasoning wrong.

Answer 84

please go read the thing I posted, and you can see exactly what I am talking about.

Answer 85

look for this line

which would imply 0<0, likewise impossible.

Answer 86

You can pm me if you want some more clarification, but you are simply wrong here. with radius 0 every open ball is the empty set.

Answer 87

your inequality is out of line
because the open ball B0(-1) contains x=-1

Answer 88

with radius 0 every closed ball is a single element set that is the center of the ball

Answer 89

please look at the definition of open set

Answer 90

IT DOES NOT DUDE. Go read a book man. You obviously are not listening and just saying the same thing over and over and it makes no sense.

Answer 91

an open set is a neighborhood of all its points

Answer 92

There is no general definition of an open set man....

Answer 93

im done, you dont know what you are talking about.

Answer 94

I have studied two textbook on topology

Answer 95

I took the masters exam 2 weeks ago.

Answer 96

I TA topology and start a phd level computation topology course in 1 month

Answer 97

dude google the definition of an open ball.