- anonymous

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

- jamiebookeater

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- zzr0ck3r

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.

- zzr0ck3r

or circle...

- ikram002p

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

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## More answers

- zzr0ck3r

3, he didnt ask about metrics...

- zzr0ck3r

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.

- zzr0ck3r

I do not know what a spare is

- zzr0ck3r

repost that

- anonymous

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)

- zzr0ck3r

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?

- zzr0ck3r

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

- ganeshie8

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

- zzr0ck3r

how many points have a distance of less than 1?

- zzr0ck3r

not everything :)

- ganeshie8

Ah except the self

- zzr0ck3r

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

- anonymous

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??

- zzr0ck3r

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

- zzr0ck3r

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.

- zzr0ck3r

because the distance from 1 to 1 is 0.

- zzr0ck3r

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

- ganeshie8

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

- zzr0ck3r

I don't know what you mean.

- zzr0ck3r

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

- ganeshie8

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

- ganeshie8

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

- zzr0ck3r

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

- zzr0ck3r

err not a binary operation but from XxX to R.

- ganeshie8

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

- zzr0ck3r

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

- zzr0ck3r

But yes.

- zzr0ck3r

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

- zzr0ck3r

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

- zzr0ck3r

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 :)

- ganeshie8

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

- zzr0ck3r

How would we define a square with a normal metric?

- ganeshie8

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

- zzr0ck3r

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

- zzr0ck3r

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.

- ganeshie8

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

- zzr0ck3r

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

- zzr0ck3r

So most shapes will give everything.

- zzr0ck3r

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

- ganeshie8

Haha that is really weird to visualize! xD

- anonymous

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

- anonymous

https://en.wikipedia.org/wiki/Discrete_space#Definitions
this is because all the points are isolated

- anonymous

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

- anonymous

- anonymous

@oldrin.bataku

- ikram002p

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

- anonymous

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

- ikram002p

where is it ? :D

- anonymous

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

- anonymous

- anonymous

- anonymous

- triciaal

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

- anonymous

- dan815

what is the definition of a ball

- anonymous

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)

- anonymous

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

- anonymous

but did not define a ball

- anonymous

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: $$B(p;1/2)=\{p\}$$

- anonymous

ok, so what nxt

- anonymous

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

- Kainui

@GIL.ojei Explain what you understand.

- anonymous

yes i remember

- anonymous

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 \(d(p,p)=0<1/2\)

- anonymous

yes, i know that

- anonymous

okay, and that's it

- anonymous

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

- anonymous

and @zzr0ck3r

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

HELLO

- anonymous

can some one please answer

- anonymous

hello

- zzr0ck3r

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

- anonymous

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

- zzr0ck3r

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

- anonymous

yes that is 1

- zzr0ck3r

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

- zzr0ck3r

no with the standard metric it would be
0.5

- zzr0ck3r

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

- anonymous

yes, e - neighborhood of 1

- zzr0ck3r

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

- zzr0ck3r

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

- zzr0ck3r

because everything has distance 1

- anonymous

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

- zzr0ck3r

yes exactly

- zzr0ck3r

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

- anonymous

waw

- zzr0ck3r

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

- anonymous

ok

- zzr0ck3r

So now you tell me the answer to this question and I will know you understand
\(B(56, 0.7)=?\)

- anonymous

1 following my definition of d(x,y)

- zzr0ck3r

I am asking for all the points within 0.7 of 56.

- zzr0ck3r

not 1 but 56

- zzr0ck3r

every open ball contains only its center point in this metric

- zzr0ck3r

if the radius is less than 1

- anonymous

ok

- zzr0ck3r

ok what about \(B(1, 3)\)?

- anonymous

1

- zzr0ck3r

no, we want to know all the things with less than distance 3 of 1, and everything has distance 1 from 1.

- zzr0ck3r

1<3, so it contains every point

- zzr0ck3r

If I tell you everything is 0 or 1, and then I ask you what is < 3. you say everything.

- zzr0ck3r

Sorry if I sound condescending in the way I explain things, I am not trying to. :)

- anonymous

no sir, its ok . as far as i understand it. you are great. please continue

- zzr0ck3r

That is about it. Everything has distance 1 from each other. So there will be only two outcomes for open balls
\(B(x, r)=\{x\}\) if \(0

- zzr0ck3r

So what is \(B(a, 3)\)?

- anonymous

waw, am thinking

- anonymous

i don't know because the first condition seems not to be the answer because r>1 so it is not a

- anonymous

is it 3? please don't be angry

- zzr0ck3r

lol I would never get angry.
Ok so the question is this.
What is the set of points that are < 3 distance from a
Everything has distance 1 from a, and 1<3. So everything has distance <3.
So the answer is ?

- anonymous

everything

- anonymous

or 1

- zzr0ck3r

everything.

- anonymous

B(1,5) will be what?

- anonymous

you there?

- zzr0ck3r

everything has distance 1 from 1, we want all the things with distance less than 5
what is the answer?

- anonymous

everything

- zzr0ck3r

correct, if you change the radius to something smaller than 1, you will get only the center, which in this case is 1

- anonymous

so, what do you think are some important important points to note down about the balls?

- zzr0ck3r

If you understand the concept, the rest will follow.

- zzr0ck3r

What class is this for?

- zzr0ck3r

Have you dealt with the \(\epsilon -\delta\) definition of a limit?

- anonymous

finals in my university. i am facing a very big challenge

- anonymous

i have done limit but i don't know if i did indept

- zzr0ck3r

This is a hard topic and most people struggle with it.

- anonymous

- anonymous

so, if i was to see that question in exam, what will be my steps to solving it?

- zzr0ck3r

First what is the ball asking for?

- zzr0ck3r

all the points that are at least distance ? from point ?

- anonymous

point 5 and 1

- anonymous

so what next?

- zzr0ck3r

We want all the points that within \(5\) of \(1\) but everything is distance \(0\) or \(1\) from \(1\)
So EVERYTHING is distance less than \(5\).

- anonymous

so which means that there are three points to note,
if r<1, the point becomes 1 and if x=r, the point becomes zero but if r>1, then the point becomes everything , right?

- anonymous

hello sir

- anonymous

hello

- anonymous

hello

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