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)

- anonymous

- chestercat

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## More answers

- anonymous

- anonymous

PLEASE HELP

- anonymous

- anonymous

@oldrin.bataku

- anonymous

- anonymous

- Michele_Laino

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

- anonymous

don't know.that is how the quation came

- zzr0ck3r

Bo(-1)? What is this?

- zzr0ck3r

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

- zzr0ck3r

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

- zzr0ck3r

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

- anonymous

yes.

- zzr0ck3r

why @Michele_Laino ?

- Michele_Laino

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

- Michele_Laino

sorry I have made a typo

- Michele_Laino

since we have the discrete metrics

- zzr0ck3r

So radius \(0\)?

- zzr0ck3r

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

- anonymous

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

- Michele_Laino

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

- zzr0ck3r

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

- anonymous

i think B0(-1)

- anonymous

B zero(-1)

- zzr0ck3r

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

- Michele_Laino

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

- anonymous

ok but can radius be= 0?

- zzr0ck3r

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

- zzr0ck3r

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.

- Michele_Laino

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

- zzr0ck3r

It makes no sense.

- zzr0ck3r

An open ball is all the points

- zzr0ck3r

Because a metric is never negative.

- Michele_Laino

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

- zzr0ck3r

yes, but not if the radius is 0.

- zzr0ck3r

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.

- Michele_Laino

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

- zzr0ck3r

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

- zzr0ck3r

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

- zzr0ck3r

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.

- Michele_Laino

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

- zzr0ck3r

yes

- zzr0ck3r

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

- Michele_Laino

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

- zzr0ck3r

NOOOOOOOOOOOOOOOOOOOOO

- zzr0ck3r

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

- anonymous

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

- zzr0ck3r

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

- zzr0ck3r

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

- Michele_Laino

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

- zzr0ck3r

every set it open

- zzr0ck3r

that is the definition

- Michele_Laino

since discrete topology exists as you know

- zzr0ck3r

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

- Michele_Laino

we have a discrete topology when we use the discrete metric

- zzr0ck3r

Do you agree that this is the definition of a ball???"
\(B_r(a)=\{x|d(x,y)

- zzr0ck3r

of an open ball on metric \(d\)

- Michele_Laino

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

- zzr0ck3r

that is a CLOSED ball....

- zzr0ck3r

else [3,4] is open in R.

- zzr0ck3r

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

- Michele_Laino

that is the definition of a neighborhood

- zzr0ck3r

there are closed and open neighborhoods man

- zzr0ck3r

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

- zzr0ck3r

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.

- zzr0ck3r

with the standard metric...

- zzr0ck3r

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

- zzr0ck3r

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.

- Michele_Laino

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

- zzr0ck3r

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...

- zzr0ck3r

A constant metric will always give the descrete topology.

- Michele_Laino

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

- zzr0ck3r

what is not true?

- Michele_Laino

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

- zzr0ck3r

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

- zzr0ck3r

\(d(x,y)=a\) 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...

- zzr0ck3r

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

- Michele_Laino

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}

- Michele_Laino

namely the discrete topology

- zzr0ck3r

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.

- zzr0ck3r

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

- zzr0ck3r

look for this line
which would imply 0<0, likewise impossible.

- zzr0ck3r

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.

- Michele_Laino

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

- zzr0ck3r

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

- Michele_Laino

please look at the definition of open set

- zzr0ck3r

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.

- Michele_Laino

an open set is a neighborhood of all its points

- zzr0ck3r

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

- zzr0ck3r

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

- Michele_Laino

I have studied two textbook on topology

- zzr0ck3r

I took the masters exam 2 weeks ago.

- zzr0ck3r

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

- zzr0ck3r

dude google the definition of an open ball.

- zzr0ck3r

please. then tell me what you think

- Michele_Laino

no, you are wrong on all line, since you have not the discrete metric on your mind

- zzr0ck3r

meterizable topologies are only a small part of topology

- zzr0ck3r

dude....omg
GOOGLE BRO GOOGLE

- Michele_Laino

I don't need of google, since I have my textbooks on topology

- zzr0ck3r

The open (metric) ball of radius r > 0 centered at a point p in M, usually denoted by Br(p) or B(p; r), is defined by

- zzr0ck3r

Dude topology has nothing to do with open balls.

- zzr0ck3r

you are dangerous to this site imo. You wont go read the things that I am showing you. You wont listen to logic. You say {-1} is open in the descrete topology so it must be {-1} lol/ well the empty set is open, and {1.-2,4} is also open.

- zzr0ck3r

Tell me one book that backs your claim. Just one.

- zzr0ck3r

THIS QUESTION IS ABOUT METRICS ^^^ LOOK AT THE QUESTION
BY THE DEFINITION OF A METRIC \(d(y,y)=0\)
This is the definition in ANY book in the entire WORLD.
You are saying \(d(y,y)<0\) YOU ARE WRONG.
Please msg one person you respect on here and ask them if you are wrong.

- zzr0ck3r

I will take my apology in pm.

- zzr0ck3r

goodnight

- Michele_Laino

dear rude user:
first) I am a theoretical physicist, I have been an external member of the famous Normal High School of Pisa, namely a place where a person like YOU is not able to be admitted
second )
I have an international publication, an article not published, and I got the maximum rating at my degree thesis
third) so, as your little brain can understan I'm not a danger for this site YOU ARE A DANGEROUS AND RUDE USER FOR OPENSTUDY
fourth) you are a complete ignorant in mathematics, so if you want to learn some correct mathematics, I can give you my help for free.
fifth) it is better if you come back to elementary schools, since there IS THE RIGHT PLACE FOR YOU
sixth) Unlike you, I respect all Open Study people.
@zzr0ck3r

- Michele_Laino

so, please measure your words when you speak to an other person
as you can saw when I someone insult me, like you have made with me, I answer in the appropriate way

- zzr0ck3r

You banned my ip and I will bring this to the moderators
Because you wrong
All of that means nothing. You are wrong here. Again, tag anyone here, like me, that has a degree in pure math and they can simply prove you are wrong, like I have. I can write you a formal proof of why you are wrong, but you wont read it/understand it from what I see.
I don't respect someone giving wrong answers, and then claiming they are right. I have proved you wrong.
proof
Suppose \(B_0(-1)=\{-1\}\). Then we have BY THE DEFINITION OF AN OPEN BALL
\[B_0(-1) = \{x|d(-1,x)<0\}=\{-1\}\implies d(-1,1-)=0<0 \]
This contradiction proves the desired result which states that your wrong.
I can get ips all day son.

- Michele_Laino

WHAT ARE YOU SAYING???
I HAVE NOT BANNED YOU!!!!!!

- zzr0ck3r

It is un banned now. I will talk to moderators. It was banned.

- Michele_Laino

listen now I writte in my language: Italian
tu continui ad insultrare, questo vuol dire che sei un maleducato e siccome ti nascondi dietro ai moderatori, significa che sei una persona in malafede, inoltre ti ricordo che sei stato tu ad iniziare ad insultarmi dicendo che sono pericoloso per OpenStudy

- zzr0ck3r

I can't listen to my screen

- zzr0ck3r

I enjoyed Rome.

- Michele_Laino

ok! read then

- Michele_Laino

I don't like Rome

- zzr0ck3r

You wrote a bunch of stuff that is wrong, so why would I read more?

- Michele_Laino

io non ho iniziato ad insultare!!!

- zzr0ck3r

thomaster you are gonna have to push enter soon. I got to sleep :)

- Michele_Laino

good go to sleep it is better

- zzr0ck3r

you don't worry about me man, just go read some stuff. I left you plenty to catch up on, so as you man stop confusing yourself

- thomaster

First: users who do NOT have a purple name can NOT ban anyone's IP.
I do not have knowledge in this subject so I can't tell who's right or wrong, but you should not engage in arguments like this. Always stay respectful to eachother. You won't get anywhere with insults.
I think it's best to get someone who might know who's right or wrong here.
It may take some time for replies since it's nighttime in most of the US.
@Ganeshie8 @Hero @amistre64 @satellite73

- zzr0ck3r

@oldrin.bataku

- Michele_Laino

great @thomaster as you can see from the previous posts @zzr0ck3r started to insult me

- zzr0ck3r

I said you were wrong bro, that is not an insult.

- Michele_Laino

no YOU SAID ME THAT i'M DANGEROUS FOR THE SITE

- Michele_Laino

PLEASE READ YOUR POSTS

- thomaster

It doesn't matter, there's no use in heating up arguments till a point where you'll get angry at eachother. That's not the way to discuss a disagreement

- zzr0ck3r

Yes because you are wrong and don't listen to reason. I think anyone giving wrong answers and not even trying to look at the other views is bad for math and everything related.

- Michele_Laino

THAT IS A BIG OFFENSE

- zzr0ck3r

this site is related.

- zzr0ck3r

you are

- zzr0ck3r

you also need to look up insult I guess.

- Michele_Laino

IF YOU INSULT ME, THENi INSULT YOU @zzr0ck3r

- Michele_Laino

NON FARE IL FURBO! TU HAI INIZIATO AD INSULTARMI, TU SEI IL MALEDUCATO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

- zzr0ck3r

You have not insulted me, you tried, but to no avail. I don't like private high schools with kids who think they know everything but are wrong often. Just not into it man.

- Michele_Laino

SEI UN GRANDE MALEDUCATO, SEI TU CHE DEVI CRESCERE!!!!

- thomaster

STOP.
you insult me so I insult you? sorry to say but that's plain childish behavior...

- zzr0ck3r

Again, I never insulted anyone. Simply read.

- Michele_Laino

I defend me @thomaster

- zzr0ck3r

@Michele_Laino thinks \(0<0\) and I simply told him it was not.

- Michele_Laino

no, you think that 0<0 I never said that

- zzr0ck3r

goodnight all especially you @Jhannybean just kicking back and watching/laughing I hope.

- zzr0ck3r

you did say that, you just don't understand that you did.

- thomaster

@Michele_Laino You can defend yourself without insults, just by proving that you are right. And since you two both can't do that, we need to bring in external expertise.
Now both stop talking and let's wait.

- zzr0ck3r

you said it over and over and over.

- Michele_Laino

ok! @thomaster nevertheless you have to understand me, I have received a big offense

- Jhannybean

@thomaster are you capable of deleting all these posts and starting anew without the OP having to close this thread and create another post? The argument between @Michele_Laino and @zzr0ck3r does not help the user in the least, so rather just leave the post alone if nothing can come out of it.

- zzr0ck3r

Also, just to be very clear. I said "you are bad for this site imo" that means
In My Opinion. Where I am from we are allowed to have those :)

- zzr0ck3r

The user is fine, I explained it all to him in pm.

- zzr0ck3r

And actually if the user could read all of the post and understand why he is wrong, then it would help him much.

- thomaster

@Jhannybean I am capable of that, but that would take a lot of time ;)
Plus I strive to resolve disputes instead of putting them under the carpet.

- zzr0ck3r

ps I see that looks like a fascism joke but it was not.

- Jhannybean

It was just a suggestion :P Oh well

- Jhannybean

@phi perhaps can help?

- zzr0ck3r

pps I feel real stupid for "bragging" that I TA a topology class (which again, topology has nothing to do with this question) and that I am taking phd level courses in computational topology. I only said that because of what was said right before that. I just wanted my position to be known. I hate when people brag....but I thought it was relevant.

- Michele_Laino

@GIL.ojei
here are the right definition: fixed x_0 a point of a metric space or a topological space, and a positive number \[\Huge r > 0\]
then we say the spherical neighborhood, centered at x_0 whose radius is r, the subsequent set:
\[\huge B\left( {{x_0},r} \right) = \left\{ {x \in X:d\left( {x,{x_0}} \right) < r} \right\}\]

- Michele_Laino

namely the case d=0 is allowed
so the point x_0 belongs to the spherical neighborhood

- Michele_Laino

it is another way to see that x_0 belongs to the neighborhood, since we can get the same conclusion using the definitions with common sense, and without computational topology. Of course at x_0 correspond d=0

- Michele_Laino

@zzr0ck3r gave you a wrong or incomplete definition
I didn't brag myself
I see that you continue to not to understand, I didn't brag myself, what I said is only to let understand you that I'm not dangerous @zzr0ck3r
anyway I hate the persons who offend other persons, since that means they have no arguments to make a civil discussion @zzr0ck3r

- Michele_Laino

so the person who was wrong is the one who gave you the PhD title

- anonymous

thank you all the same but i am yet to get the answer. is it ( ) or (-1) ?

- anonymous

@GIL.ojei i would recommend you get a proper text on metric spaces rather than that very poorly composed pdf

- anonymous

there are plenty of text books you can get for free using Library Genesis:
http://gen.lib.rus.ec/

- anonymous

also in the future, please attach an image of your questions, since they are usually written in a difficult to follow manner

- anonymous

also @Michele_Laino I'm sorry to say but you are incorrect; the very definition you've provided stipulates \(r>0\), so the original post was nonsense to start with. if we extend this idea to open balls of radius \(0\) we will find they are a neighborhood for all of its points, sure, but this is a *vacuous truth* as by definition a metric \(d\) must be positive semidefinite, only assuming \(d(x,y)=0\) in the case \(x=y\). this is because \(d\ge0\) and yet the definition of an open ball tells us \(x\in B_0(p)\Longleftrightarrow d(p,x)<0\), which is impossible -- this means \(B_0(p)=\emptyset\). note that this idea of an open ball of radius \(0\) is useless and for most purposes impossible by definition, which is why you'll find on any respectable site on metric spaces that \(r>0\). in the case we were talking about a generalization of *closed* balls, then, sure, we would instead have \(x\in\bar B_0(p)\Longleftrightarrow d(p,x)\le 0\), which would then be satisfied by \(p\) and \(p\) alone, giving the singleton \(\{p\}\).
as a side note, in the future, if I were you I would avoid bragging about credentials and take criticism in a less defensive fashion just to be on the safe side.

- anonymous

note that a lot of these types of debates stem from poorly posed questions, which is why I suggest that for future questions @GIL.ojei provide exact definitions and images of the problems from his text to avoid situations like these

- anonymous

note the claim that \(B_r(p)\) for a non-degenerate open ball (i.e. the standard definition of an open ball, with \(\color{red}{r>0}\)) is an open neighborhood of \(p\) is trivially true, since \(B_r(p)\subseteq B_r(p)\) is open by definition. but this does not hold for extending the definition to \(r=0\), since \(B_0(p)=\emptyset\)

- anonymous

thank you very much. in my own opinion, they are both correct AND ALL WHAT THEY DID WAS A debate . and we have all learnt something today. @Michele_Laino thank you and @zzr0ck3r , thank you too. you both are genius.

- anonymous

so i learnt that in open ball, if r equal to 0, then it is the empty set from @zzr0ck3r and if r=0 in a closed ball, it is = (x) from @Michele_Laino

- anonymous

they were not both correct, though

- anonymous

but yes, @GIL.ojei that is true -- although that is not what @Michele_Laino was saying

- zzr0ck3r

Exactly @GIL.ojei There is no point at all in defining it for r=0 in either case; you will always get the same thing.

- Michele_Laino

@oldrin.bataku the claim "you are dangerous for the site" is not a criticism, it is a big offense, so I shown my credentials to your friends @zzr0ck3r in order to give him the possibility to understand that I'm not dangerous, so I have not bragged myself.
On the other hand the question is I have never said 0<0 as your friend said.
What I said is:
The metric space is a Hausdorff space, so we have not an open ball, namely B0(-1) is a closed ball, so, in other words, the answer is:
\[\Large {B_0}\left( { - 1} \right) = \left\{ { - 1} \right\}\]
infact we can show, that in a Hausdorff space the one point sets are closed sets. That's I try to say yesterday, evidently your friend @zzr0ck3r was not able to understand my words

- zzr0ck3r

You dont know what you are talking about.
\[\Large {B_0}\left( { - 1} \right) = \left\{ { - 1} \right\}\implies 0<0\]

- zzr0ck3r

you actually just wrote 0<0 in your response. You said
I did not write 0<0, I wrote 0<0. You just dont understand that you said that. '

- Michele_Laino

I NEVERSAID THAT 0<0 BECAUSE R CAN NOT BE 0, DO YOU UNDERSTAND MY WORDS, OR YOU NEED FOR A SECOND BRAIN

- Michele_Laino

r CAN NOT BE ZERO PLESE STUDY THE DEFINITIONS BEFORE TO ANSWER TO OTHER PEOPLE

- Michele_Laino

\[\Huge {B_0}\left( { - 1} \right)\] is A CLOSED BALL!! DO YOU UNDERSTAND IT?

- Michele_Laino

GO TO STUDY TOPOLOGY IT IS BETTER FOR ALL

- zzr0ck3r

Listen man
\(\Large {B_0}\left( { - 1} \right) = \left\{ { - 1} \right\}\)
you wrote that. Now follow this and only respond to this
What you are saying, is that \(d(-1,-1)<0\) and by definition of a mertric \(d(-1,-1) = 0\)
So put that all together and what you have said is
\[0=d(-1,-1)<0\]
Now unless you respond with math, I will not read anything else from you.

- zzr0ck3r

And again, this is not topology. It is metric spaces.

- zzr0ck3r

You are a dumb dumb

- zzr0ck3r

goodbye

- Michele_Laino

agai a metric space is a topological space, please study mathematics, if you are able to do it

- Michele_Laino

you are able to offend others, that means your life a very miserable life

- zzr0ck3r

Even that is wrong man. Not every metric space is a topological space. lol
Every metric space induces a topological space.
Dumb dumb

- zzr0ck3r

You dont know what your talking about and every comment you add proves it .

- zzr0ck3r

lol

- Michele_Laino

BUT THE REAL LINE IS A TOPOLOGICAL SPACE IGNORANT!!!!

- zzr0ck3r

http://math.stackexchange.com/questions/275614/are-all-metric-spaces-topological-spaces

- zzr0ck3r

lol omg dude just stop.

- Michele_Laino

as you can see you have not idea about topology, please go back to study it if you are able

- zzr0ck3r

Show me how anything you say relates to the topic. Or can you even tell me what I claim you are doing wrong? You cant because you dont listen...

- Michele_Laino

please go to read this book, I have studied it, there you will find al notion that you don't know:
"INTRODUCTION TO TOPOLOGICAL SPACE"
by J.M.Lee, Sprionger

- Michele_Laino

go to study it is better for all!!

- zzr0ck3r

"as you can see you have not idea about topology, " -Michele_Laino
"agai a metric space is a topological space, please study mathematics," -Michele_Laino
http://math.stackexchange.com/questions/275614/are-all-metric-spaces-topological-spaces

- zzr0ck3r

dumb dumb bad for the site dumb dumb

- Michele_Laino

BUT THE REAL LINE IS A TOPOLOGICAL SPACE, DO YOU UNDERSTAN MY WORDS?

- zzr0ck3r

so what?

- Michele_Laino

IT IS A HAUSDORFF SPACE, PLEASE STUDY IT

- zzr0ck3r

and 3< 4, what does that have to do with what we are talking about?

- zzr0ck3r

yes it is both T_0, T_1, T_2. So what?

- zzr0ck3r

SO WHAT?

- zzr0ck3r

what does that have to do with anything man?

- Michele_Laino

YOU HAVE NOT STUDY TO SUFFICE, i'M SAID TO SEE THAT IN YOU COUNTRY GIVE THE pHd TITLE TO ALL

- zzr0ck3r

answer the one question

- zzr0ck3r

I dont have a phd dude.

- zzr0ck3r

You change subject when you have nothing to say

- zzr0ck3r

B"UT THE REAL LINE IS A TOPOLOGICAL SPACE, DO YOU UNDERSTAN MY WORDS?"
WHAT DOES THIS HAVE TO DO WITH ANYTHING WE ARE TALKING ABOUT ?

- zzr0ck3r

"BUT THE REAL LINE IS A TOPOLOGICAL SPACE, DO YOU UNDERSTAN MY WORDS?"
What does this have to do with what we are talking about. If you know anything then you should at least know what you yourself are talking about. So answer the question.
What does this have to do with what we are talking about?

- zzr0ck3r

OK man I have had enough. You clearly are not a dumb person but you are simply wrong here. You think that an open ball and a closed ball are the same thing, or you think that an open ball does not have strict inequality, which would lead to a one element set. You keep saying something about topologies, yes in the descrete topology all one point sets are open but this does not rule out the fact that there unions and finite intersections are also open by the definition of a topology but you don't seem to understand that only open sets are in a topology by definition(a topology is the set of all open sets).
So you keep saying it must be a one element set, because one element sets are open and you refer to an open ball as a neighborhood, and since you only relate a ball to a neighbor hood, and the fact that you think only one point sets are open are leading you to the wrong conclusion. If you can argue why I am wrong please do. But don't reply with more talk about that it is a topology, we know it is and that does not matter at all here.
If you would like to show me why I am wrong, not just by repeating the word topology, then I would love to listen.
As for your repeated insults about me and topology. I love topology, it is what I study most outside of abstract algebra, and I am soon to mix the two. I have taken 2 terms of intro analysis, and then a full year of analysis, and another full year of topology.
Again, if you write a proof I will respond, else we are done here.

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