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anonymous

  • one year ago

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)

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  1. anonymous
    • one year ago
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    @zzr0ck3r

  2. anonymous
    • one year ago
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    @Loser66

  3. anonymous
    • one year ago
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    @dan815

  4. anonymous
    • one year ago
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    @misty1212

  5. anonymous
    • one year ago
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    PLEASE HELP

  6. anonymous
    • one year ago
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    @Kainui

  7. anonymous
    • one year ago
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    @oldrin.bataku

  8. anonymous
    • one year ago
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    @Hero

  9. anonymous
    • one year ago
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    @ganeshie8

  10. Michele_Laino
    • one year ago
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    sorry, how have you defined Bo(-1) ?

  11. anonymous
    • one year ago
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    don't know.that is how the quation came

  12. zzr0ck3r
    • one year ago
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    Bo(-1)? What is this?

  13. zzr0ck3r
    • one year ago
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    Usually with a ball we have a center and a radius.

  14. zzr0ck3r
    • one year ago
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    sometimes you will see notation \(B_{\delta}(a)\) which means a ball of radius \(\delta\) about \(a\).

  15. zzr0ck3r
    • one year ago
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    You sure it says \(Bo(-1)\)?

  16. anonymous
    • one year ago
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    yes.

  17. zzr0ck3r
    • one year ago
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    why @Michele_Laino ?

  18. Michele_Laino
    • one year ago
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    n that case the answer is the subsequent set: {-1}

  19. Michele_Laino
    • one year ago
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    sorry I have made a typo

  20. Michele_Laino
    • one year ago
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    since we have the discrete metrics

  21. zzr0ck3r
    • one year ago
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    So radius \(0\)?

  22. zzr0ck3r
    • one year ago
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    What is \(Bo(-1)\)?

  23. anonymous
    • one year ago
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    well i have note seen that king of quation before, i only know the one b(x,r) so explain sir

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

  25. zzr0ck3r
    • one year ago
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    Did it say \(B_0(-1)\) or \(Bo(-1)\)???

  26. anonymous
    • one year ago
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    i think B0(-1)

  27. anonymous
    • one year ago
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    B zero(-1)

  28. zzr0ck3r
    • one year ago
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    I think radius must be positive in every definition I have seen.

  29. Michele_Laino
    • one year ago
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    so, it is an open ball whose center is located at x=-1 and whose radius is zero

  30. anonymous
    • one year ago
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    ok but can radius be= 0?

  31. zzr0ck3r
    • one year ago
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    Yeah, that does not make sense in any book I have used. Even wiki has that \(r>0\)/

  32. zzr0ck3r
    • one year ago
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    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.

  33. Michele_Laino
    • one year ago
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    the point x=-1 belongs to B0(-1)

  34. zzr0ck3r
    • one year ago
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    It makes no sense.

  35. zzr0ck3r
    • one year ago
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    An open ball is all the points <r, we cant have <0.

  36. zzr0ck3r
    • one year ago
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    Because a metric is never negative.

  37. Michele_Laino
    • one year ago
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    In my textbook I see that an open ball centerd at x_0 contains x_0

  38. zzr0ck3r
    • one year ago
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    yes, but not if the radius is 0.

  39. zzr0ck3r
    • one year ago
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    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.

  40. Michele_Laino
    • one year ago
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    radius is given by the metric, and we have the discrete metric

  41. zzr0ck3r
    • one year ago
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    lol ok, you agree that given a metric \(\rho\) we have \(B_{\delta}(a):=\{x|\rho(a, x) < \delta\}\)?

  42. zzr0ck3r
    • one year ago
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    Then what you are saying is \(\rho(-1,-1) =0<0\) YIKES!

  43. zzr0ck3r
    • one year ago
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    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.

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

  45. zzr0ck3r
    • one year ago
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    yes

  46. zzr0ck3r
    • one year ago
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    A ball with \(r=0\) makes no sense. I have said exactly nothing about the definition of this metric.

  47. Michele_Laino
    • one year ago
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    so radius=0 means that B0(-1)={-1}

  48. zzr0ck3r
    • one year ago
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    NOOOOOOOOOOOOOOOOOOOOO

  49. zzr0ck3r
    • one year ago
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    that implies \(d(-1,-1)<0\) and it is not!

  50. anonymous
    • one year ago
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    please, my battery is low. i willl look at the solving and solution in the next 30mins.. thank you sirs

  51. zzr0ck3r
    • one year ago
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    http://math.stackexchange.com/questions/787892/open-ball-of-radius-r-0-is-empty

  52. zzr0ck3r
    • one year ago
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    if you are going to accept a zero radius you will ALWAYS get the empty set for ANY ball.

  53. Michele_Laino
    • one year ago
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    let me know, for you what is the discrete topology? @zzr0ck3r

  54. zzr0ck3r
    • one year ago
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    every set it open

  55. zzr0ck3r
    • one year ago
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    that is the definition

  56. Michele_Laino
    • one year ago
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    since discrete topology exists as you know

  57. zzr0ck3r
    • one year ago
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    sure, infinite of them, but that has nothing to do with what we are talking about.

  58. Michele_Laino
    • one year ago
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    we have a discrete topology when we use the discrete metric

  59. zzr0ck3r
    • one year ago
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    Do you agree that this is the definition of a ball???" \(B_r(a)=\{x|d(x,y)<r\}\) Yes or no?

  60. zzr0ck3r
    • one year ago
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    of an open ball on metric \(d\)

  61. Michele_Laino
    • one year ago
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    no, since we have to use \[\Large ...\leqslant r\]

  62. zzr0ck3r
    • one year ago
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    that is a CLOSED ball....

  63. zzr0ck3r
    • one year ago
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    else [3,4] is open in R.

  64. zzr0ck3r
    • one year ago
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    Here is our problem. :) I have said about 3 times we have strict inequality.

  65. Michele_Laino
    • one year ago
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    that is the definition of a neighborhood

  66. zzr0ck3r
    • one year ago
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    there are closed and open neighborhoods man

  67. zzr0ck3r
    • one year ago
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    just google what I am saying, google open ball....

  68. zzr0ck3r
    • one year ago
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    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.

  69. zzr0ck3r
    • one year ago
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    with the standard metric...

  70. zzr0ck3r
    • one year ago
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    or go read the thing I posted. you pick :)

  71. zzr0ck3r
    • one year ago
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    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.

  72. Michele_Laino
    • one year ago
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    [2,5] is a closed set, of course, nevertheless I beleve that using the discrete metric we get a discrete topology

  73. zzr0ck3r
    • one year ago
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    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...

  74. zzr0ck3r
    • one year ago
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    A constant metric will always give the descrete topology.

  75. Michele_Laino
    • one year ago
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    it is not true, since a discrete topology is made by isolated points

  76. zzr0ck3r
    • one year ago
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    what is not true?

  77. Michele_Laino
    • one year ago
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    a costant metric doesn't gives a discrete topology, we have to see how is defined

  78. zzr0ck3r
    • one year ago
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    Not constant out side of d(x,x)=0

  79. zzr0ck3r
    • one year ago
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    \(d(x,y)=a\) 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...

  80. zzr0ck3r
    • one year ago
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    I still don't know your point and the answer \(\{-1\}\) is not right for an open ball. I have proved why.

  81. Michele_Laino
    • one year ago
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    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}

  82. Michele_Laino
    • one year ago
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    namely the discrete topology

  83. zzr0ck3r
    • one year ago
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    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.

  84. zzr0ck3r
    • one year ago
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    please go read the thing I posted, and you can see exactly what I am talking about.

  85. zzr0ck3r
    • one year ago
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    look for this line which would imply 0<0, likewise impossible.

  86. zzr0ck3r
    • one year ago
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    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.

  87. Michele_Laino
    • one year ago
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    your inequality is out of line because the open ball B0(-1) contains x=-1

  88. zzr0ck3r
    • one year ago
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    with radius 0 every closed ball is a single element set that is the center of the ball

  89. Michele_Laino
    • one year ago
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    please look at the definition of open set

  90. zzr0ck3r
    • one year ago
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    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.

  91. Michele_Laino
    • one year ago
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    an open set is a neighborhood of all its points

  92. zzr0ck3r
    • one year ago
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    There is no general definition of an open set man....

  93. zzr0ck3r
    • one year ago
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    im done, you dont know what you are talking about.

  94. Michele_Laino
    • one year ago
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    I have studied two textbook on topology

  95. zzr0ck3r
    • one year ago
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    I took the masters exam 2 weeks ago.

  96. zzr0ck3r
    • one year ago
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    I TA topology and start a phd level computation topology course in 1 month

  97. zzr0ck3r
    • one year ago
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    dude google the definition of an open ball.

  98. zzr0ck3r
    • one year ago
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    please. then tell me what you think

  99. Michele_Laino
    • one year ago
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    no, you are wrong on all line, since you have not the discrete metric on your mind

  100. zzr0ck3r
    • one year ago
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    meterizable topologies are only a small part of topology

  101. zzr0ck3r
    • one year ago
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    dude....omg GOOGLE BRO GOOGLE

  102. Michele_Laino
    • one year ago
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    I don't need of google, since I have my textbooks on topology

  103. zzr0ck3r
    • one year ago
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    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

  104. zzr0ck3r
    • one year ago
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    Dude topology has nothing to do with open balls.

  105. zzr0ck3r
    • one year ago
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    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.

  106. zzr0ck3r
    • one year ago
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    Tell me one book that backs your claim. Just one.

  107. zzr0ck3r
    • one year ago
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    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.

  108. zzr0ck3r
    • one year ago
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    I will take my apology in pm.

  109. zzr0ck3r
    • one year ago
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    goodnight

  110. Michele_Laino
    • one year ago
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    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

  111. Michele_Laino
    • one year ago
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    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

  112. zzr0ck3r
    • one year ago
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    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.

  113. Michele_Laino
    • one year ago
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    WHAT ARE YOU SAYING??? I HAVE NOT BANNED YOU!!!!!!

  114. zzr0ck3r
    • one year ago
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    It is un banned now. I will talk to moderators. It was banned.

  115. Michele_Laino
    • one year ago
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    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

  116. zzr0ck3r
    • one year ago
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    I can't listen to my screen

  117. zzr0ck3r
    • one year ago
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    I enjoyed Rome.

  118. Michele_Laino
    • one year ago
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    ok! read then

  119. Michele_Laino
    • one year ago
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    I don't like Rome

  120. zzr0ck3r
    • one year ago
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    You wrote a bunch of stuff that is wrong, so why would I read more?

  121. Michele_Laino
    • one year ago
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    io non ho iniziato ad insultare!!!

  122. zzr0ck3r
    • one year ago
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    thomaster you are gonna have to push enter soon. I got to sleep :)

  123. Michele_Laino
    • one year ago
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    good go to sleep it is better

  124. zzr0ck3r
    • one year ago
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    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

  125. thomaster
    • one year ago
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    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

  126. zzr0ck3r
    • one year ago
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    @oldrin.bataku

  127. Michele_Laino
    • one year ago
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    great @thomaster as you can see from the previous posts @zzr0ck3r started to insult me

  128. zzr0ck3r
    • one year ago
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    I said you were wrong bro, that is not an insult.

  129. Michele_Laino
    • one year ago
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    no YOU SAID ME THAT i'M DANGEROUS FOR THE SITE

  130. Michele_Laino
    • one year ago
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    PLEASE READ YOUR POSTS

  131. thomaster
    • one year ago
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    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

  132. zzr0ck3r
    • one year ago
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    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.

  133. Michele_Laino
    • one year ago
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    THAT IS A BIG OFFENSE

  134. zzr0ck3r
    • one year ago
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    this site is related.

  135. zzr0ck3r
    • one year ago
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    you are

  136. zzr0ck3r
    • one year ago
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    you also need to look up insult I guess.

  137. Michele_Laino
    • one year ago
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    IF YOU INSULT ME, THENi INSULT YOU @zzr0ck3r

  138. Michele_Laino
    • one year ago
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    NON FARE IL FURBO! TU HAI INIZIATO AD INSULTARMI, TU SEI IL MALEDUCATO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  139. zzr0ck3r
    • one year ago
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    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.

  140. Michele_Laino
    • one year ago
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    SEI UN GRANDE MALEDUCATO, SEI TU CHE DEVI CRESCERE!!!!

  141. thomaster
    • one year ago
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    STOP. you insult me so I insult you? sorry to say but that's plain childish behavior...

  142. zzr0ck3r
    • one year ago
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    Again, I never insulted anyone. Simply read.

  143. Michele_Laino
    • one year ago
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    I defend me @thomaster

  144. zzr0ck3r
    • one year ago
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    @Michele_Laino thinks \(0<0\) and I simply told him it was not.

  145. Michele_Laino
    • one year ago
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    no, you think that 0<0 I never said that

  146. zzr0ck3r
    • one year ago
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    goodnight all especially you @Jhannybean just kicking back and watching/laughing I hope.

  147. zzr0ck3r
    • one year ago
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    you did say that, you just don't understand that you did.

  148. thomaster
    • one year ago
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    @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.

  149. zzr0ck3r
    • one year ago
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    you said it over and over and over.

  150. Michele_Laino
    • one year ago
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    ok! @thomaster nevertheless you have to understand me, I have received a big offense

  151. Jhannybean
    • one year ago
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    @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.

  152. zzr0ck3r
    • one year ago
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    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 :)

  153. zzr0ck3r
    • one year ago
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    The user is fine, I explained it all to him in pm.

  154. zzr0ck3r
    • one year ago
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    And actually if the user could read all of the post and understand why he is wrong, then it would help him much.

  155. thomaster
    • one year ago
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    @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.

  156. zzr0ck3r
    • one year ago
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    ps I see that looks like a fascism joke but it was not.

  157. Jhannybean
    • one year ago
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    It was just a suggestion :P Oh well

  158. Jhannybean
    • one year ago
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    @phi perhaps can help?

  159. zzr0ck3r
    • one year ago
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    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.

  160. Michele_Laino
    • one year ago
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    @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: \[\huge B\left( {{x_0},r} \right) = \left\{ {x \in X:d\left( {x,{x_0}} \right) < r} \right\}\]

  161. Michele_Laino
    • one year ago
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    namely the case d=0 is allowed so the point x_0 belongs to the spherical neighborhood

  162. Michele_Laino
    • one year ago
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    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

  163. Michele_Laino
    • one year ago
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    @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

  164. Michele_Laino
    • one year ago
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    so the person who was wrong is the one who gave you the PhD title

  165. anonymous
    • one year ago
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    thank you all the same but i am yet to get the answer. is it ( ) or (-1) ?

  166. anonymous
    • one year ago
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    @GIL.ojei i would recommend you get a proper text on metric spaces rather than that very poorly composed pdf

  167. anonymous
    • one year ago
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    there are plenty of text books you can get for free using Library Genesis: http://gen.lib.rus.ec/

  168. anonymous
    • one year ago
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    also in the future, please attach an image of your questions, since they are usually written in a difficult to follow manner

  169. anonymous
    • one year ago
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    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.

  170. anonymous
    • one year ago
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    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

  171. anonymous
    • one year ago
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    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 \(B_0(p)=\emptyset\)

  172. anonymous
    • one year ago
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    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.

  173. anonymous
    • one year ago
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    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

  174. anonymous
    • one year ago
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    they were not both correct, though

  175. anonymous
    • one year ago
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    but yes, @GIL.ojei that is true -- although that is not what @Michele_Laino was saying

  176. zzr0ck3r
    • one year ago
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    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.

  177. Michele_Laino
    • one year ago
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    @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

  178. zzr0ck3r
    • one year ago
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    You dont know what you are talking about. \[\Large {B_0}\left( { - 1} \right) = \left\{ { - 1} \right\}\implies 0<0\]

  179. zzr0ck3r
    • one year ago
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    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. '

  180. Michele_Laino
    • one year ago
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    I NEVERSAID THAT 0<0 BECAUSE R CAN NOT BE 0, DO YOU UNDERSTAND MY WORDS, OR YOU NEED FOR A SECOND BRAIN

  181. Michele_Laino
    • one year ago
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    r CAN NOT BE ZERO PLESE STUDY THE DEFINITIONS BEFORE TO ANSWER TO OTHER PEOPLE

  182. Michele_Laino
    • one year ago
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    \[\Huge {B_0}\left( { - 1} \right)\] is A CLOSED BALL!! DO YOU UNDERSTAND IT?

  183. Michele_Laino
    • one year ago
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    GO TO STUDY TOPOLOGY IT IS BETTER FOR ALL

  184. zzr0ck3r
    • one year ago
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    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.

  185. zzr0ck3r
    • one year ago
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    And again, this is not topology. It is metric spaces.

  186. zzr0ck3r
    • one year ago
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    You are a dumb dumb

  187. zzr0ck3r
    • one year ago
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    goodbye

  188. Michele_Laino
    • one year ago
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    agai a metric space is a topological space, please study mathematics, if you are able to do it

  189. Michele_Laino
    • one year ago
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    you are able to offend others, that means your life a very miserable life

  190. zzr0ck3r
    • one year ago
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    Even that is wrong man. Not every metric space is a topological space. lol Every metric space induces a topological space. Dumb dumb

  191. zzr0ck3r
    • one year ago
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    You dont know what your talking about and every comment you add proves it .

  192. zzr0ck3r
    • one year ago
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    lol

  193. Michele_Laino
    • one year ago
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    BUT THE REAL LINE IS A TOPOLOGICAL SPACE IGNORANT!!!!

  194. zzr0ck3r
    • one year ago
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    http://math.stackexchange.com/questions/275614/are-all-metric-spaces-topological-spaces

  195. zzr0ck3r
    • one year ago
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    lol omg dude just stop.

  196. Michele_Laino
    • one year ago
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    as you can see you have not idea about topology, please go back to study it if you are able

  197. zzr0ck3r
    • one year ago
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    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...

  198. Michele_Laino
    • one year ago
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    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

  199. Michele_Laino
    • one year ago
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    go to study it is better for all!!

  200. zzr0ck3r
    • one year ago
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    "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

  201. zzr0ck3r
    • one year ago
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    dumb dumb bad for the site dumb dumb

  202. Michele_Laino
    • one year ago
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    BUT THE REAL LINE IS A TOPOLOGICAL SPACE, DO YOU UNDERSTAN MY WORDS?

  203. zzr0ck3r
    • one year ago
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    so what?

  204. Michele_Laino
    • one year ago
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    IT IS A HAUSDORFF SPACE, PLEASE STUDY IT

  205. zzr0ck3r
    • one year ago
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    and 3< 4, what does that have to do with what we are talking about?

  206. zzr0ck3r
    • one year ago
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    yes it is both T_0, T_1, T_2. So what?

  207. zzr0ck3r
    • one year ago
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    SO WHAT?

  208. zzr0ck3r
    • one year ago
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    what does that have to do with anything man?

  209. Michele_Laino
    • one year ago
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    YOU HAVE NOT STUDY TO SUFFICE, i'M SAID TO SEE THAT IN YOU COUNTRY GIVE THE pHd TITLE TO ALL

  210. zzr0ck3r
    • one year ago
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    answer the one question

  211. zzr0ck3r
    • one year ago
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    I dont have a phd dude.

  212. zzr0ck3r
    • one year ago
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    You change subject when you have nothing to say

  213. zzr0ck3r
    • one year ago
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    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 ?

  214. zzr0ck3r
    • one year ago
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    "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?

  215. zzr0ck3r
    • one year ago
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    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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