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anonymous

  • one year ago

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

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  1. zzr0ck3r
    • one year ago
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    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.

  2. zzr0ck3r
    • one year ago
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    or circle...

  3. ikram002p
    • one year ago
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    They are two important conditions in metric I think we should mention them right ?

  4. zzr0ck3r
    • one year ago
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    3, he didnt ask about metrics...

  5. zzr0ck3r
    • one year ago
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    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.

  6. zzr0ck3r
    • one year ago
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    I do not know what a spare is

  7. zzr0ck3r
    • one year ago
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    repost that

  8. anonymous
    • one year ago
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    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)

  9. zzr0ck3r
    • one year ago
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    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?

  10. zzr0ck3r
    • one year ago
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    Do(1,1.4)=1 Do(1,1.2)=1 Do(1, 0.8)=1 Do(1,1)=0 Do(1,0.6)=1

  11. ganeshie8
    • one year ago
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    lol it is a weird metric, everything is at 1 unit away from everything

  12. zzr0ck3r
    • one year ago
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    how many points have a distance of less than 1?

  13. zzr0ck3r
    • one year ago
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    not everything :)

  14. ganeshie8
    • one year ago
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    Ah except the self

  15. zzr0ck3r
    • one year ago
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    Right, so here is a metric that gives an open ball that is a singleton. Does this make sense? @GIL.ojei ?

  16. anonymous
    • one year ago
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    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??

  17. zzr0ck3r
    • one year ago
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    \(B_{Do}(1;1/2)=\{x\in R \mid Do(1,x)<\frac{1}{2}\}=\{1\}\)

  18. zzr0ck3r
    • one year ago
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    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.

  19. zzr0ck3r
    • one year ago
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    because the distance from 1 to 1 is 0.

  20. zzr0ck3r
    • one year ago
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    if this was the euclidean metric we would have the interval (0.5, 1.5)

  21. ganeshie8
    • one year ago
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    i think "n" points require "n-1" dimensions for this metric to be valid/used

  22. zzr0ck3r
    • one year ago
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    I don't know what you mean.

  23. zzr0ck3r
    • one year ago
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    It passes all the rules of a metric on a set.

  24. ganeshie8
    • one year ago
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    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

  25. ganeshie8
    • one year ago
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    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

  26. zzr0ck3r
    • one year ago
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    A metric is a binary operation on the set. It takes only two elements as an argument .

  27. zzr0ck3r
    • one year ago
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    err not a binary operation but from XxX to R.

  28. ganeshie8
    • one year ago
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    XxX to R is a binary operation which takes two operands as input and spits out one real number as output right

  29. zzr0ck3r
    • one year ago
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    I think a binary operation on X has to have X itself as the codomain

  30. zzr0ck3r
    • one year ago
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    But yes.

  31. zzr0ck3r
    • one year ago
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    \(\circ : X \times X \rightarrow X\)

  32. zzr0ck3r
    • one year ago
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    That's a binary operation... but anyway.

  33. zzr0ck3r
    • one year ago
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    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 :)

  34. ganeshie8
    • one year ago
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    Exactly! I am trying to visualize, which is forbidden sometimes in real analysis haha!

  35. zzr0ck3r
    • one year ago
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    How would we define a square with a normal metric?

  36. ganeshie8
    • one year ago
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    I see your point, taxicab metric works well i think ?

  37. zzr0ck3r
    • one year ago
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    I want to think about a square in this metric with that definition.

  38. zzr0ck3r
    • one year ago
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    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.

  39. ganeshie8
    • one year ago
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    there are only two possible values for distances here : {0, 1}

  40. zzr0ck3r
    • one year ago
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    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

  41. zzr0ck3r
    • one year ago
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    So most shapes will give everything.

  42. zzr0ck3r
    • one year ago
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    I think the only purpose of this metric is to ask this question :) ok 5am good night

  43. ganeshie8
    • one year ago
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    Haha that is really weird to visualize! xD

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

  45. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Discrete_space#Definitions this is because all the points are isolated

  46. anonymous
    • one year ago
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    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

  47. anonymous
    • one year ago
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    @ikram002p

  48. anonymous
    • one year ago
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    @oldrin.bataku

  49. ikram002p
    • one year ago
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    i need to see the definition in your book of metric space in your book + which class is this to help you more ^_^

  50. anonymous
    • one year ago
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    i am in my finals in national open university. here is a link to the book

  51. ikram002p
    • one year ago
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    where is it ? :D

  52. anonymous
    • one year ago
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    http://www.nou.edu.ng/uploads/NOUN_OCL/pdf/SST/MTH%20401.pdf

  53. anonymous
    • one year ago
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    @zzr0ck3r

  54. anonymous
    • one year ago
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    @triciaal

  55. anonymous
    • one year ago
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    @dan815

  56. triciaal
    • one year ago
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    sorry can't help on this I don't know

  57. anonymous
    • one year ago
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    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)

  58. dan815
    • one year ago
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    what is the definition of a ball

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

  60. anonymous
    • one year ago
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    yes that was the definition in my book, they gave just open balls, closed balls and sphare

  61. anonymous
    • one year ago
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    but did not define a ball

  62. anonymous
    • one year ago
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    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: $$B(p;1/2)=\{p\}$$

  63. anonymous
    • one year ago
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    ok, so what nxt

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

  65. Kainui
    • one year ago
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    @GIL.ojei Explain what you understand.

  66. anonymous
    • one year ago
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    yes i remember

  67. anonymous
    • one year ago
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    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 \(d(p,p)=0<1/2\)

  68. anonymous
    • one year ago
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    yes, i know that

  69. anonymous
    • one year ago
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    okay, and that's it

  70. anonymous
    • one year ago
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    that's the problem you asked about, @ganeshie8 answered it hours ago

  71. anonymous
    • one year ago
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    and @zzr0ck3r

  72. anonymous
    • one year ago
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    so, how did he get does points like d(1;0.8)=1 I MEAN THE 0.8

  73. anonymous
    • one year ago
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    OK, WHAT ABOUT THE COMPUTATION OF THIS AND SOLVE COMPLETELY WITH STEPS, PLEASE

  74. anonymous
    • one year ago
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    OK, WHAT ABOUT THE COMPUTATION OF THIS AND SOLVE COMPLETELY WITH STEPS, PLEASE

  75. anonymous
    • one year ago
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    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)

  76. anonymous
    • one year ago
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    HELLO

  77. anonymous
    • one year ago
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    can some one please answer

  78. anonymous
    • one year ago
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    hello

  79. zzr0ck3r
    • one year ago
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    You need to tell us when you don't understand us. We were not arguing we were discussing math. What do you not understand?

  80. anonymous
    • one year ago
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    does it mean that hat they told us to do is to find points from 1 to <5?

  81. zzr0ck3r
    • one year ago
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    Do you understand that \(B(1, \frac{1}{2})\) is the set of all points that are within one half of 1?

  82. anonymous
    • one year ago
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    yes that is 1<x<1/2 right?

  83. zzr0ck3r
    • one year ago
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    So with the normal metric we would get stuff like 0.6,0.7,0.8,0.9,1.1,1.2,1.3

  84. zzr0ck3r
    • one year ago
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    no with the standard metric it would be 0.5<x<1.5

  85. zzr0ck3r
    • one year ago
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    Do you see that? \(1\pm0.5\)

  86. anonymous
    • one year ago
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    yes, e - neighborhood of 1

  87. zzr0ck3r
    • one year ago
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    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

  88. zzr0ck3r
    • one year ago
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    So now there are no points within 1/2 of 1 (except 1 itself)

  89. zzr0ck3r
    • one year ago
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    because everything has distance 1

  90. anonymous
    • one year ago
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    So now there are no points within 1/2 of 1 (except 1 itself) ,, please give mare example on it

  91. zzr0ck3r
    • one year ago
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    yes exactly

  92. zzr0ck3r
    • one year ago
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    the distance between 1 and 3 is 1 the distance between 1 and 7 is 1 the distance between 1 and 900000000000 is 1

  93. anonymous
    • one year ago
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    waw

  94. zzr0ck3r
    • one year ago
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    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

  95. anonymous
    • one year ago
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    ok

  96. zzr0ck3r
    • one year ago
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    So now you tell me the answer to this question and I will know you understand \(B(56, 0.7)=?\)

  97. anonymous
    • one year ago
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    1 following my definition of d(x,y)

  98. zzr0ck3r
    • one year ago
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    I am asking for all the points within 0.7 of 56.

  99. zzr0ck3r
    • one year ago
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    not 1 but 56

  100. zzr0ck3r
    • one year ago
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    every open ball contains only its center point in this metric

  101. zzr0ck3r
    • one year ago
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    if the radius is less than 1

  102. anonymous
    • one year ago
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    ok

  103. zzr0ck3r
    • one year ago
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    ok what about \(B(1, 3)\)?

  104. anonymous
    • one year ago
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    1

  105. zzr0ck3r
    • one year ago
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    no, we want to know all the things with less than distance 3 of 1, and everything has distance 1 from 1.

  106. zzr0ck3r
    • one year ago
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    1<3, so it contains every point

  107. zzr0ck3r
    • one year ago
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    If I tell you everything is 0 or 1, and then I ask you what is < 3. you say everything.

  108. zzr0ck3r
    • one year ago
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    Sorry if I sound condescending in the way I explain things, I am not trying to. :)

  109. anonymous
    • one year ago
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    no sir, its ok . as far as i understand it. you are great. please continue

  110. zzr0ck3r
    • one year ago
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    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 \(0<r<1\) \(B(x, r)\) if \(r\ge 1\)

  111. zzr0ck3r
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    So what is \(B(a, 3)\)?

  112. anonymous
    • one year ago
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    waw, am thinking

  113. anonymous
    • one year ago
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    i don't know because the first condition seems not to be the answer because r>1 so it is not a

  114. anonymous
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    is it 3? please don't be angry

  115. zzr0ck3r
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    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 ?

  116. anonymous
    • one year ago
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    everything

  117. anonymous
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    or 1

  118. zzr0ck3r
    • one year ago
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    everything.

  119. anonymous
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    B(1,5) will be what?

  120. anonymous
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    you there?

  121. zzr0ck3r
    • one year ago
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    everything has distance 1 from 1, we want all the things with distance less than 5 what is the answer?

  122. anonymous
    • one year ago
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    everything

  123. zzr0ck3r
    • one year ago
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    correct, if you change the radius to something smaller than 1, you will get only the center, which in this case is 1

  124. anonymous
    • one year ago
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    so, what do you think are some important important points to note down about the balls?

  125. zzr0ck3r
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    If you understand the concept, the rest will follow.

  126. zzr0ck3r
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    What class is this for?

  127. zzr0ck3r
    • one year ago
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    Have you dealt with the \(\epsilon -\delta\) definition of a limit?

  128. anonymous
    • one year ago
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    finals in my university. i am facing a very big challenge

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

  130. zzr0ck3r
    • one year ago
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    This is a hard topic and most people struggle with it.

  131. anonymous
    • one year ago
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    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)

  132. anonymous
    • one year ago
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    so, if i was to see that question in exam, what will be my steps to solving it?

  133. zzr0ck3r
    • one year ago
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    First what is the ball asking for?

  134. zzr0ck3r
    • one year ago
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    all the points that are at least distance ? from point ?

  135. anonymous
    • one year ago
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    point 5 and 1

  136. anonymous
    • one year ago
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    so what next?

  137. zzr0ck3r
    • one year ago
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    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 \(5\).

  138. anonymous
    • one year ago
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    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?

  139. anonymous
    • one year ago
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    hello sir

  140. anonymous
    • one year ago
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    hello

  141. anonymous
    • one year ago
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    hello

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