A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

please help @zzrock3r

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Dully noted, there is no question here.

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmmm -.- what did you do to the question! put your pencils up!

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Okay so that is password protected.

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whats the password?!

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    help us. help you!

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i need helo on those questions 2,3,4 and the password is osho

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ow! my head HuRtS! i dont know.. sorry

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @pooja195

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r

  13. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am not sure how to open that, nor do I really want to. But just ask the question here.

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what have i done wrong ? i am so sorry

  15. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ?

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know you are angry

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please just forgive .

  18. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lol why would I be angry?

  19. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I just said to ask the question :)

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the question is in the file ,

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and the password is osho

  22. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I don't know how to open the file, and I don't know what the file is, and I don't like opening things when I don't know where they come from. If it is a math question, there is no reason you can't post it here?

  23. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    It does not ask for a password when I try and open it, it says its corrupt or protected, but there is no place for a password.

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (a) Let X ε Rn. Show that the set B(X, ε ) is open. (b) Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that On is not empty.

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r

  26. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What is your definition of open?

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2)Let (X, d) and (Y, d) be metric spaces and f a mapping of X into Y. Let τ1 and τ2 be the topologies determined by d and d1 respectively. Then f(X, τ) (y, τ) is continuous if and only if ; that is if x1, x2, . . . , xn, . . . , is a sequence of points in (X, d) converging to x, show that the sequence of points f(x1), f(x2), . . . , f(xn), . . . in (Y, d) converges to x.

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    an open interval is an open set. interval without its boundary points

  29. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    How do you define a boundary point?

  30. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we must show that your ball does not contain any boundary points

  31. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What would it mean if it did?

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    boundary point to me is the being or the end of a thing . in inequality, it is denoted by < or =, and > or = , or [ ]

  33. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    well you are kind of defining what we are asked to prove.

  34. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A point \(x\) is a boundary point for set \(A\) if \(\forall\) \(\delta>0\) we have \(B_{\delta}(x)\cap A\ne \emptyset\) and \(B_{\delta}(x)\cap A^C\ne \emptyset\)

  35. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Does this make sense? This says for any neighborhood of \(x\) we have a point in A and a point out of A.

  36. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but should it be out of a?

  38. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Well what you want to show is that \(X\) contains non of its boundary points. So suppose it does. What does that mean?

  39. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it means it is close

  40. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right?

  41. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    brb I need to take my wife to a party. I might be half hour or so.

  42. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Its only closed if it contains all of them. Suppose it contains one, then by the def I gave you, what does that mean?

  43. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  44. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it means it is have open and half close

  45. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what is?

  46. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am here now.

  47. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Suppose to the contrary that \(B(x, \epsilon)\) contains one of its boundary points, lets call it \(a\). Then \(a\in B(x,\epsilon)\). Now consider \(\delta=\min(d(a, x), d(a, \epsilon))\). Then \(B(a, \delta)\subseteq B(x, \epsilon)\) contradicting the assumption that \(x\) was a boundary point.

  48. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thus \(B(x, \epsilon)\) is open because it does NOT contain any of it's boundary points.

  49. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What does it mean for a metric to generate a topology?

  50. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For you to understand the second question. You need to know what a metric space is and what does it mean to generate a topology from a metric, in order to understand that you will need to understand what a topology is, and what a basis and sub basis are for a topological space and what it means to be an arbitrary union, in order to understand that you will need to understand what a ray is. Then you will need to know what it means for a sequence of points in some product space to converge, and you must understand the metric definition of continuity, which means you must understand a metric. Do you understand all of these things?

  51. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i know a bit of them but not that good at it

  52. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Feel free to ask me anything

  53. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    5(a) Prove that for any y, z ε , max(y, z) = ½[y + z+ |y-z|], min(y, z) = ½[y + z- |y-z|]. can you try to prove that for me ?

  54. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r

  55. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    any \(y,z\in?\)

  56. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    of R

  57. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Suppose w.l.o.g. that \(x<y\). Then \[\max(x,y) = y = \dfrac{x}{2}+\dfrac{y}{2}-\dfrac{x}{2}+\dfrac{y}{2}=\dfrac{1}{2}[x+y-x+y]=\dfrac{1}{2}[x+y+|y-x|]\] Similarly for \(\min\)

  58. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now you write up a similar proof for the min case, and show it here.

  59. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok but please just help me so that i don't do any errors

  60. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but sir, why did you use x,y. thought the given function is y,z e R

  61. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @zzr0ck3r

  62. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The fact that you don't know that it does not matter if I call it \(z\) or \(x\) really scares me and again begs the question. What are you doing?

  63. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i am trying to know what you did

  64. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    please see the complete solution @GIL.ojei

    1 Attachment
  65. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    by definition of absolute value, we can write this: \[\Large \begin{gathered} y \geqslant z, \Rightarrow \left| {y - z} \right| = y - z \hfill \\ \hfill \\ y \leqslant z \Rightarrow \left| {y - z} \right| = - \left( {y - z} \right) = - y + z \hfill \\ \end{gathered} \]

  66. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    furthermore, those subsequent expressions, namely "max" and "min" are defined for all elements of R, so there is no need for further restrictions.

  67. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    as I said before, we have to distinguish two cases, since R is a totally ordered set. First case: \[y \geqslant z\]

  68. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then \[y - z \geqslant 0\], so by definition of absolute value, we get: \[\left| {y - z} \right| = y - z\]

  69. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes, i know that is true

  70. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then I can replace \[\left| {y - z} \right|\] with \[y - z\]

  71. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  72. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so I can write this: \[\max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z + y - z} \right\} = \frac{{2y}}{2} = y\]

  73. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  74. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    furthermore I can write this: \[\begin{gathered} \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( {y - z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \end{gathered} \]

  75. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    now we have to consider the other case, namely: \[y \leqslant z\]

  76. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    by definition of absolute value, we have: \[y - z \leqslant 0 \Rightarrow \left| {y - z} \right| = - \left( {y - z} \right) = - y + z\]

  77. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok. but i thought you have applied that already

  78. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no, since in this case is y-z less or equal to zero, whereas in the first case is y-z greater or equal to zero

  79. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    in other words, in this second case, we can replace: \[\left| {y - z} \right|\] with: \[ - y + z\]

  80. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    are you saying that we ought to have four proves?

  81. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and then we can write this: \[\begin{gathered} \max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z + \left( { - y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \hfill \\ \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( { - y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z + y - z} \right\} = \frac{{2y}}{2} = y \hfill \\ \end{gathered} \]

  82. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no, I say that we have to prove your statement for two different cases for value of y-z

  83. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that's all!

  84. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    which is the main prove ?

  85. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    because all you stated are corect

  86. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok. now i understand sir

  87. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    :)

  88. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.