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manjuthottam

could someone help me understand Topology of the Reals?!

  • one year ago
  • one year ago

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  1. manjuthottam
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    attached is the solutions for some question relating to it. the first 5 pages deal with it

    • one year ago
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  2. manjuthottam
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    i keep getting int and bd wrong

    • one year ago
  3. wio
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    I wish I could help, dunno enough about topology yet

    • one year ago
  4. manjuthottam
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    ok thank you though :)

    • one year ago
  5. wio
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    I can help reason it out if you tell me definitions

    • one year ago
  6. wio
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    Seems like interior is just stripping away the boundery.

    • one year ago
  7. manjuthottam
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    Defn of int and bd: Let S be a subset of R. a point x in R is an interior point of S if there exists a neighborhood N of x such that N is a subset of S. if for every neighborhood N of x, N (junction) S does not = empty set. and N junction of (R\S) does not equal empty set.

    • one year ago
  8. manjuthottam
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    then x is called a boundary point of S.

    • one year ago
  9. wio
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    You understand what neighborhood is?

    • one year ago
  10. manjuthottam
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    well for first problem a) why is the interior empty set while for b) it is the whole set with open parentheses

    • one year ago
  11. manjuthottam
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    and yes i understand what neighborhood is

    • one year ago
  12. manjuthottam
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    what made 1/n interior pt be special to be empty while the union of two sets interior pt is the complete whole thing

    • one year ago
  13. manjuthottam
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    i would have guessed the ans for b) to be 3 or something

    • one year ago
  14. wio
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    Well, the problem is that you have: \[ 1/1,1/2,1/3,1/4 \]Which is just rational a subset of rational numbers. The rational numbers are no continuous, so neither would be this subset. You can find an irrational number between any two elements, for example. Thus they are all boundary points.

    • one year ago
  15. wio
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    Despite not understanding neighborhood rigorously, that is my reasoning for b)

    • one year ago
  16. wio
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    I mean for a)

    • one year ago
  17. wio
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    And it applies to c) and d) apparently

    • one year ago
  18. wio
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    @manjuthottam make sense?

    • one year ago
  19. manjuthottam
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    so the answer is focusing on rational numbers?

    • one year ago
  20. wio
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    Okay, @manjuthottam do you know what continuity is?

    • one year ago
  21. manjuthottam
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    is it just that the limit exists?

    • one year ago
  22. wio
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    yeah the limit exists at the function

    • one year ago
  23. wio
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    But actually, let's talk about neighborhoods a bit ok?

    • one year ago
  24. manjuthottam
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    ok so neighborhood is x in R. while there is an epsilon greater than zero where |x-y|< epsilon?

    • one year ago
  25. wio
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    Yeah, notice how that looks kinda like a limit.... Anyway...

    • one year ago
  26. wio
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    Consider 1/2, 1/3, and 1/4 Do they have a neighborhood?

    • one year ago
  27. manjuthottam
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    i'm guessing no?

    • one year ago
  28. manjuthottam
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    at least its not that specific of a number

    • one year ago
  29. wio
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    If you think about it, they're equal to 12/24, 8/24, and 6/24 Since you don't have 10/24 or 4/24 in there, you can't really say they make up a neighborhood.

    • one year ago
  30. wio
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    So can you understand how these three points would be boundary points?

    • one year ago
  31. manjuthottam
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    so since there are values in between like 10/24 = (5/12) that are not part of the set,, there is no neighborhoood? ... because it is not a continuous set?

    • one year ago
  32. wio
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    Well, consider if we were to make a function: \[ f:\mathbb{R}\mapsto\{0,1\} \]We let it be \(0\) if \(x\in S\) and \(1\) if \(x\notin S\) In this case, a neighborhood is just an interval for which \(f\) is continuous.

    • one year ago
  33. wio
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    In this way, neighborhoods are analogous to continuity.

    • one year ago
  34. manjuthottam
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    hmm ok

    • one year ago
  35. wio
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    Does my analogy sound consistent to you understanding of what a neighborhood is?

    • one year ago
  36. manjuthottam
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    yes it does

    • one year ago
  37. wio
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    Okay so suppose we did have two points that were directly next to each other... they'd have to be \[ \frac{1}{n}, \frac{1}{n+1} \]

    • one year ago
  38. manjuthottam
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    yes i see

    • one year ago
  39. wio
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    Notice how they are equal to: \[ \frac{2}{2n}, \frac{2}{2n+2} \]And the thing is: \[ \frac{2}{2n+1} \]is between these two points.

    • one year ago
  40. wio
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    And since the numerator is even and denominator is odd, we can't reduce this fraction anymore, so it can't be expressed as \[ \frac{1}{k},k\in\mathbb{N} \]

    • one year ago
  41. wio
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    thus it is not in our set

    • one year ago
  42. wio
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    So no matter how big you make \(n, n+1\), we can still find a number between them... you'll never have any continuity for our function \(f\) so you'll never find a neighborhood.

    • one year ago
  43. manjuthottam
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    hmm ok that makes sense!

    • one year ago
  44. wio
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    This is sort of a proof by contradiction as to why there can't be any neighborhoods.

    • one year ago
  45. manjuthottam
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    ok, so then based on the defn of int and bd, we have to first see about continuity and neighborhood

    • one year ago
  46. manjuthottam
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    if it is continuous then we can see that int and bd are not an empty set

    • one year ago
  47. wio
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    The continuity thing was more to help me understand neighborhoods than it was to help you. =)

    • one year ago
  48. manjuthottam
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    =D kk but it helped me learn too! lol!

    • one year ago
  49. wio
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    I think what you need to understand is that rational numbers do NOT make any neighborhoods.

    • one year ago
  50. wio
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    you'll always be able to find some irrational number between two rational numbers.

    • one year ago
  51. manjuthottam
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    ok

    • one year ago
  52. wio
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    Thus when they give you a set involving rational numbers or natural numbers, etc. Remember that it must all be boundaries, no neighborhoods and no interior sets.

    • one year ago
  53. wio
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    If you get something in interval notation, just remember that changing [ ] to ( ) should give you interior set.

    • one year ago
  54. manjuthottam
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    ok so even union of sets

    • one year ago
  55. wio
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    There is a trick question in there where they use \(\cap\) which is the set intersection... in those cases it has to be points in BOTH sets.

    • one year ago
  56. wio
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    Yeah, it should work for unions of sets well too Just not for intersections of sets.

    • one year ago
  57. manjuthottam
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    so like e)

    • one year ago
  58. manjuthottam
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    automatically the int is empty set

    • one year ago
  59. wio
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    For example: \[ [1,4]\cap [2,6] \]This gives us: \[ [2,4] \]this you can change them... \[ int([2,4]) = (2,4)\quad bd([2,4]) = \{2,4\} \]

    • one year ago
  60. wio
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    So for intersections... just compute the intersection. It shouldn't be too bad.

    • one year ago
  61. manjuthottam
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    oh ok, so in sets that have a intersection we apply that method you mentioned but if the intersection is just one point like in e) then int is empty set because there is no neighborhood

    • one year ago
  62. wio
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    yeah

    • one year ago
  63. wio
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    You'd have to give me some \(\epsilon>0\) but clearly \(2+\epsilon\) is NOT in the set \(\{2\}\)

    • one year ago
  64. manjuthottam
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    kk thank you so much for helping me !!!!

    • one year ago
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