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manjuthottam

  • one year ago

could someone help me understand Topology of the Reals?!

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

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

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

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

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

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

  7. manjuthottam
    • one year ago
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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.

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

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

  10. manjuthottam
    • one year ago
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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

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

  12. manjuthottam
    • one year ago
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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

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

  14. wio
    • one year ago
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  29. wio
    • one year ago
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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.

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

  31. manjuthottam
    • one year ago
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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?

  32. wio
    • one year ago
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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.

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

  34. manjuthottam
    • one year ago
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    hmm ok

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

  36. manjuthottam
    • one year ago
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    yes it does

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

  38. manjuthottam
    • one year ago
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    yes i see

  39. wio
    • one year ago
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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.

  40. wio
    • one year ago
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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 \[ \frac{1}{k},k\in\mathbb{N} \]

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

  42. wio
    • one year ago
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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.

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

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

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

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

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

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

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

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

  51. manjuthottam
    • one year ago
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    ok

  52. wio
    • one year ago
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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.

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

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

  55. wio
    • one year ago
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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.

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

  57. manjuthottam
    • one year ago
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    so like e)

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

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

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

  61. manjuthottam
    • one year ago
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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

  62. wio
    • one year ago
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    yeah

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

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

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