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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    If [xn] be a sequenceinR endowed with the finite complement topology. If [xn] converges,In R then

  2. anonymous
    • one year ago
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    The limit is unique

  3. anonymous
    • one year ago
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    @zzr0ck3r

  4. anonymous
    • one year ago
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    (xn) converges to only two points

  5. anonymous
    • one year ago
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    (xn) converges to every element in R

  6. anonymous
    • one year ago
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    what do you think sir?

  7. zzr0ck3r
    • one year ago
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    I am thinking and playing with it. I have only done basic work in that space. But I am sure I can figure it out...

  8. anonymous
    • one year ago
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    the last option

  9. anonymous
    • one year ago
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    (xn) converges to one point in R and one point outside R

  10. zzr0ck3r
    • one year ago
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    there is no outside of R

  11. anonymous
    • one year ago
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    i have not heard anything outside R

  12. zzr0ck3r
    • one year ago
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    R is the universal set.

  13. anonymous
    • one year ago
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    yes that was y i din't put that option first

  14. zzr0ck3r
    • one year ago
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    I think since we are assuming it converges, it must have some repeated element infinite times, and thus will converge to all x

  15. zzr0ck3r
    • one year ago
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    actually I think it would be unique in this case.

  16. anonymous
    • one year ago
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    so, i should go with unique ?

  17. anonymous
    • one year ago
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    @zzr0ck3r

  18. zzr0ck3r
    • one year ago
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    well really you should convince yourself...

  19. zzr0ck3r
    • one year ago
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    what does it mean for a sequence to converge?

  20. anonymous
    • one year ago
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    A sequence is said to be convergent if it approaches some limit

  21. zzr0ck3r
    • one year ago
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    lol, what does that mean?

  22. anonymous
    • one year ago
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    in real analysis , a sequence that has limit is convergent

  23. anonymous
    • one year ago
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    or explain more

  24. zzr0ck3r
    • one year ago
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    but what does that mean. If you are going to show that a sequence converges, then what do you need to show?

  25. anonymous
    • one year ago
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    please explain . thats all i know

  26. zzr0ck3r
    • one year ago
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    then there is no way you can understand this question

  27. anonymous
    • one year ago
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    please help me understand

  28. zzr0ck3r
    • one year ago
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    Just google converging sequences.

  29. zzr0ck3r
    • one year ago
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    What is the finite compliment topology?

  30. anonymous
    • one year ago
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    ok i will do that later but help with more question

  31. zzr0ck3r
    • one year ago
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    you mean. Will I answer your questions so that you don't have to study ? lol

  32. zzr0ck3r
    • one year ago
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    ok sure.

  33. anonymous
    • one year ago
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    a cofinite subset of a set X is a subset A whose complement in X is a finite set.

  34. anonymous
    • one year ago
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    sir, its because i have less time

  35. zzr0ck3r
    • one year ago
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    but you are not learning anything...

  36. anonymous
    • one year ago
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    and you are not always online

  37. zzr0ck3r
    • one year ago
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    ok, ill stop telling you how to learn. Just keep asking questions.

  38. anonymous
    • one year ago
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    Let X be a topological space. Then X is regular if

  39. anonymous
    • one year ago
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    XisbothT1andT2 XisT3only XisT2only XisbothT2andT

  40. zzr0ck3r
    • one year ago
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    read that and tell me what is wrong

  41. zzr0ck3r
    • one year ago
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    this is a definition. I bet you could figure this out with a look in a book

  42. anonymous
    • one year ago
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    is it T3?

  43. anonymous
    • one year ago
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    ok, give me a book to look into

  44. zzr0ck3r
    • one year ago
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    I already have

  45. zzr0ck3r
    • one year ago
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    https://en.wikipedia.org/wiki/Regular_space

  46. anonymous
    • one year ago
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    waw. T3 then?

  47. anonymous
    • one year ago
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    Which of these spaces is not Hausdorff? R with stanard topology R with lower limit topology R metric toplogy R with the finite complement topology

  48. zzr0ck3r
    • one year ago
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    Can you rule out any cases?

  49. anonymous
    • one year ago
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    i think b

  50. anonymous
    • one year ago
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    what do you think?

  51. zzr0ck3r
    • one year ago
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    why?

  52. anonymous
    • one year ago
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    i can't really explain but i fink it is the answer. please correct me if am wrong

  53. anonymous
    • one year ago
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    @zzr0ck3r

  54. anonymous
    • one year ago
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    @zzr0ck3r

  55. zzr0ck3r
    • one year ago
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    But that is the entire point, you need to be able to explain.

  56. zzr0ck3r
    • one year ago
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    why would you want to know the answer, if you don't understand it. You are not correct

  57. zzr0ck3r
    • one year ago
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    I can't tell you why you are wrong unless you tell me why you thought you were right.

  58. anonymous
    • one year ago
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    first i know what Hausdorff

  59. anonymous
    • one year ago
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    but i understand Hausdorff more with set than real line

  60. zzr0ck3r
    • one year ago
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    ok, the lower limit topology has open sets in the form [a,b) Given two points \(x,y\) we can always find disjoint sets around each even if one of the points is a

  61. zzr0ck3r
    • one year ago
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    well to show something is NOT Hausdorff, what do you need to show?

  62. anonymous
    • one year ago
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    that there intersection does not give empty set

  63. zzr0ck3r
    • one year ago
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    you are almost correct You need to show that there is at least two points \(x,y\) where \(x\ne y\) such that there are no open sets \(O\) and \(U\) where \(x\in O\) and \(y\in U\) and \(O\cap U=\emptyset\)

  64. anonymous
    • one year ago
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    ok.

  65. zzr0ck3r
    • one year ago
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    Ok, let \(R\) be endowed with the finite compliment topology. Suppose there exists \(x,y\) where \(x\ne y\) and there is open sets \(O,U\) with \(x\in O\) and \(y\in Y\) where \(U\cap O=\emptyset\). Since \(U\) is open, we have that \(U^C\) is finite and since \(U\cap O=\emptyset\) it must be that \(O\subset U^C\) this means that \(O\) must be finite, but every finite set in \(R\) has uncountably infinite compliment, and thus \(O\) can not be open and thus a contradiction. It is obvious that neither \(U\) or \(O\) is the emptyset.

  66. zzr0ck3r
    • one year ago
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    This proves that non infinite set endowed with the finite complement topology can not be Hausdorff.

  67. anonymous
    • one year ago
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    waw. thank you sir.

  68. anonymous
    • one year ago
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    Let Y be a subset of a topological space (X,τ) . If Y is a union of a countable number of nowhere dense subset of X, then Y is called a set of the first category, otherwise it is said to be ________ Meager Third Category fourth Category Second Category

  69. zzr0ck3r
    • one year ago
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    I do not know this terminology but since it is a definition, it could not be any easier...

  70. anonymous
    • one year ago
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    i think is second

  71. zzr0ck3r
    • one year ago
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    why?

  72. anonymous
    • one year ago
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    i have something like that in my book but not sure

  73. zzr0ck3r
    • one year ago
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    @GIL.ojei why?

  74. zzr0ck3r
    • one year ago
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    do you think that is a good reason?

  75. zzr0ck3r
    • one year ago
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    the first thing I get when I google This observation motivates the introduction of a larger class of sets: A subset A ⊆ X is called meager (or of first category) in X if it can be written as a countable union of nowhere dense sets. Any set that is not meager is said to be nonmeager (or of second category). The complement of a meager set is called residual

  76. anonymous
    • one year ago
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    ok. sp am i correct?

  77. zzr0ck3r
    • one year ago
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    seriously?

  78. anonymous
    • one year ago
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    ok.so its the second category

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