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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    In the finite complement topology of R, let the sequence [x_{n} be defined by x_{n} = n, for n\epsilon N. If the limit of the sequence is x, then x must be ∞ 0 A unique constant Arbitrary in R

  2. anonymous
    • one year ago
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    @zzr0ck3r

  3. anonymous
    • one year ago
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    what is the complement of R?

  4. zzr0ck3r
    • one year ago
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    what is the set of all elements that are not in R For the 6th times, please stop using epsilon. IT DOES NOT MEAN IN

  5. anonymous
    • one year ago
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    i do not know

  6. anonymous
    • one year ago
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    i thought there are two types of numbers

  7. anonymous
    • one year ago
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    the imaginary and the real

  8. zzr0ck3r
    • one year ago
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    when we talk about the real numbers as the universe, we do not consider them as a subset of the complex numbers. The compliment of the real numbers is the emptyset

  9. anonymous
    • one year ago
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    ok. thanks so much

  10. anonymous
    • one year ago
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    so will that option b empty since is the complement of R?

  11. zzr0ck3r
    • one year ago
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    In the cofinite topology, 1) a sequence that does not take an infinite many of the same value will converge to every element in the space. 2) A sequence that does(but only one number), will converge to that number 3) A sequence that has more than one number for which it obtains infinite times does not converge

  12. zzr0ck3r
    • one year ago
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    So what is the answer?

  13. anonymous
    • one year ago
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    Arbitrary in R

  14. zzr0ck3r
    • one year ago
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    correct

  15. anonymous
    • one year ago
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    i wonder how \(\{ x_n,\varnothing ,R \}\) is considered cofinite :O

  16. anonymous
    • one year ago
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    what does arbitrary really mean. because i see it as finite or infinite

  17. zzr0ck3r
    • one year ago
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    um, it means if you close your eye and pick one

  18. anonymous
    • one year ago
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    Arbitrary = any

  19. zzr0ck3r
    • one year ago
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    @halmos? \(\{x_n, \emptyset, \mathbb{R}\}\) what is this?

  20. anonymous
    • one year ago
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    In the finite complement topology of R, let the sequence [x_{n} be defined by x_{n} = n was trying to interpret in notations

  21. zzr0ck3r
    • one year ago
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    Arbitrary size means we are not given anything about the set, so as far as order, a set of arbitrary order is either finite or infinite. But we use the term this other way as well :)

  22. zzr0ck3r
    • one year ago
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    he sucks with latex :)

  23. zzr0ck3r
    • one year ago
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    he also usees \(\epsilon\) when he means \(\in\) so you got to be careful with that as well :)

  24. anonymous
    • one year ago
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    oh i see now :)

  25. zzr0ck3r
    • one year ago
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    Sorry @GIL.ojei I am just playing with you...

  26. zzr0ck3r
    • one year ago
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    but seriously start using \(\in\) when you mean in. It is coded as `\(\in)`

  27. zzr0ck3r
    • one year ago
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    \in

  28. zzr0ck3r
    • one year ago
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    \(\epsilon\), in general, means a positive real number

  29. anonymous
    • one year ago
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    @GIL.ojei u can use this next time for math notations :) http://prntscr.com/8e7ekc and yes epsilon used to denote very small values

  30. anonymous
    • one year ago
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    thank u sir

  31. zzr0ck3r
    • one year ago
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    or "arbitrarily small" positive number :)

  32. anonymous
    • one year ago
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    ok

  33. anonymous
    • one year ago
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    Let (X,τ) be a topological space, and let A be a subset of X. A is dense in X if and only if every non-empty open subset U of X, _________________ A⋂U=0 A⋃U=ϕ A⋂U=ϕ A⋂U=X

  34. zzr0ck3r
    • one year ago
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    What is that weird symbol?

  35. zzr0ck3r
    • one year ago
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    @GIL.ojei ?

  36. anonymous
    • one year ago
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    this is just a definition, and i think the notation should be empty set u can use `\varnothing` \( \varnothing \) instead

  37. anonymous
    • one year ago
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    empty set

  38. zzr0ck3r
    • one year ago
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    then what is the 0?

  39. anonymous
    • one year ago
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    i think he copied it from a place which typed it as \(\emptyset \) first (we see it 0) then as \(\phi\)

  40. anonymous
    • one year ago
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    its just zero

  41. zzr0ck3r
    • one year ago
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    that makes no sense

  42. zzr0ck3r
    • one year ago
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    It should be intuitively clear from the definition of a dense set, that if \(U\) is open then \(A\cap U\ne \emptyset\). But I do not see this option, and I dont know what the weird \(\phi \) looking thing is

  43. zzr0ck3r
    • one year ago
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    or `\emptyset`

  44. zzr0ck3r
    • one year ago
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    The union of any non empty set and anything else is non empty

  45. zzr0ck3r
    • one year ago
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    these options don't make sense.

  46. zzr0ck3r
    • one year ago
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    @GIL.ojei please reread the question and see if it shuold say \(\neq \emptyset\)

  47. anonymous
    • one year ago
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    Let (X,τ) be a topological space, and let A be a subset of X. A is dense in X if and only if every non-empty open subset U of X, _________________

  48. anonymous
    • one year ago
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    A⋂U=0

  49. anonymous
    • one year ago
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    that is option a

  50. anonymous
    • one year ago
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    A⋃U=\[ϕ \]

  51. anonymous
    • one year ago
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    that is option b

  52. anonymous
    • one year ago
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    A⋂U=\[ϕ \]

  53. anonymous
    • one year ago
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    thats option C

  54. anonymous
    • one year ago
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    A⋂U=X

  55. anonymous
    • one year ago
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    option D

  56. anonymous
    • one year ago
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    @zzr0ck3r

  57. anonymous
    • one year ago
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    i think is option c

  58. anonymous
    • one year ago
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    or option D

  59. zzr0ck3r
    • one year ago
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    option a makes no sense the union of subsets of some general topology are not necessarily equal to a number option b is wrong because U is non empty option c is wrong because of what I wrote about the non empty intersection option d is only true sometimes

  60. zzr0ck3r
    • one year ago
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    \(A\cap U \neq \emptyset\) is always true

  61. anonymous
    • one year ago
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    Which of these is not true about T1 - spaces? a)Every singleton set is closed b)Every finite set is closed c)Every Hausdorff space is T1 d)Y1isinR

  62. anonymous
    • one year ago
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    this is the last question on topology for today. please, i will tell you what to teach me today . please

  63. anonymous
    • one year ago
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    i know option A and B are correct

  64. anonymous
    • one year ago
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    but what about C and D? i fink C is also correct

  65. zzr0ck3r
    • one year ago
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    what does Y1isinR mean and why do you keep writing things like this?

  66. anonymous
    • one year ago
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    that was how i saw it

  67. anonymous
    • one year ago
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    \[Y_1 \] is in R

  68. zzr0ck3r
    • one year ago
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    you look at the page and it looks like this ? Y1isinR ?

  69. anonymous
    • one year ago
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    i think that was what they wanted to write

  70. zzr0ck3r
    • one year ago
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    what is \(Y_1\)?

  71. anonymous
    • one year ago
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    do not have idea but is Every Hausdorff space is T1?

  72. zzr0ck3r
    • one year ago
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    Tell me the definitions of both

  73. anonymous
    • one year ago
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    X is a T1 space if any two distinct points in X are separated.

  74. anonymous
    • one year ago
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    i know that Hausdorff space has to do with intersection. but what is really separated?

  75. zzr0ck3r
    • one year ago
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    what does separated mean?

  76. anonymous
    • one year ago
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    i asked the question

  77. zzr0ck3r
    • one year ago
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    So if it is Hausdorff, then for all \(x,y\) we have open set \(O,U\) such that \(x\in O, y\in U, O\cap U=\emptyset\) Does this imply that there is a open set \(A\) that contains \(x\) and does not contain \(y\)? Does this imply that there is a open set \(B\) that contains \(y\) and does not contain \(x\)? IF the answer is yes, then we must be \(T_1\).

  78. anonymous
    • one year ago
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    waw. so, T1 space is also Hausdorff space

  79. zzr0ck3r
    • one year ago
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    we are not showing that T1 is Hausdorff, we are showing that Hausdorff is T1.

  80. zzr0ck3r
    • one year ago
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    every horse has 4 legs but not every 4 legged animal is a horse i.e. if a implies b it is not necessarily true that b implies a

  81. anonymous
    • one year ago
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    ok

  82. anonymous
    • one year ago
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    i want to close the tab and open another. i think there are thinks i want to understand

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