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anonymous

  • one year ago

here

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  1. anonymous
    • one year ago
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    Let R be endowed with the usual standard topology. Consider Y = [-1,1] as a subspace of R. Which one of the following sets is closed in Y

  2. anonymous
    • one year ago
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    @zzr0ck3r

  3. zzr0ck3r
    • one year ago
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    ?

  4. anonymous
    • one year ago
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    {x: \[\frac{1}{2}\ < \] |x| < 1}

  5. zzr0ck3r
    • one year ago
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    not closed

  6. anonymous
    • one year ago
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    is that open?

  7. anonymous
    • one year ago
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    {x:\[ \frac{1}{2} \] <\[ |x|\leqslant \]1}

  8. zzr0ck3r
    • one year ago
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    yes, since it is open in R and and it equals the intersection with the subspace.

  9. zzr0ck3r
    • one year ago
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    its open

  10. anonymous
    • one year ago
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    \[{x: \frac{1}{2}\leqslant|x| < 1} \]

  11. zzr0ck3r
    • one year ago
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    neither open nor closed

  12. anonymous
    • one year ago
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    \[{x: \frac{1}{2}\leqslant|x| < 1} \]

  13. zzr0ck3r
    • one year ago
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    thats the same. please use parentheses. bbl

  14. anonymous
    • one year ago
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    \[{x: \frac{1}{2} < |x|\leqslant1} \]

  15. anonymous
    • one year ago
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    \[{x: \frac{1}{2}\leqslant|x|\leqslant1} \]

  16. anonymous
    • one year ago
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    i presume this is close

  17. zzr0ck3r
    • one year ago
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    yep

  18. zzr0ck3r
    • one year ago
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    closed sets in a subspace S, are of the form \(A\cap S\) where \(A\) is open in the parent topology.

  19. zzr0ck3r
    • one year ago
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    ok walking out the door

  20. anonymous
    • one year ago
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    and which is open in Y with same options ?

  21. anonymous
    • one year ago
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    is it \[{x: \frac{1}{2}\ < |x| < 1} \]

  22. anonymous
    • one year ago
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    2 With the standard topology on R,which one of the sets in question (1) above is open in R? with same options given above?

  23. anonymous
    • one year ago
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    With the standard topology on R,which one of the sets in question (1) above is closed in R? with same options given above

  24. zzr0ck3r
    • one year ago
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    hi

  25. anonymous
    • one year ago
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    sir, i am here @zzr0ck3r

  26. zzr0ck3r
    • one year ago
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    ok

  27. anonymous
    • one year ago
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    so, can you help with the asked question sir?

  28. zzr0ck3r
    • one year ago
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    \((-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\) is open \((-1, -\dfrac{1}{2}]\cup [\dfrac{1}{2}, 1)\) is not open \([-1, -\dfrac{1}{2})\cup (\dfrac{1}{2}, 1]\) is not open \([-1, -\dfrac{1}{2}]\cup [\dfrac{1}{2}, 1]\) is not open

  29. anonymous
    • one year ago
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    so which means that \[ x:1/2 <|x|<1 \] is open in the usual standard toplogy

  30. zzr0ck3r
    • one year ago
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    correct

  31. anonymous
    • one year ago
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    what about in the standard topology on R

  32. anonymous
    • one year ago
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    2 With the standard topology on R,which one of the sets in question (1) above is open in R with same options given above?

  33. anonymous
    • one year ago
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    you here sir?

  34. anonymous
    • one year ago
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    @zzr0ck3r

  35. anonymous
    • one year ago
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    hello sir, @zzr0ck3r

  36. anonymous
    • one year ago
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    let me post the question in full . please don't get mad at me

  37. anonymous
    • one year ago
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    With the standard topology on R,which one of the sets is open in R?

  38. anonymous
    • one year ago
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    \[ {x: \frac{1}{2}\ < |x| < 1} \] \[{x: \frac{1}{2} < |x|\leqslant1} \] \[{x: \frac{1}{2}\leqslant|x| < 1} \] \[{x: \frac{1}{2}\leqslant|x|\leqslant1}\]

  39. anonymous
    • one year ago
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    is the answer still the first here sir?

  40. anonymous
    • one year ago
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    i guess from what you thought, A is the only open set while B,C, D are close sets

  41. anonymous
    • one year ago
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    am i correct?

  42. anonymous
    • one year ago
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    another question

  43. anonymous
    • one year ago
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    A topological space is called a Hausdorff space, if for each x, y of distinct points of X, there exists nbds \[U_{x} \] and \[U_{y} \] of x and y respectively that are disjoint. This implies X is Hausdorff wiith these properties except one.

  44. anonymous
    • one year ago
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    A) if \[ \forall x, y \epsilon\mathbb X; x \neg y \]

  45. anonymous
    • one year ago
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    B)\[There \exists U_x \epsilon N(x). U_y \epsilon N(y) \]

  46. anonymous
    • one year ago
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    C) \[U_x \bigcap U-y = \phi \]

  47. anonymous
    • one year ago
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    D)\[\forall x, y \epsilon X, x \bigcap y = 0 \]

  48. zzr0ck3r
    • one year ago
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    what is the question?

  49. anonymous
    • one year ago
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    should i post it in anew question?

  50. zzr0ck3r
    • one year ago
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    yes

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