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anonymous

  • one year ago

With the standard topology on R,which one of the sets is open in R?

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

  2. anonymous
    • one year ago
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    i fink this is the answer but i have other options

  3. zzr0ck3r
    • one year ago
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    so why is this open? Can you write it in interval notation?

  4. anonymous
    • one year ago
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    (1/2,1)

  5. zzr0ck3r
    • one year ago
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    but the absolute value sign means you also get (-1,-1/2). Do you see that? 1/2 < |-0.8|<1

  6. anonymous
    • one year ago
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    ok,

  7. zzr0ck3r
    • one year ago
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    So the set, in interval notation, is \[(-1,- \dfrac{1}{2})\cup (\dfrac{1}{2}, 1)\]

  8. anonymous
    • one year ago
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    please why -0.8?

  9. zzr0ck3r
    • one year ago
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    I was just showing you that you get negative number in there as well

  10. zzr0ck3r
    • one year ago
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    \(\{x\mid \frac{1}{2}< |x|<1\}=(-1,-\frac{1}{2})\cup (\frac{1}{2}, 1)\)

  11. anonymous
    • one year ago
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    ok. now i got it. because of the absolute value .

  12. anonymous
    • one year ago
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    should i post the other options?

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

  14. anonymous
    • one year ago
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    now this is neither open or close in R . right ?

  15. zzr0ck3r
    • one year ago
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    Can you write it in interval notation?

  16. anonymous
    • one year ago
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    please don't shout on me. i will try

  17. zzr0ck3r
    • one year ago
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    what?

  18. anonymous
    • one year ago
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    i mean please do not get angry if i fail

  19. zzr0ck3r
    • one year ago
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    I don't even know how to respond to that.

  20. anonymous
    • one year ago
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    what if i say it can not be written as union of open sets

  21. anonymous
    • one year ago
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    (-1,-1/2)u(1/2,1]

  22. anonymous
    • one year ago
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    you there?

  23. zzr0ck3r
    • one year ago
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    almost, the -1 gets a [ also

  24. anonymous
    • one year ago
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    i thought as much

  25. zzr0ck3r
    • one year ago
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    \([-1, -0.4)\cup(0.5,1]\)

  26. zzr0ck3r
    • one year ago
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    |-1|<=1

  27. zzr0ck3r
    • one year ago
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    ok what is a closed set?

  28. zzr0ck3r
    • one year ago
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    you gone?

  29. anonymous
    • one year ago
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    a close set is a close interval. meaning that every set that has it boundary involve is close . because if we take a small radius around any boundary, it will not be in the set

  30. anonymous
    • one year ago
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    no. i am here,network is getting bad

  31. anonymous
    • one year ago
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    @zzr0ck3r

  32. zzr0ck3r
    • one year ago
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    Ok, closed intervals are closed sets, but that is not all of them. R is both open and closed, and so is the empty set, and neither of those are closed intervals. A set is closed if its compliment is open. Do you know what a compliment is?

  33. zzr0ck3r
    • one year ago
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    Ok, we can go with boundary, does this set contain all of its boundary points? If not, which 2 does it not contain?

  34. zzr0ck3r
    • one year ago
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    Aight man, I am sorry about the internet, but this is taking way to long. It took about 35 minutes to get replies. I will be back tomorrow.

  35. anonymous
    • one year ago
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    am sorry sir. my network is really poor

  36. anonymous
    • one year ago
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    it does not contain all its boundary point. it contains only one which is 1 but does not contain 1/2. so it is half open and half close

  37. zzr0ck3r
    • one year ago
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    Its boundary points are -1, -1/2, 1/2, and 1. While it is not a great idea to think of every open/closed set as an interval, loosely said, the boundary points are the endpoints of any broken up union of sets.

  38. anonymous
    • one year ago
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    so, x:1/2<|x|⩽1 is neither open nor close . so is it right to say that the only open sets in R in the options given above are \[x:1/2<|x|<1 \] and the close set in the option above in R is \[x:1/2\le|x|⩽1 \] right?

  39. anonymous
    • one year ago
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    @zzr0ck3r

  40. zzr0ck3r
    • one year ago
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    yes, unless I did not see all the options

  41. anonymous
    • one year ago
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    ok sir

  42. anonymous
    • one year ago
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    let me close this question and ask another . can i?

  43. zzr0ck3r
    • one year ago
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    of course

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