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anonymous

  • one year ago

here sir

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  1. anonymous
    • one year ago
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    @zzr0ck3r

  2. zzr0ck3r
    • one year ago
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    I dont understand the last sentence.

  3. zzr0ck3r
    • one year ago
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    I know what Hausdorff is.

  4. 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 U_{x} and U_{y} of x and y respectively that are disjoint. This implies X is Hausdorff with these properties except one.

  5. zzr0ck3r
    • one year ago
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    It gives a definition of Hausdorff, and then said it implies something. There is no question here.

  6. anonymous
    • one year ago
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    If \[\forall x, y \epsilon\mathbb X; x \neg y \]

  7. zzr0ck3r
    • one year ago
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    that says the following: For all x and y in some space x, x is related to y.

  8. zzr0ck3r
    • one year ago
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    Do you know what your question is?

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

  10. zzr0ck3r
    • one year ago
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    Are you listening to me?

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

  12. zzr0ck3r
    • one year ago
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    ok man, I am gonna go then... this is pointless and a waste of time.

  13. anonymous
    • one year ago
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    they said i should bring out the odd one

  14. anonymous
    • one year ago
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    it is an option question

  15. anonymous
    • one year ago
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    i am only trying to list all the options sir

  16. anonymous
    • one year ago
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    @zzr0ck3r

  17. zzr0ck3r
    • one year ago
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    for the 5th time, what is the question?

  18. zzr0ck3r
    • one year ago
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    and please dont just retype what you already typed...

  19. anonymous
    • one year ago
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    they defined the Hausdorff space and gave some properties , asking me to point out the odd property

  20. zzr0ck3r
    • one year ago
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    do you mean \(U_x \bigcap U-\{y\} = \phi\) ?

  21. zzr0ck3r
    • one year ago
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    or \(U_x \bigcap U_y-\{y\} = \phi\)

  22. anonymous
    • one year ago
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    yes and i know that option A and B are properties of Hausdorff space but the C and D is where u am confused

  23. anonymous
    • one year ago
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    the last option is \[\forall x, y \epsilon X, x \bigcap y = 0 \]

  24. anonymous
    • one year ago
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    ok, i think option C is a typo error

  25. zzr0ck3r
    • one year ago
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    Is there a reason you are not using parentheses when I keep asking you to?

  26. anonymous
    • one year ago
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    may be they wanted to state\[ U_x \bigcap U_y = \phi \]

  27. zzr0ck3r
    • one year ago
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    \(\forall x, y \epsilon\mathbb X; x \neg y\) makes no sense.

  28. anonymous
    • one year ago
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    sir, i am so sorry about the parentheses but that was how the question came

  29. zzr0ck3r
    • one year ago
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    But then you say yes when I say "should it be like this"

  30. zzr0ck3r
    • one year ago
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    ok they all make sense except the equivalence one. \(\forall x, y \epsilon\mathbb X; x \neg y\)

  31. zzr0ck3r
    • one year ago
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    I again, don't see how you are learning here. but what ever...

  32. anonymous
    • one year ago
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    what about the last option. is it a property of Hausdorff space?

  33. anonymous
    • one year ago
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    i mean this option \[∀x,yϵX,x⋂y=0 \]

  34. zzr0ck3r
    • one year ago
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    sort of, it should say \(x\ne y\)

  35. zzr0ck3r
    • one year ago
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    \(∀x,yϵX\text{ where } x\ne y, \{x\}⋂\{y\}=0\)

  36. anonymous
    • one year ago
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    ok, so sorry again sir for the parenthesis , its because that the way it came . i am so sorry

  37. anonymous
    • one year ago
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    A topological space X satisfies the first separation axiom T_{1 } if each one of any two points of X has a nbd that does not contain the other point. Thus,X is called a T_{1} - space otherwise known as what?

  38. anonymous
    • one year ago
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    i think it is Hausdorff Space. am i correct ? @zzr0ck3r

  39. zzr0ck3r
    • one year ago
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    no, it is not necessarily Hausdorff. I think they sometimes call \(T_1\) "accessible space". But I have never heard it referred to as anything but \(T_1\).

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