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anonymous

  • one year ago

A topological space X satifies the first separation axiom T_{1 }if each one of any two points of X has a nbd that does not containthe other point. Thus,X is called a T_{1} - space otherwise known as ______________ Metric Space Hausdorff Space Frechet Space Topological Space

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  1. anonymous
    • one year ago
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    i think frechet space

  2. anonymous
    • one year ago
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    @zzr0ck3r

  3. zzr0ck3r
    • one year ago
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    correct

  4. anonymous
    • one year ago
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    One of the conditions below is not a condition fot a topological space X to satisfy the first separation axiom If and only if all one point set in X is closed If and only if all one point set in X is open If and only if every finite set in is closed If and only if for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y}

  5. anonymous
    • one year ago
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    @zzr0ck3r

  6. zzr0ck3r
    • one year ago
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    chill, reading

  7. zzr0ck3r
    • one year ago
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    1) and 3) are true, I cant tell what you mean by the last line

  8. anonymous
    • one year ago
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    \[Iff for x, y \epsilon X x\epsilon U_{x}and y\epsilon U_{y} \] the last statement

  9. zzr0ck3r
    • one year ago
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    \[\forall x,y\in X \ x\in U_x\text{ and } y\in U_y\] What are \(U_x, U_y\) ?

  10. zzr0ck3r
    • one year ago
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    It has to be the open one.

  11. zzr0ck3r
    • one year ago
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    If every one point set was open, then we would just get the discrete topology every time, and thus the T_1 definition would be redundant.

  12. zzr0ck3r
    • one year ago
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    What open sets?

  13. zzr0ck3r
    • one year ago
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    \[\epsilon \neq \in\]

  14. zzr0ck3r
    • one year ago
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    Don't une `\epsilon` to mean "in" :)

  15. anonymous
    • one year ago
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    Ux and Uy are open sets not given the open sets sir . ok. what about option 2. is it right?

  16. anonymous
    • one year ago
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    thanks for tell me that sir

  17. zzr0ck3r
    • one year ago
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    2) is not true

  18. zzr0ck3r
    • one year ago
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    unless it is finite.

  19. zzr0ck3r
    • one year ago
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    lol

  20. anonymous
    • one year ago
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    Let X and Y be topological spaces; let f: X\rightarrow Y be a bijection. If both f and its inverse f^{-1} are continuous, then f is called _____________________ Epimorphism Homomorphism Homeomorphism Endomorpism

  21. zzr0ck3r
    • one year ago
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    homeomorphism

  22. zzr0ck3r
    • one year ago
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    homomorphism is a group concept

  23. anonymous
    • one year ago
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    sorry C

  24. zzr0ck3r
    • one year ago
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    correct

  25. 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 with these properties except one.

  26. anonymous
    • one year ago
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    sorry wrong option sir

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

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

  29. anonymous
    • one year ago
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    c \[ \forall x, y \epsilon X, x \bigcap y = 0 \]

  30. zzr0ck3r
    • one year ago
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    what is \(\neg\)?

  31. anonymous
    • one year ago
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    i think they want to confuse me with that or its an error

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

  33. anonymous
    • one year ago
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    to me C is true

  34. anonymous
    • one year ago
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    B is also true

  35. anonymous
    • one year ago
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    but a and d, i am confuse

  36. zzr0ck3r
    • one year ago
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    these do not make sense The first one: What the hell is \(\neg\) The second one: \(U_x\in N(x)\) makes no sense, Sets are not elements in the neighborhood, they are subsets if anything. The third one: Are we to assume \(x\ne y\)? Because if they are equal then this is of course true. The last one should say \(U_x\cap U_y=\emptyset\) and this is true.

  37. anonymous
    • one year ago
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    ok sir, what if option A was \[If \forall x, y \epsilon\mathbb X; x \rightarrow y \] is it correct?

  38. zzr0ck3r
    • one year ago
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    can you take a screen shot?

  39. zzr0ck3r
    • one year ago
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    you latex work needs some love :)

  40. anonymous
    • one year ago
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    lol

  41. anonymous
    • one year ago
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    i will but its not always clear

  42. zzr0ck3r
    • one year ago
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    here is one more thing for latest. Dont use words in the math part Like if the following: If `\(x\in X\)` then we have `\(\forall x \)`

  43. zzr0ck3r
    • one year ago
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    keep the text outside of the inline math. or else it ill look like this \(if x\in X then \forall x\) Which was given by the following code `\(if x\in X then \forall x\)`

  44. zzr0ck3r
    • one year ago
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    or put the text in `\text{stuff here}` Example \(\text{if }x\in X\text{ then } \forall x\)

  45. zzr0ck3r
    • one year ago
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    ok close this and ask a new one if you have something else.

  46. zzr0ck3r
    • one year ago
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    tag me if you need me...

  47. anonymous
    • one year ago
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    There exists U_x \epsilon N(x). U_y \epsilon N(y) here is how option B is

  48. anonymous
    • one year ago
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    @zzr0ck3r

  49. anonymous
    • one year ago
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    is epsilon not same as in?

  50. zzr0ck3r
    • one year ago
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    nope. epsilon is a greek letter, in is just means ... in :)

  51. zzr0ck3r
    • one year ago
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    That option makes no sense...

  52. anonymous
    • one year ago
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    it din't show fully but thats how it is.ok. thank you sir

  53. zzr0ck3r
    • one year ago
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    or it is trivially true. can you say in normal words what that option says?

  54. zzr0ck3r
    • one year ago
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    Like, generally \(U_X\) means the following: An open set \(U\) that contains \(x\). But this is always true in a topological space..

  55. anonymous
    • one year ago
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    hmmm

  56. zzr0ck3r
    • one year ago
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    Are you on windows?

  57. anonymous
    • one year ago
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    but, so, that option is wrong since a set containing X can not be in the neighborhood of x

  58. anonymous
    • one year ago
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    no sir.

  59. zzr0ck3r
    • one year ago
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    Why cant a set containing \(x\) not be in \(N(x)\)?

  60. zzr0ck3r
    • one year ago
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    \(N(x)\subseteq N(x)\). Unless you mean \(\in\) in which case it is weird....

  61. anonymous
    • one year ago
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    N(x) means elements around X and not sets around X. i might be wrong

  62. anonymous
    • one year ago
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    ok. can we try another?

  63. zzr0ck3r
    • one year ago
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    yep, yes then, if you mean \(\in\) and not \(\epsilon\), then yes it does not make sense. But it actually could This will be a little weird but you elements could actually be sets. \(\{x, \{x\}\}\) is a thing. and \(\{x\}\in \{x,\{x\}\}\) But lets move on.

  64. zzr0ck3r
    • one year ago
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    close this please, it takes to long to scroll on this pc

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