1. anonymous

@zzr0ck3r

2. anonymous

Dully noted, there is no question here.

3. anonymous

hmmm -.- what did you do to the question! put your pencils up!

4. anonymous

lol

5. anonymous

6. anonymous

Okay so that is password protected.

7. anonymous

8. anonymous

9. anonymous

i need helo on those questions 2,3,4 and the password is osho

10. anonymous

ow! my head HuRtS! i dont know.. sorry

11. anonymous

@pooja195

12. anonymous

@zzr0ck3r

13. zzr0ck3r

I am not sure how to open that, nor do I really want to. But just ask the question here.

14. anonymous

what have i done wrong ? i am so sorry

15. zzr0ck3r

?

16. anonymous

i know you are angry

17. anonymous

18. zzr0ck3r

lol why would I be angry?

19. zzr0ck3r

I just said to ask the question :)

20. anonymous

the question is in the file ,

21. anonymous

22. zzr0ck3r

I don't know how to open the file, and I don't know what the file is, and I don't like opening things when I don't know where they come from. If it is a math question, there is no reason you can't post it here?

23. zzr0ck3r

It does not ask for a password when I try and open it, it says its corrupt or protected, but there is no place for a password.

24. anonymous

(a) Let X ε Rn. Show that the set B(X, ε ) is open. (b) Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that On is not empty.

25. anonymous

@zzr0ck3r

26. zzr0ck3r

What is your definition of open?

27. anonymous

2)Let (X, d) and (Y, d) be metric spaces and f a mapping of X into Y. Let τ1 and τ2 be the topologies determined by d and d1 respectively. Then f(X, τ) (y, τ) is continuous if and only if ; that is if x1, x2, . . . , xn, . . . , is a sequence of points in (X, d) converging to x, show that the sequence of points f(x1), f(x2), . . . , f(xn), . . . in (Y, d) converges to x.

28. anonymous

an open interval is an open set. interval without its boundary points

29. zzr0ck3r

How do you define a boundary point?

30. zzr0ck3r

we must show that your ball does not contain any boundary points

31. zzr0ck3r

What would it mean if it did?

32. anonymous

boundary point to me is the being or the end of a thing . in inequality, it is denoted by < or =, and > or = , or [ ]

33. zzr0ck3r

well you are kind of defining what we are asked to prove.

34. zzr0ck3r

A point $$x$$ is a boundary point for set $$A$$ if $$\forall$$ $$\delta>0$$ we have $$B_{\delta}(x)\cap A\ne \emptyset$$ and $$B_{\delta}(x)\cap A^C\ne \emptyset$$

35. zzr0ck3r

Does this make sense? This says for any neighborhood of $$x$$ we have a point in A and a point out of A.

36. anonymous

yes

37. anonymous

but should it be out of a?

38. zzr0ck3r

Well what you want to show is that $$X$$ contains non of its boundary points. So suppose it does. What does that mean?

39. anonymous

it means it is close

40. anonymous

right?

41. zzr0ck3r

brb I need to take my wife to a party. I might be half hour or so.

42. zzr0ck3r

Its only closed if it contains all of them. Suppose it contains one, then by the def I gave you, what does that mean?

43. anonymous

44. anonymous

it means it is have open and half close

45. zzr0ck3r

what is?

46. zzr0ck3r

I am here now.

47. zzr0ck3r

Suppose to the contrary that $$B(x, \epsilon)$$ contains one of its boundary points, lets call it $$a$$. Then $$a\in B(x,\epsilon)$$. Now consider $$\delta=\min(d(a, x), d(a, \epsilon))$$. Then $$B(a, \delta)\subseteq B(x, \epsilon)$$ contradicting the assumption that $$x$$ was a boundary point.

48. zzr0ck3r

Thus $$B(x, \epsilon)$$ is open because it does NOT contain any of it's boundary points.

49. zzr0ck3r

What does it mean for a metric to generate a topology?

50. zzr0ck3r

For you to understand the second question. You need to know what a metric space is and what does it mean to generate a topology from a metric, in order to understand that you will need to understand what a topology is, and what a basis and sub basis are for a topological space and what it means to be an arbitrary union, in order to understand that you will need to understand what a ray is. Then you will need to know what it means for a sequence of points in some product space to converge, and you must understand the metric definition of continuity, which means you must understand a metric. Do you understand all of these things?

51. anonymous

i know a bit of them but not that good at it

52. zzr0ck3r

Feel free to ask me anything

53. anonymous

5(a) Prove that for any y, z ε , max(y, z) = ½[y + z+ |y-z|], min(y, z) = ½[y + z- |y-z|]. can you try to prove that for me ?

54. anonymous

@zzr0ck3r

55. zzr0ck3r

any $$y,z\in?$$

56. anonymous

of R

57. zzr0ck3r

Suppose w.l.o.g. that $$x<y$$. Then $\max(x,y) = y = \dfrac{x}{2}+\dfrac{y}{2}-\dfrac{x}{2}+\dfrac{y}{2}=\dfrac{1}{2}[x+y-x+y]=\dfrac{1}{2}[x+y+|y-x|]$ Similarly for $$\min$$

58. zzr0ck3r

now you write up a similar proof for the min case, and show it here.

59. anonymous

ok but please just help me so that i don't do any errors

60. anonymous

but sir, why did you use x,y. thought the given function is y,z e R

61. anonymous

@zzr0ck3r

62. zzr0ck3r

The fact that you don't know that it does not matter if I call it $$z$$ or $$x$$ really scares me and again begs the question. What are you doing?

63. anonymous

i am trying to know what you did

64. Michele_Laino

please see the complete solution @GIL.ojei

65. Michele_Laino

by definition of absolute value, we can write this: $\Large \begin{gathered} y \geqslant z, \Rightarrow \left| {y - z} \right| = y - z \hfill \\ \hfill \\ y \leqslant z \Rightarrow \left| {y - z} \right| = - \left( {y - z} \right) = - y + z \hfill \\ \end{gathered}$

66. Michele_Laino

furthermore, those subsequent expressions, namely "max" and "min" are defined for all elements of R, so there is no need for further restrictions.

67. Michele_Laino

as I said before, we have to distinguish two cases, since R is a totally ordered set. First case: $y \geqslant z$

68. Michele_Laino

then $y - z \geqslant 0$, so by definition of absolute value, we get: $\left| {y - z} \right| = y - z$

69. anonymous

yes, i know that is true

70. Michele_Laino

then I can replace $\left| {y - z} \right|$ with $y - z$

71. anonymous

yes

72. Michele_Laino

so I can write this: $\max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z + y - z} \right\} = \frac{{2y}}{2} = y$

73. anonymous

yes

74. Michele_Laino

furthermore I can write this: $\begin{gathered} \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( {y - z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \end{gathered}$

75. Michele_Laino

now we have to consider the other case, namely: $y \leqslant z$

76. Michele_Laino

by definition of absolute value, we have: $y - z \leqslant 0 \Rightarrow \left| {y - z} \right| = - \left( {y - z} \right) = - y + z$

77. anonymous

ok. but i thought you have applied that already

78. Michele_Laino

no, since in this case is y-z less or equal to zero, whereas in the first case is y-z greater or equal to zero

79. Michele_Laino

in other words, in this second case, we can replace: $\left| {y - z} \right|$ with: $- y + z$

80. anonymous

are you saying that we ought to have four proves?

81. Michele_Laino

and then we can write this: $\begin{gathered} \max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z + \left( { - y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \hfill \\ \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( { - y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z + y - z} \right\} = \frac{{2y}}{2} = y \hfill \\ \end{gathered}$

82. Michele_Laino

no, I say that we have to prove your statement for two different cases for value of y-z

83. Michele_Laino

that's all!

84. anonymous

which is the main prove ?

85. anonymous

because all you stated are corect

86. anonymous

ok. now i understand sir

87. Michele_Laino

:)