Question

please help @zzrock3r

Answer 1

@zzr0ck3r

Answer 2

Dully noted, there is no question here.

Answer 3

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

Answer 4

lol

Answer 5

Okay so that is password protected.

Answer 6

whats the password?!

Answer 7

help us. help you!

Answer 8

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

Answer 9

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

Answer 10

@pooja195

Answer 11

@zzr0ck3r

Answer 12

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

Answer 13

what have i done wrong ? i am so sorry

Answer 14

?

Answer 15

i know you are angry

Answer 16

please just forgive .

Answer 17

lol why would I be angry?

Answer 18

I just said to ask the question :)

Answer 19

the question is in the file ,

Answer 20

and the password is osho

Answer 21

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?

Answer 22

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.

Answer 23

(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.

Answer 24

@zzr0ck3r

Answer 25

What is your definition of open?

Answer 26

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.

Answer 27

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

Answer 28

How do you define a boundary point?

Answer 29

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

Answer 30

What would it mean if it did?

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

yes

Answer 36

but should it be out of a?

Answer 37

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

Answer 38

it means it is close

Answer 39

right?

Answer 40

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

Answer 41

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

Answer 42

it means it is have open and half close

Answer 43

what is?

Answer 44

I am here now.

Answer 45

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.

Answer 46

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

Answer 47

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

Answer 48

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?

Answer 49

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

Answer 50

Feel free to ask me anything

Answer 51

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 ?

Answer 52

@zzr0ck3r

Answer 53

any \(y,z\in?\)

Answer 54

of R

Answer 55

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 \(\min\)

Answer 56

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

Answer 57

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

Answer 58

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

Answer 59

@zzr0ck3r

Answer 60

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?

Answer 61

i am trying to know what you did

Answer 62

please see the complete solution @GIL.ojei

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

yes, i know that is true

Answer 68

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

Answer 69

yes

Answer 70

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

Answer 71

yes

Answer 72

furthermore I can write this:

\[\begin{gathered}
\min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( {y - z} \right)} \right\} = \hfill \\
\hfill \\
= \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\
\end{gathered} \]

Answer 73

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

Answer 74

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

Answer 75

ok. but i thought you have applied that already

Answer 76

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

Answer 77

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

Answer 78

are you saying that we ought to have four proves?

Answer 79

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

Answer 80

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

Answer 81

that's all!

Answer 82

which is the main prove ?

Answer 83

because all you stated are corect

Answer 84

ok. now i understand sir

Answer 85

:)