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anonymous
 one year ago
please help @zzrock3r
anonymous
 one year ago
please help @zzrock3r

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Dully noted, there is no question here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm . what did you do to the question! put your pencils up!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so that is password protected.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whats the password?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i need helo on those questions 2,3,4 and the password is osho

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ow! my head HuRtS! i dont know.. sorry

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I am not sure how to open that, nor do I really want to. But just ask the question here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what have i done wrong ? i am so sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know you are angry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please just forgive .

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1lol why would I be angry?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I just said to ask the question :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the question is in the file ,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the password is osho

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I 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?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1It 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(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.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1What is your definition of open?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02)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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0an open interval is an open set. interval without its boundary points

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1How do you define a boundary point?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1we must show that your ball does not contain any boundary points

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1What would it mean if it did?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0boundary point to me is the being or the end of a thing . in inequality, it is denoted by < or =, and > or = , or [ ]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1well you are kind of defining what we are asked to prove.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1A point \(x\) is a boundary point for set \(A\) if \(\forall\) \(\delta>0\) we have \(B_{\delta}(x)\cap A\ne \emptyset\) and \(B_{\delta}(x)\cap A^C\ne \emptyset\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Does this make sense? This says for any neighborhood of \(x\) we have a point in A and a point out of A.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but should it be out of a?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Well what you want to show is that \(X\) contains non of its boundary points. So suppose it does. What does that mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it means it is close

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1brb I need to take my wife to a party. I might be half hour or so.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Its only closed if it contains all of them. Suppose it contains one, then by the def I gave you, what does that mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it means it is have open and half close

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Suppose 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.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Thus \(B(x, \epsilon)\) is open because it does NOT contain any of it's boundary points.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1What does it mean for a metric to generate a topology?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1For 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know a bit of them but not that good at it

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Feel free to ask me anything

anonymous
 one year ago
Best ResponseYou've already chosen the best response.05(a) Prove that for any y, z ε , max(y, z) = ½[y + z+ yz], min(y, z) = ½[y + z yz]. can you try to prove that for me ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Suppose w.l.o.g. that \(x<y\). Then \[\max(x,y) = y = \dfrac{x}{2}+\dfrac{y}{2}\dfrac{x}{2}+\dfrac{y}{2}=\dfrac{1}{2}[x+yx+y]=\dfrac{1}{2}[x+y+yx]\] Similarly for \(\min\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1now you write up a similar proof for the min case, and show it here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok but please just help me so that i don't do any errors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but sir, why did you use x,y. thought the given function is y,z e R

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1The 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am trying to know what you did

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please see the complete solution @GIL.ojei

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1by definition of absolute value, we can write this: \[\Large \begin{gathered} y \geqslant z, \Rightarrow \left {y  z} \right = y  z \hfill \\ \hfill \\ y \leqslant z \Rightarrow \left {y  z} \right =  \left( {y  z} \right) =  y + z \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1furthermore, those subsequent expressions, namely "max" and "min" are defined for all elements of R, so there is no need for further restrictions.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1as I said before, we have to distinguish two cases, since R is a totally ordered set. First case: \[y \geqslant z\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then \[y  z \geqslant 0\], so by definition of absolute value, we get: \[\left {y  z} \right = y  z\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, i know that is true

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then I can replace \[\left {y  z} \right\] with \[y  z\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so I can write this: \[\max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left {y  z} \right} \right\} = \frac{1}{2}\left\{ {y + z + y  z} \right\} = \frac{{2y}}{2} = y\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1furthermore I can write this: \[\begin{gathered} \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z  \left {y  z} \right} \right\} = \frac{1}{2}\left\{ {y + z  \left( {y  z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z  y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we have to consider the other case, namely: \[y \leqslant z\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1by definition of absolute value, we have: \[y  z \leqslant 0 \Rightarrow \left {y  z} \right =  \left( {y  z} \right) =  y + z\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. but i thought you have applied that already

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since in this case is yz less or equal to zero, whereas in the first case is yz greater or equal to zero

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in other words, in this second case, we can replace: \[\left {y  z} \right\] with: \[  y + z\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you saying that we ought to have four proves?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and then we can write this: \[\begin{gathered} \max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left {y  z} \right} \right\} = \frac{1}{2}\left\{ {y + z + \left( {  y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z  y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \hfill \\ \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z  \left {y  z} \right} \right\} = \frac{1}{2}\left\{ {y + z  \left( {  y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z + y  z} \right\} = \frac{{2y}}{2} = y \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, I say that we have to prove your statement for two different cases for value of yz

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is the main prove ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because all you stated are corect

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. now i understand sir
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