anonymous
  • anonymous
please help @zzrock3r
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
Dully noted, there is no question here.
anonymous
  • anonymous
hmmm -.- what did you do to the question! put your pencils up!

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anonymous
  • anonymous
lol
anonymous
  • anonymous
anonymous
  • anonymous
Okay so that is password protected.
anonymous
  • anonymous
whats the password?!
anonymous
  • anonymous
help us. help you!
anonymous
  • anonymous
i need helo on those questions 2,3,4 and the password is osho
anonymous
  • anonymous
ow! my head HuRtS! i dont know.. sorry
anonymous
  • anonymous
@pooja195
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
I am not sure how to open that, nor do I really want to. But just ask the question here.
anonymous
  • anonymous
what have i done wrong ? i am so sorry
zzr0ck3r
  • zzr0ck3r
?
anonymous
  • anonymous
i know you are angry
anonymous
  • anonymous
please just forgive .
zzr0ck3r
  • zzr0ck3r
lol why would I be angry?
zzr0ck3r
  • zzr0ck3r
I just said to ask the question :)
anonymous
  • anonymous
the question is in the file ,
anonymous
  • anonymous
and the password is osho
zzr0ck3r
  • 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?
zzr0ck3r
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
What is your definition of open?
anonymous
  • 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.
anonymous
  • anonymous
an open interval is an open set. interval without its boundary points
zzr0ck3r
  • zzr0ck3r
How do you define a boundary point?
zzr0ck3r
  • zzr0ck3r
we must show that your ball does not contain any boundary points
zzr0ck3r
  • zzr0ck3r
What would it mean if it did?
anonymous
  • anonymous
boundary point to me is the being or the end of a thing . in inequality, it is denoted by < or =, and > or = , or [ ]
zzr0ck3r
  • zzr0ck3r
well you are kind of defining what we are asked to prove.
zzr0ck3r
  • 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 \(B_{\delta}(x)\cap A^C\ne \emptyset\)
zzr0ck3r
  • zzr0ck3r
Does this make sense? This says for any neighborhood of \(x\) we have a point in A and a point out of A.
anonymous
  • anonymous
yes
anonymous
  • anonymous
but should it be out of a?
zzr0ck3r
  • zzr0ck3r
Well what you want to show is that \(X\) contains non of its boundary points. So suppose it does. What does that mean?
anonymous
  • anonymous
it means it is close
anonymous
  • anonymous
right?
zzr0ck3r
  • zzr0ck3r
brb I need to take my wife to a party. I might be half hour or so.
zzr0ck3r
  • 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?
anonymous
  • anonymous
anonymous
  • anonymous
it means it is have open and half close
zzr0ck3r
  • zzr0ck3r
what is?
zzr0ck3r
  • zzr0ck3r
I am here now.
zzr0ck3r
  • 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.
zzr0ck3r
  • zzr0ck3r
Thus \(B(x, \epsilon)\) is open because it does NOT contain any of it's boundary points.
zzr0ck3r
  • zzr0ck3r
What does it mean for a metric to generate a topology?
zzr0ck3r
  • 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?
anonymous
  • anonymous
i know a bit of them but not that good at it
zzr0ck3r
  • zzr0ck3r
Feel free to ask me anything
anonymous
  • 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 ?
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
any \(y,z\in?\)
anonymous
  • anonymous
of R
zzr0ck3r
  • zzr0ck3r
Suppose w.l.o.g. that \(x
zzr0ck3r
  • zzr0ck3r
now you write up a similar proof for the min case, and show it here.
anonymous
  • anonymous
ok but please just help me so that i don't do any errors
anonymous
  • anonymous
but sir, why did you use x,y. thought the given function is y,z e R
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • 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?
anonymous
  • anonymous
i am trying to know what you did
Michele_Laino
  • Michele_Laino
please see the complete solution @GIL.ojei
1 Attachment
Michele_Laino
  • Michele_Laino
by definition of absolute value, we can write this: \[\Large \begin{gathered} y \geqslant z, \Rightarrow \left| {y - z} \right| = y - z \hfill \\ \hfill \\ y \leqslant z \Rightarrow \left| {y - z} \right| = - \left( {y - z} \right) = - y + z \hfill \\ \end{gathered} \]
Michele_Laino
  • 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.
Michele_Laino
  • Michele_Laino
as I said before, we have to distinguish two cases, since R is a totally ordered set. First case: \[y \geqslant z\]
Michele_Laino
  • Michele_Laino
then \[y - z \geqslant 0\], so by definition of absolute value, we get: \[\left| {y - z} \right| = y - z\]
anonymous
  • anonymous
yes, i know that is true
Michele_Laino
  • Michele_Laino
then I can replace \[\left| {y - z} \right|\] with \[y - z\]
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
so I can write this: \[\max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z + y - z} \right\} = \frac{{2y}}{2} = y\]
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
furthermore I can write this: \[\begin{gathered} \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( {y - z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now we have to consider the other case, namely: \[y \leqslant z\]
Michele_Laino
  • Michele_Laino
by definition of absolute value, we have: \[y - z \leqslant 0 \Rightarrow \left| {y - z} \right| = - \left( {y - z} \right) = - y + z\]
anonymous
  • anonymous
ok. but i thought you have applied that already
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
in other words, in this second case, we can replace: \[\left| {y - z} \right|\] with: \[ - y + z\]
anonymous
  • anonymous
are you saying that we ought to have four proves?
Michele_Laino
  • Michele_Laino
and then we can write this: \[\begin{gathered} \max \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z + \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z + \left( { - y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z - y + z} \right\} = \frac{{2z}}{2} = z \hfill \\ \hfill \\ \min \left( {y,z} \right) = \frac{1}{2}\left\{ {y + z - \left| {y - z} \right|} \right\} = \frac{1}{2}\left\{ {y + z - \left( { - y + z} \right)} \right\} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {y + z + y - z} \right\} = \frac{{2y}}{2} = y \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
no, I say that we have to prove your statement for two different cases for value of y-z
Michele_Laino
  • Michele_Laino
that's all!
anonymous
  • anonymous
which is the main prove ?
anonymous
  • anonymous
because all you stated are corect
anonymous
  • anonymous
ok. now i understand sir
Michele_Laino
  • Michele_Laino
:)

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