please help @zzrock3r

- anonymous

please help @zzrock3r

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

@zzr0ck3r

- anonymous

Dully noted, there is no question here.

- anonymous

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

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## More answers

- anonymous

lol

- anonymous

##### 1 Attachment

- anonymous

Okay so that is password protected.

- anonymous

whats the password?!

- anonymous

help us. help you!

- anonymous

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

- anonymous

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

- anonymous

@pooja195

- anonymous

@zzr0ck3r

- zzr0ck3r

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

- anonymous

what have i done wrong ? i am so sorry

- zzr0ck3r

?

- anonymous

i know you are angry

- anonymous

please just forgive .

- zzr0ck3r

lol why would I be angry?

- zzr0ck3r

I just said to ask the question :)

- anonymous

the question is in the file ,

- anonymous

and the password is osho

- 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

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

(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

@zzr0ck3r

- zzr0ck3r

What is your definition of open?

- 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

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

- zzr0ck3r

How do you define a boundary point?

- zzr0ck3r

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

- zzr0ck3r

What would it mean if it did?

- 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

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

- 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

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

- anonymous

yes

- anonymous

but should it be out of a?

- 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

it means it is close

- anonymous

right?

- zzr0ck3r

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

- 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

##### 1 Attachment

- anonymous

it means it is have open and half close

- zzr0ck3r

what is?

- zzr0ck3r

I am here now.

- 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

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

- zzr0ck3r

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

- 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

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

- zzr0ck3r

Feel free to ask me anything

- 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

@zzr0ck3r

- zzr0ck3r

any \(y,z\in?\)

- anonymous

of R

- zzr0ck3r

Suppose w.l.o.g. that \(x

- zzr0ck3r

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

- anonymous

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

- anonymous

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

- anonymous

@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

i am trying to know what you did

- Michele_Laino

please see the complete solution @GIL.ojei

##### 1 Attachment

- 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

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

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

- Michele_Laino

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

- anonymous

yes, i know that is true

- Michele_Laino

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

- anonymous

yes

- 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

yes

- 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

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

- 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

ok. but i thought you have applied that already

- 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

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

- anonymous

are you saying that we ought to have four proves?

- 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

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

- Michele_Laino

that's all!

- anonymous

which is the main prove ?

- anonymous

because all you stated are corect

- anonymous

ok. now i understand sir

- Michele_Laino

:)

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