i have another question in metric topology.

- anonymous

i have another question in metric topology.

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

let E=R be endowed with the euclidean metric \[d {2}(X,Y)= \sum_{k=1}^{2}(Xk -Yk)^{1/2} for all X=(x1,x2),y=(y1,y2)\in R ^{2}.. descibe the set B _{0}(0,0);1). and also (2) discribe the open ball B _{0}( (0,0);1)\]

- anonymous

@oldrin.bataku

- anonymous

@zzr0ck3r

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

@zzr0ck3r

- zzr0ck3r

I cant see the full question

- zzr0ck3r

describe the .....
It goes off the side of the page.

- anonymous

describe the set \[B _{0}( (0,0);1)

- zzr0ck3r

\(\{x\in \mathbb{R}^2 \mid \sqrt{x_1^2+x_2^2} <1\ \}\)

- zzr0ck3r

It is a circle around the origin of radius 1. We include everything in the circle, but not the "border" of the circle

- anonymous

2) describe the open ball \[B _{0}( (0,0);1)\]

- anonymous

The Euclidean metric is the normal metric you are used to. We are looking at the open set $$\{ (x,y) : x^2 + y^2 \le 1\}$$, which is just the set of things within distance \(1\) of our point. This is an open disk of radius \(1\)

- anonymous

Er the inequality should read less than

- zzr0ck3r

<1 @oldrin.bataku

- zzr0ck3r

word*

- anonymous

I'm on my phone and OpenStudy lags

- anonymous

Okay, it seems open balls of radius \(r\) centered at \(p\) are notated \(B_o(p;r)\)

- anonymous

so, what is the description?

- zzr0ck3r

yes, in general open is \(B(a,b)\) and closed is \(B[a,b]\)

- zzr0ck3r

When we give the answer as a set, that is a description
@GIL.ojei this one is very basic. Can you explain exactly what you are having a problem with.

- anonymous

you know my book did not explain that but this is just the question.

- anonymous

so, number the answers for me because it is two questions

- anonymous

I have never seen the closed ball notated as \(B[p;r]\), only as \(\bar B(p;r),\operatorname{cl}B(p;r),[B(p;r)]\), but I imagine it looks like \(B_c(p;r)\) in @GIL.ojei's book

- anonymous

ok . thank you sirs

- anonymous

show that the mapping f;R---.R+ defined by f(x)=e^x is homeomorphism

- anonymous

do you know what a homeomorphism is?

- anonymous

that is bijective funtiom f such that f and f^-1 are both continuous

- anonymous

it's bicontinuous, i.e. it's an invertible map that is continuous in both directions

- anonymous

what definition of continuity are you working with? sequential continuity? the Cauchy definition in terms of \(\epsilon,\delta\)?

- anonymous

its topological space question

- anonymous

yes, but are we dealing with \(\mathbb R,\mathbb R^+\) as metric spaces? there are several equivalent definitions of continuity in this case

- anonymous

any easy one will be ok

- anonymous

you need to tell us which one you are using, it's not a matter of one being easy or not :p

- anonymous

Cauchy definition

- zzr0ck3r

Which is?

- anonymous

@zzr0ck3r i do not know how to prove it

- zzr0ck3r

If it is a topological question. then I assume we are working with the definition pre image of open sets is open.

- zzr0ck3r

For sure it is a bijection. Do you know how to prove that?

- anonymous

no

- zzr0ck3r

onto: Let \(y\in \mathbb{R}^{+}\), then \(f(\ln(y))= e^{\ln(y)}=y\). Note that since \(y\in \mathbb{R}^{+}\) we have that \(\ln(y) \in \mathbb{R}\).
one-to-one. Suppose \(f(x) = f(y) \). Then \(e^x=e^y\) and as a result we have \(\ln(e^x)=\ln(e^y)\implies x=y\).
The function is a bijection. Do you follow?

- anonymous

yes

- anonymous

so , that is the prove

- zzr0ck3r

This proves that it is a bijection. We now need to show that \(\ln(x)\) is continuous on its domain. Also I guess you might want to show that \(e^x\) is continuous on \(\mathbb{R}\).

- zzr0ck3r

Do you need to show they are continuous?

- zzr0ck3r

Or just go from the known fact that they both are?

- anonymous

just go from the known fact that they both are

- zzr0ck3r

Then you are done.

- zzr0ck3r

You can google how to prove they are continuous, and if I prove it, it will look exactly like any proof you find online.

- anonymous

waw

- zzr0ck3r

what is waw? I see you make that comment before.

- anonymous

i mean you are a genius

- zzr0ck3r

not at all...

- anonymous

i wish i am good like you

- zzr0ck3r

keep at it.
Go back and look at the questions I asked 2 years ago. They are the same questions you are asking

- zzr0ck3r

1 year ago*

- anonymous

please show they are continuous but before then, the last quation for discribing a set , which is open ball and which is set B_{0}

- anonymous

?

- zzr0ck3r

I am not sure if they want a description like you and I think of when we describe things, or a mathematical description which is a term we use to when referring to a set and a relation on that set.
Case 1: visual description: Consider a circle at the origin with radius 1, it is all the points inside the circle.
|dw:1438991405420:dw|
In this picture, it is all the "white" inside the circle.
Case 2: Mathematical description: \(B_0((0,0), 1)=\{(x_1, x_2)\in \mathbb{R}^2 \mid x_1^2+x_2^2<1\}\)

- zzr0ck3r

Here is a good read on showing things are continuous using \(\epsilon-\delta\) proofs.
http://people.bath.ac.uk/sej20/docs/epsilondelta.pdf
Let me know if you have any questions.

- anonymous

ok. got it . i have another quation

- zzr0ck3r

close this and ask a new one.

- anonymous

ok

Looking for something else?

Not the answer you are looking for? Search for more explanations.