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anonymous
 one year ago
i have another question in metric topology.
anonymous
 one year ago
i have another question in metric topology.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let E=R be endowed with the euclidean metric \[d {2}(X,Y)= \sum_{k=1}^{2}(Xk Yk)^{1/2} for all X=(x1,x2),y=(y1,y2)\in R ^{2}.. descibe the set B _{0}(0,0);1). and also (2) discribe the open ball B _{0}( (0,0);1)\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I cant see the full question

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2describe the ..... It goes off the side of the page.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0describe the set \[B _{0}( (0,0);1)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(\{x\in \mathbb{R}^2 \mid \sqrt{x_1^2+x_2^2} <1\ \}\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2It is a circle around the origin of radius 1. We include everything in the circle, but not the "border" of the circle

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02) describe the open ball \[B _{0}( (0,0);1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0Er the inequality should read less than

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm on my phone and OpenStudy lags

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, it seems open balls of radius \(r\) centered at \(p\) are notated \(B_o(p;r)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, what is the description?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2yes, in general open is \(B(a,b)\) and closed is \(B[a,b]\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2When 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
 one year ago
Best ResponseYou've already chosen the best response.0you know my book did not explain that but this is just the question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, number the answers for me because it is two questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0show that the mapping f;R.R+ defined by f(x)=e^x is homeomorphism

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know what a homeomorphism is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is bijective funtiom f such that f and f^1 are both continuous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's bicontinuous, i.e. it's an invertible map that is continuous in both directions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what definition of continuity are you working with? sequential continuity? the Cauchy definition in terms of \(\epsilon,\delta\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its topological space question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, but are we dealing with \(\mathbb R,\mathbb R^+\) as metric spaces? there are several equivalent definitions of continuity in this case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0any easy one will be ok

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need to tell us which one you are using, it's not a matter of one being easy or not :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r i do not know how to prove it

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2If it is a topological question. then I assume we are working with the definition pre image of open sets is open.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2For sure it is a bijection. Do you know how to prove that?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2onto: Let \(y\in \mathbb{R}^{+}\), then \(f(\ln(y))= e^{\ln(y)}=y\). Note that since \(y\in \mathbb{R}^{+}\) we have that \(\ln(y) \in \mathbb{R}\). onetoone. 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
 one year ago
Best ResponseYou've already chosen the best response.0so , that is the prove

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2This 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
 one year ago
Best ResponseYou've already chosen the best response.2Do you need to show they are continuous?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Or just go from the known fact that they both are?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just go from the known fact that they both are

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2You can google how to prove they are continuous, and if I prove it, it will look exactly like any proof you find online.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what is waw? I see you make that comment before.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean you are a genius

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wish i am good like you

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2keep at it. Go back and look at the questions I asked 2 years ago. They are the same questions you are asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please show they are continuous but before then, the last quation for discribing a set , which is open ball and which is set B_{0}

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2Here 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
 one year ago
Best ResponseYou've already chosen the best response.0ok. got it . i have another quation

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2close this and ask a new one.
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