A community for students.
Here's the question you clicked on:
 0 viewing
amilapsn
 one year ago
This is about real analysis continuity:
(jpeg below in comments)
amilapsn
 one year ago
This is about real analysis continuity: (jpeg below in comments)

This Question is Closed

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0@Luigi0210 @UnkleRhaukus

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0the question is incomplete .... Show that if {A}, and {B}. ?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0could you please, check the link again

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i think you've cropper out some words after ".... and discontinues at each point of [ something missing] Q n (0,1)"

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0No I checked it again with the paper...

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0I can understand the proposition and I feel it's correct... But I don't know how to convert those feelings into mathematical language..

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0We just have to prove \(\Large\sf{\lim_{x\rightarrow a}{f(x)}=f(a) \ \forall\ a\in\mathbb{I}\cap(0,1) }\) and \(\Large\sf{\lim_{x\rightarrow a}f(x) }\) doesn't exist \(\Large\sf{\forall a\in\mathbb{Q}\cap(0,1)}\) The problem is I don't know how to put that mathematically....

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2ur try is correct, but without finding limts can u write the definition of continuous in ur book here ? i think its the one with " f is continuous at x_0 if for every open interval (a,b) in f(x) containing f(x_0) there is an open interval (c,d) in x(as whole) contains x_0 such that f is continuous" also f is discontinuous at x_0 if there is an open interval in(a,b) in f(x) contains f(x_0) such that for every open interval in x , x_0 in (c,d) f((c,d)) is not sub interval of (a,b)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2huh i need ur definition or i'll show u how with mine

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0our definition for continuity is just the delta epsilon definition for the limit... same idea as your definition... But there's no explicit definition for discontinuity for us as for you. We just have to assume there's a limit and get a contradiction...(contradiction method)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2can u take snap for example ? idk what do u mean but getting contradiction with assuming there is a limit

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2does I represent imaginary ?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2and Q rational number ?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0Iirrational Qrational

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433147116393:dw

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2for part a: we took irrational btw (0,1) let x_0 be any point in I and (0,1) and x_0 in (c,d) f(x_0) in {0} , and f((c,d)) in {0} according to the function definition so its continuous through its domain

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0shouldn't we have to define (c.d)?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2yes u should its the open interval which contains x_0 and subset of I and (a,b) u only have to write in in epsilon delta way

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0but how can we explain about the rationals contained in (c,d)?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2we dont need them see (c,d) is subset of I and (0,1) so it does not have rational at all

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0can we make an interval without rationals?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2yes sure this is the main idea of the question :)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0Do we have to prove that we can make an interval without rationals?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2its given in the question :O I and (0,1) means all real btw 0 and 1 without rational

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0no no I'm talking about the interval (c,d)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0although we took (c,d) as a subset of I and (0,1) don't we have to show that we "can" REALLY take an interval (c,d) out of the set I and (0,1)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2well if u wanna take c,d as subset of real then more work u need to disjoint rational :\ why u wanna do that when u already can define it as subset of (0,1) without rational ?

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2i have to go but for second part show a contradiction by search of continuity about cretin point for example take ,10/24 ,11/24 and 12/24

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3For showing \(f\) is discontinuous at each point of \(\mathbb{Q}\cap (0,1) \) : Consider a sequence \(\{x_n\}\) of irrational numbers in the interval \((0, 1)\) that converge to a rational number \(\frac{a}{b}\) in the interval \((0,1)\). Then, as \(x_n\to \frac{a}{b}\), we have \(f(x_n)\to 0\) because \(f(\text{irrational}) \) is \(0\). But \(f(\frac{a}{b}) = \frac{1}{b^2} \ne 0\) proving the discontinuity.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2oh that really works

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah that works because rational numbers are dense or something

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 thanks... Would telling that be sufficient in proving our second proposition rigorously?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, I think that is a rigorous proof for second part.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3may be add this also : Such a sequence \(\{x_n\}\) exists because rational numbers are dense in reals.

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand it... I think I have to study this section more.... After studying I'll medal one of you... Thanks @ikram002p and @ganeshie8 for your time and effort.. :)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.2plz medal ganesh for second idea he deserves it :)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0You know better than me, I'll do it. :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.