amilapsn
  • amilapsn
This is about real analysis continuity: (jpeg below in comments)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amilapsn
  • amilapsn
1 Attachment
amilapsn
  • amilapsn
@Luigi0210 @UnkleRhaukus
UnkleRhaukus
  • UnkleRhaukus
the question is incomplete .... Show that if {A}, and {B}. ?

Looking for something else?

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

More answers

amilapsn
  • amilapsn
No it's complete
amilapsn
  • amilapsn
could you please, check the link again
UnkleRhaukus
  • UnkleRhaukus
i think you've cropper out some words after ".... and discontinues at each point of [ something missing] Q n (0,1)"
amilapsn
  • amilapsn
No I checked it again with the paper...
amilapsn
  • amilapsn
I can understand the proposition and I feel it's correct... But I don't know how to convert those feelings into mathematical language..
amilapsn
  • amilapsn
We 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....
amilapsn
  • amilapsn
@Loser66
amilapsn
  • amilapsn
@ikram002p
ikram002p
  • ikram002p
ur 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
  • ikram002p
huh i need ur definition or i'll show u how with mine
amilapsn
  • amilapsn
wait a sec
amilapsn
  • amilapsn
our 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
  • ikram002p
can u take snap for example ? idk what do u mean but getting contradiction with assuming there is a limit
amilapsn
  • amilapsn
hmmm
ikram002p
  • ikram002p
does I represent imaginary ?
ikram002p
  • ikram002p
and Q rational number ?
amilapsn
  • amilapsn
I-irrational Q-rational
ikram002p
  • ikram002p
hehehe ok
ikram002p
  • ikram002p
|dw:1433147116393:dw|
ikram002p
  • ikram002p
for 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
  • amilapsn
shouldn't we have to define (c.d)?
ikram002p
  • ikram002p
yes 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
  • amilapsn
but how can we explain about the rationals contained in (c,d)?
ikram002p
  • ikram002p
we dont need them see (c,d) is subset of I and (0,1) so it does not have rational at all
amilapsn
  • amilapsn
can we make an interval without rationals?
ikram002p
  • ikram002p
yes sure this is the main idea of the question :)
amilapsn
  • amilapsn
Do we have to prove that we can make an interval without rationals?
ikram002p
  • ikram002p
its given in the question :O I and (0,1) means all real btw 0 and 1 without rational
amilapsn
  • amilapsn
no no I'm talking about the interval (c,d)
amilapsn
  • amilapsn
although 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)
amilapsn
  • amilapsn
@ikram002p
amilapsn
  • amilapsn
@ganeshie8
ikram002p
  • ikram002p
well 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
  • ikram002p
i 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
  • ganeshie8
For 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
  • ikram002p
oh that really works
ganeshie8
  • ganeshie8
yeah that works because rational numbers are dense or something
ikram002p
  • ikram002p
hehehe yeah woah :O
amilapsn
  • amilapsn
@ganeshie8 thanks... Would telling that be sufficient in proving our second proposition rigorously?
ganeshie8
  • ganeshie8
Yes, I think that is a rigorous proof for second part.
ganeshie8
  • ganeshie8
may be add this also : Such a sequence \(\{x_n\}\) exists because rational numbers are dense in reals.
amilapsn
  • amilapsn
I 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
  • ikram002p
plz medal ganesh for second idea he deserves it :)
amilapsn
  • amilapsn
You know better than me, I'll do it. :)
amilapsn
  • amilapsn
\[\href{https://en.wikipedia.org/wiki/Radix}{hey}\]
amilapsn
  • amilapsn
\[\href{ hey }{ https://en.wikipedia.org/wiki/Radix}\]

Looking for something else?

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