This is about real analysis continuity:
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- amilapsn

This is about real analysis continuity:
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- schrodinger

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- amilapsn

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- amilapsn

@Luigi0210 @UnkleRhaukus

- UnkleRhaukus

the question is incomplete
.... Show that if {A}, and {B}. ?

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## More answers

- amilapsn

No it's complete

- amilapsn

could you please, check the link again

- UnkleRhaukus

i think you've cropper out some words after
".... and discontinues at each point of [ something missing]
Q n (0,1)"

- amilapsn

No I checked it again with the paper...

- 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

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

@Loser66

- amilapsn

@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

huh i need ur definition or i'll show u how with mine

- amilapsn

wait a sec

- 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

can u take snap for example ?
idk what do u mean but getting contradiction with assuming there is a limit

- amilapsn

hmmm

- ikram002p

does I represent imaginary ?

- ikram002p

and Q rational number ?

- amilapsn

I-irrational Q-rational

- ikram002p

hehehe ok

- ikram002p

|dw:1433147116393:dw|

- 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

shouldn't we have to define (c.d)?

- 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

but how can we explain about the rationals contained in (c,d)?

- ikram002p

we dont need them see (c,d) is subset of I and (0,1) so it does not have rational at all

- amilapsn

can we make an interval without rationals?

- ikram002p

yes sure this is the main idea of the question :)

- amilapsn

Do we have to prove that we can make an interval without rationals?

- ikram002p

its given in the question :O
I and (0,1) means all real btw 0 and 1 without rational

- amilapsn

no no I'm talking about the interval (c,d)

- 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

@ikram002p

- amilapsn

@ganeshie8

- 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

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

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

oh that really works

- ganeshie8

yeah that works because rational numbers are dense or something

- ikram002p

hehehe yeah woah :O

- amilapsn

@ganeshie8 thanks... Would telling that be sufficient in proving our second proposition rigorously?

- ganeshie8

Yes, I think that is a rigorous proof for second part.

- ganeshie8

may be add this also :
Such a sequence \(\{x_n\}\) exists because rational numbers are dense in reals.

- 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

plz medal ganesh for second idea he deserves it :)

- amilapsn

You know better than me, I'll do it. :)

- amilapsn

\[\href{https://en.wikipedia.org/wiki/Radix}{hey}\]

- amilapsn

\[\href{
hey
}{ https://en.wikipedia.org/wiki/Radix}\]

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