This is about real analysis continuity: (jpeg below in comments)

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This is about real analysis continuity: (jpeg below in comments)

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

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