Question

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

Answer 1

@Luigi0210 @UnkleRhaukus

Answer 2

the question is incomplete

.... Show that if {A}, and {B}. ?

Answer 3

No it's complete

Answer 4

could you please, check the link again

Answer 5

i think you've cropper out some words after


".... and discontinues at each point of [ something missing]
Q n (0,1)"

Answer 6

No I checked it again with the paper...

Answer 7

I can understand the proposition and I feel it's correct... But I don't know how to convert those feelings into mathematical language..

Answer 8

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....

Answer 9

@Loser66

Answer 10

@ikram002p

Answer 11

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)

Answer 12

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

Answer 13

wait a sec

Answer 14

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)

Answer 15

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

Answer 16

hmmm

Answer 17

does I represent imaginary ?

Answer 18

and Q rational number ?

Answer 19

I-irrational Q-rational

Answer 20

hehehe ok

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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