1 if x rational, 0 if x irrational?

Question

so can someone go over all these types with me. I know there is one function that is continous at just the irrationals, then there are others that are continous when x get close to one of the numbers and the last are the ones that are continous nowhere. Can someone give intutive examples of each? Are there any more types then the ones i mention. There is none for continous on the rationals right?

Not a homework problem. Oh, and i do physics, so don't use less than sigma in the answer please... explain it intutively so i don't get a headache from the mathspeak.

you mean continous at irrationals? i don't think there is one continous at the rationals. If someone has an example please give me one.

here is the one that is continous at all irrationals

f(x) = (p/q) = 1/q for rationals
f(x) = 0 for irrationals

ok, i will write all the ones i know of

f(x) = 1 when x is rational
f(x) = 0 when x is irrational

i think this is nowhere

then there are these types

f(x) = x when irrational
f(x) = 0 when rational

then you can switch it around

f(x) = x when rational
f(x) = 0 when irrational

and the one i gave in the example already.

Are there any more of these types and can someone go over each one, and tell me where they are cont.

and a real complex one

f(x) = e^x when x is irrational
f(x) = 0 when x is rational

this would be nowhere because e^x doesn't cross 0?

Answer 1

I don't feel like doing the review, but I can offer something. It turns of that there are functions can are LeBesgue (sp?) integrable but no Riemann. The f(x) = 0 x rational, =1 x irrational is an example. You can cook up a set that lands on just the rationals so that it has a limit at a given rational, and hence every rational even though both the rationals and irrationals are infinitely dense in any small neighborhood of any point. I don't think that you can do the same with irrationals. If I'm correct then the rationals would be continuous and the irrationals not. But I don't think that I'm correct.

You are essentially talking about what's known as point set topology, a part of an undergraduate, or graduate, real analysis course.

I've found Wikipedia to be very helpful for math concepts.

I just found something that said

f(x) = 0, x irrational
f(x) = x, x rational

is continuous at one point, x = 0.

Answer 2

This really is very confusing. Let me say what I understood of the question and maybe you can refine it.

The "function" is

f(x) = 1 if x is rational
.........0 if x is irrational

The rational numbers are not a continuous set, but a discrete set (they have to be expressible as a quotient of two integers), therefore, the rational domain is represented by "dots on a line".

f(x) = 0 on these dots and 1 otherwise, therefore it is not continuous when x is rational and is continuous when x is irrational (or better yet x + or minus XI is irrational for a positive XI as small as necessary).

Answer 3

Well, the set of rationals has measure zero (it's countable). Therefore F is = 1 except on a set of measure zero, hence measurable.

Answer 4

I agree with the previous answerer. But out of curiosity, what is this function that is continuos at just the rationals?

Answer 5

I have never read a more confusing question in my whole life.