EPTDoes there exist a function f:R -> R that is continuous at each rational but discontinuous at each irrational?
Does anyone have an idea of what to do? I swear math wasn't this hard in high school
[H] Calculus (realy!)
| Blogs > infinity21 |
|
infinity21
Canada6683 Posts
Does there exist a function f:R -> R that is continuous at each rational but discontinuous at each irrational? Does anyone have an idea of what to do? I swear math wasn't this hard in high school | ||
|
fight_or_flight
United States3988 Posts
| ||
|
infinity21
Canada6683 Posts
R = ![[image loading]](http://upload.wikimedia.org/math/6/9/a/69a45f1e602cd2b2c2e67e41811fd226.png) | ||
|
infinity21
Canada6683 Posts
speaking of which, can a mod change that? thanks x_x | ||
|
fight_or_flight
United States3988 Posts
http://www.google.com/search?hl=en&q=continuous at each rational but discontinuous at each irrational?&btnG=Google Search http://www.jstor.org/view/0025570x/di021166/02p0055m/0 The answer is no. However, if you switch the words "irrational" and "rational" then the answer is yes. edit: btw this seems pretty bizarre, and interesting. What class is this? | ||
Chill
Calgary25987 Posts
On October 18 2007 11:57 infinity21 wrote: LOL i mispelled really ;; speaking of which, can a mod change that? thanks x_x Nope =] | ||
Kau
Canada3500 Posts
Edit: So I guess I'd say that such a function doesn't exist. | ||
|
infinity21
Canada6683 Posts
| ||
|
infinity21
Canada6683 Posts
| ||
|
fight_or_flight
United States3988 Posts
On October 18 2007 12:58 infinity21 wrote: I find it interesting that when my internet's down, the only things that still work are the university website, MSN, and teamliquid.net =/ Well that sucks. Maybe this will help: + Show Spoiler + ![[image loading]](http://www.jstor.org/jstor/gifcvtdir/di001758/0025570x/di021166/02p0055m_l.1.gif?jstor) ![[image loading]](http://www.jstor.org/jstor/gifcvtdir/di001758/0025570x/di021166/02p0055m_l.2.gif?jstor) ![[image loading]](http://www.jstor.org/jstor/gifcvtdir/di001758/0025570x/di021166/02p0055m_l.3.gif?jstor) | ||
|
infinity21
Canada6683 Posts
On October 18 2007 13:04 fight_or_flight wrote: Well that sucks. Maybe this will help: + Show Spoiler + Thanks for trying but it's not working too well. I'll just take a look at it when my internet starts to work again. I'm not in a hurry -- this question is a bonus question and we are allowed to submit this at any time of the term.  | ||
|
Ender
United States294 Posts
I'll work on your particular problem later, but let's say we had the function g(x) = xf(x) where f(x) was 1 for x rational and 0 for x irrational. Let's say we wanted to prove that the limit as x approaches 0 of xf(x) = 0. This means for |xf(x)-0|<epsilon, there exists a 0<|x|<delta. Note that since f(x) can be 1 at the most, then we can choose delta = epsilon and our limit is proved. Now for continuity, we need to show that for each epsilon>0 there exists a delta>0 such that if 0<|x-c|<delta, then |f(x)-f(c)|<epsilon. Hope that starts you off a little. | ||
|
infinity21
Canada6683 Posts
On October 18 2007 13:10 Ender wrote: Ok to show a limit at a point we must show that lim as x approaches c of f(x) = L. Now in rigorous terms this means for any epsilon>0, there exists a delta>0 such that if 0<|x-c|<delta then |f(x)-L|<epsilon. I'll work on your particular problem later, but let's say we had the function g(x) = xf(x) where f(x) was 1 for x rational and 0 for x irrational. Let's say we wanted to prove that the limit as x approaches 0 of xf(x) = 0. This means for |xf(x)-0|<epsilon, there exists a 0<|x|<delta. Note that since f(x) can be 1 at the most, then we can choose delta = epsilon and our limit is proved. Now for continuity, we need to show that for each epsilon>0 there exists a delta>0 such that if 0<|x-c|<delta, then |f(x)-f(c)|<epsilon. Hope that starts you off a little. I'm not entirely sure I understood what you said. Does f(x) simply look like y = 1 and y = 0 and g(x) look like y = x and y = 0 when graphed? (albeit with infinitely many holes) Would I have to come up with a function f(x) where f(a) is defined for all a in rationals and is not defined for all a in irrationals? | ||
|
Ender
United States294 Posts
Now consider irrational numbers. Here, L = c(0) = 0. So, for any epsilon |xf(x)|<epsilon implies that |x-c|<delta. hmmm...this one's tricky...yeah according to the links other people have been giving, it seems that the irrational numbers must be continuous so there must be an epsilon that works here but i have an exam tomorrow so ill do more stuff later. | ||
|
oneofthem
Cayman Islands24199 Posts
| ||
|
r0ar
Australia24 Posts
there are no restrictions on the function first of all there is an irrational number between any two rational numbers so i dont really get what they mean by continous on the rationals, maybe defined on the rationals assuming that i would suggest x if x is rational x= 1/0 if x is irrational have you been given a rigorous defn of continuity yet? | ||
|
LxRogue
United States1415 Posts
| ||
|
infinity21
Canada6683 Posts
On October 18 2007 15:26 r0ar wrote: ok there are no restrictions on the function first of all there is an irrational number between any two rational numbers so i dont really get what they mean by continous on the rationals, maybe defined on the rationals assuming that i would suggest x if x is rational x= 1/0 if x is irrational have you been given a rigorous defn of continuity yet? fairly rigorous, I'd say. | ||
|
infinity21
Canada6683 Posts
On October 18 2007 15:54 LxRogue wrote: 2^x i think qualifies? How so? It's continuous for all real numbers, including the irrationals. | ||
|
LxRogue
United States1415 Posts
Well i guess that only means it doesn't exist at irrationals...not really saying anything about ccontinuity. | ||
| ||
