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Thread: Function Continuous on Rationals but not on Rationals

  1. #1
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    Function Continuous on Rationals but not on Rationals

    So we all know of a function that's continuous on the irrationals but discontinuous on the rationals, but what about this:

    Are there functions $\displaystyle f : [0, 1] \rightarrow \mathbb{R}$
    continuous on $\displaystyle \mathbb{Q}$ and discontinuous on $\displaystyle \mathbb{R} \backslash \mathbb{Q}$? Justify.

    I'm quite sure the answer's no, but not entirely sure on how to prove it.

    Any help would be much appreciated, thanks!
    Last edited by h2osprey; May 27th 2009 at 01:04 PM.
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by h2osprey View Post
    So we all know of a function that's continuous on the irrationals but discontinuous on the rationals, but what about this:

    Are there functions $\displaystyle f : [0, 1] \rightarrow \mathbb{R}$
    continuous on $\displaystyle \mathbb{Q}$ and discontinuous on $\displaystyle \mathbb{R} \backslash \mathbb{Q}$? Justify.

    I'm quite sure the answer's no, but not entirely sure on how to prove it.

    Any help would be much appreciated, thanks!
    Consider $\displaystyle f: [0,1] \to \mathbb{R} $ defined by $\displaystyle f(x) = \begin{cases} 0 \ \ \text{if} \ x \ \text{is irrational} \\ \frac{1}{n} \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = \frac{m}{n} \ \text{in lowest terms} \end{cases} $

    This is continuous on the irrationals and discontinuous on the rationals. You can find a sequence $\displaystyle (a_n) $ such that $\displaystyle a_n \to x $ but $\displaystyle f(a_n) \not \to f(x) $ where $\displaystyle x \in \mathbb{Q} $.
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    The answer is no.

    There is a theorem that states the points where a function f is continuous is a $\displaystyle G_{\delta}$ set.
    A $\displaystyle G_{\delta}$ set, is a set that is the countable intersection of open sets. So the irrations can be written as $\displaystyle \bigcap Q_n$, where $\displaystyle Q_n=\mathbb{R}\backslash q_n$, with $\displaystyle \{q_n\}$ being an enumeration of the rationals.

    I'll leave it to you to show that $\displaystyle \mathbb{Q}$ isn't a $\displaystyle G_{\delta}$ set.
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    Quote Originally Posted by Sampras View Post
    Consider $\displaystyle f: [0,1] \to \mathbb{R} $ defined by $\displaystyle f(x) = \begin{cases} 0 \ \ \text{if} \ x \ \text{is irrational} \\ \frac{1}{n} \ \ \text{if} \ x \in \mathbb{Q} \ \text{and} \ x = \frac{m}{n} \ \text{in lowest terms} \end{cases} $

    This is continuous on the irrationals and discontinuous on the rationals. You can find a sequence $\displaystyle (a_n) $ such that $\displaystyle a_n \to x $ but $\displaystyle f(a_n) \not \to f(x) $ where $\displaystyle x \in \mathbb{Q} $.
    Also the function in this example is continuous at x=0, which is a rational so it doesn't satisfy the necessary condition.
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    Senior Member Sampras's Avatar
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    Quote Originally Posted by putnam120 View Post
    Also the function in this example is continuous at x=0, which is a rational so it doesn't satisfy the necessary condition.
    Let $\displaystyle a_n = \frac{\sqrt{2}}{n} $. Then $\displaystyle a_n \to 0 $. And $\displaystyle f(a_n) \to 0 $. But $\displaystyle f(0) \neq 0 $. Because $\displaystyle 0 = \frac{0}{1} = \frac{0}{2} = \cdots = \frac{0}{n} $ for $\displaystyle n \neq 0 $.

    And if you fix $\displaystyle t \in \mathbb{R} $ and $\displaystyle n \in \mathbb{N} $ you can show that $\displaystyle f $ maps finitely many elements of $\displaystyle \left(t- \frac{1}{2}, t+ \frac{1}{2} \right) $ to $\displaystyle \frac{1}{n} $. And we can use this to prove that $\displaystyle f $ is continuous at every irrational number.
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    Oops, thanks for the clarification. However, I still stand by the fact that the function isn't what he is asking about. What he wants a function that is continuous on $\displaystyle \mathbb{Q}$ and discontinuous everywhere else. (The example you gave is not continuous at the $\displaystyle x=0$ so right there it fails)
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  7. #7
    Senior Member Sampras's Avatar
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    But $\displaystyle 0 $ is rational! And he edited his question. Before it was the other way around.
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    Ah sorry, didn't know that he had changed the question.
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    Quote Originally Posted by Sampras View Post
    But $\displaystyle 0 $ is rational! And he edited his question. Before it was the other way around.
    Actually, I edited it to change the Latex which was showing an error, otherwise it's unchanged. I do know about the function continuous on irrationals but discontinuous on rationals - and the title's a typo sorry.

    Thanks anyway!
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    Super Member Showcase_22's Avatar
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    I'll leave it to you to show that isn't a set.
    Can you show me how to do this? I haven't encountered these before and I would like to see what they're like!
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    Quote Originally Posted by Showcase_22 View Post
    Can you show me how to do this? I haven't encountered these before and I would like to see what they're like!
    Once you've shown that the irrationals are a G-delta set, you can prove by contradiction.

    Assume the the rationals are a G-delta set (say an intersection of countably many$\displaystyle A_n$. Since the rationals are dense in the reals, each $\displaystyle A_n$ is dense in the reals. Applying the same argument to the irrationals, we get that the intersection of the rationals and the irrationals (which is $\displaystyle \emptyset$) is a countable intersection of open dense sets. This violates the Baire category theorem, which states that any such intersection is non-empty (since the reals are complete).

    And this completes the proof!
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    Super Member Showcase_22's Avatar
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    Once you've shown that the irrationals are a G-delta set, you can prove by contradiction.
    How do you do that part?

    (sorry but i'm a first year maths student and i've only encountered these 5 mins ago when I read this post! However, they sound really interesting. Plus, I tihnk I may be able to use them to help me with some continuity proofs that i'm not super happy with).
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    putnam120 mentioned it earlier:

    Quote Originally Posted by putnam120 View Post
    The answer is no.

    There is a theorem that states the points where a function f is continuous is a $\displaystyle G_{\delta}$ set.
    A $\displaystyle G_{\delta}$ set, is a set that is the countable intersection of open sets. So the irrations can be written as $\displaystyle \bigcap Q_n$, where $\displaystyle Q_n=\mathbb{R}\backslash q_n$, with $\displaystyle \{q_n\}$ being an enumeration of the rationals.

    I'll leave it to you to show that $\displaystyle \mathbb{Q}$ isn't a $\displaystyle G_{\delta}$ set.
    Hope this helps!
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