Results 1 to 8 of 8

Math Help - Continuity mk2

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Continuity mk2

    Prove that f(x) = \begin{cases} x \ \ \ \ \text{if} \ x \ \ \text{is irrational} \\ 0 \ \ \ \ \text{if} \ x \ \text{is rational} \end{cases} is discontinuous at x_0 \neq 0.
    I need to find |f(x)-f(x_0)|< \epsilon when 0<|x-x_0|< \delta (where \delta,\epsilon>0).

    Suppose that x_0 is rational and x is rational. Therefore:

    |f(x)-f(x_0)|< \epsilon \Leftrightarrow \ |f(x)|< \epsilon \Leftrightarrow \ 0< \epsilon.
    This is valid from the definition of \epsilon.

    Suppose that x_0 is rational and x is irrational.

    |f(x)-f(x_0)|< \epsilon \Leftrightarrow \ |f(x)|<\epsilon \Leftrightarrow \ x< \delta.
    Hence take \delta=\epsilon.

    So the proof for when x_0 is rational is as follows:

    Let \epsilon>0, x_0 be rational and \delta:=\epsilon. When 0<|x-x_0|< \delta I have:

    |f(x)-f(x_0)|=|f(x)|< \epsilon as required.

    Suppose that x_0 is irrational and x is also irrational.

    |f(x)-f(x_0)|<\epsilon \Leftrightarrow \ |f(x)-x_0|< \epsilon \Leftrightarrow \ |x-x_0|<\epsilon.
    Therefore pick \delta=\epsilon as before.

    Suppose that x_0 is irrational and x is rational.

    |f(x)-f(x_0)|<\epsilon \Leftrightarrow \ |f(x)-x_0|< \epsilon \Leftrightarrow |-x_0|<\epsilon \Leftrightarrow \ |x_0|<\epsilon.

    At this point, a \delta value cannot be chosen since it isn't in the above inequality chain. Since |x_0|<\epsilon has to be valid \forall \epsilon>0, there is a clear contradiction (eg, if x_0=5 I can make \epsilon=4).

    Is this sufficient proof to show that f(x) is discontinuous at x_0 \neq 0?

    (btw, I have shown that f(x) is continuous at x_0=0).

    PS. Can someone show me a way of doing this with sequences? This \epsilon, \delta method is a little long.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Theorem: If f is continuous at a and \left| {f(a)} \right| > 0\, \Rightarrow \,\left( {\exists \delta } \right)\left[ {x \in \left( {a - \delta ,a + \delta } \right)\, \Rightarrow \,\left| {f(x)} \right| > 0} \right].
    In other words, if a continuous at a point and is not zero there then there is an open interval about the point on which the function is not zero.

    Applying that theorem to this problem quickly gives the result.
    Note in any open interval there are irrational numbers and at irrational numbers the function is zero.
    That contradicts the theorem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    hmm, I think I need a little help with this one:

    So I need to use |f(a)|>\epsilon when |x-a|<\delta to achieve a contradiction of some sort.

    Let a be irrational.

    |f(a)|>\epsilon \Leftrightarrow|a|>\epsilon \Leftrightarrow \ -\epsilon<a<\epsilon \Leftrightarrow -\epsilon<-a<\epsilon \Leftrightarrow x-\epsilon<x-a<x+\epsilon.

    What's the next step from here? Comparing with |x-a|<\delta seems to suggest \delta=x+\epsilon, which is clearly wrong, but at the same time it isn't a contradiction.
    Last edited by Showcase_22; March 19th 2009 at 06:42 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    Posts
    5
    Why can't you use open balls to prove this? Consider, in the image of f, the set A={f(x)| 1<=f(x)<=2 }. Because neither 1, nor 2, are irrational or 0, they are not members of the image of f, so they are not members of A. So A is an open set. But the preimage of A is going to be [1,2], which is closed. But f is continuous iff every preimage of an open set is open itself. Is that right?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    umm, you probably can.

    I've never learnt about open balls (to wiki!!), only about epsilon-delta proofs and sequences. For this question, continuity at 0 is proved using epsilon-delta so i'm trying to prove discontinuity at other points using this method.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by Showcase_22 View Post
    only about epsilon-delta proofs and sequences. For this question, continuity at 0 is proved using epsilon-delta so i'm trying to prove discontinuity at other points using this method.
    Use the given function f(x) = \left\{ {\begin{array}{ll}<br />
   {x,} & {x \text{ is rational}}  \\<br />
   {0,} & { x \text{ is irrational}}  \\ \end{array} } \right..
    Suppose that this function continuous at a\ne 0 and \left| {f(a)} \right| > 0.
    That means that a is rational, why?
    Using the definition of continuity, \left( {\exists \delta  > 0} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}} \right]
    It follows that \left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0.
    In other words, \left( {\forall x \in \left( {a - \delta ,a + \delta } \right)} \right)\left[ {|f(x)| > 0} \right].
    That is impossible because between any two numbers there is an irrational number at which f must be zero.

    On the other hand, suppose the function continuous at c \ne 0 and \left| {f(c)} \right| = 0.
    That means that c is irrational, why? We may assume that c > 0.
    Now 0 < \frac{c}{2} < c\;\& \,\left( {\exists r \in \mathbb{Q}} \right)\left[ {\frac{c}{2} < r < c} \right].
    Using \varepsilon  = \frac{c}{2} it is impossible to find a \delta that works.

    P.S. In any \varepsilon -neighborhood \left( {c - \varepsilon ,c + \varepsilon } \right) of c, \left( {\exists q \in \mathbb{Q}} \right)\left[ {c < q < c + \varepsilon } \right].
    That implies that c < q = f(q).
    That means that there is no \delta that works using the definition if we take {\varepsilon  = c > 0}. (if c<0 a similar trick still works).
    Last edited by Plato; March 20th 2009 at 10:23 AM. Reason: In response to a plea in a MP
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    <br />
\left( {\exists \delta > 0} \right)\left[ {\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}} \right]<br />
?
    How did you get the above?

    <br />
\left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0<br />
    I also can't see how one side implies the other. Did you minus something from both sides?

    It is impossible to find a \delta that works.
    Why is that the case?

    (sorry for the barrage of questions!)
    Last edited by Showcase_22; March 20th 2009 at 02:36 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Sorry to double post but I needed the toolbar at the top of the text box.

    I think i'm starting to get this, so i'll have a go at answering my own question (thanks for answering one!):

    <br />
\left| {\left| {f(x)} \right| - \left| {f(a)} \right|} \right| \leqslant \,\left| {f(x) - f(a)} \right| < \frac{{\left| {f(a)} \right|}}{2}\, \Rightarrow \,\left| {f(x)} \right| > \,\frac{{\left| {f(a)} \right|}}{2} > 0<br />

    I can't see how one side implies the other.
    x can either be rational or irrational.
    If x is rational we get |x-a|< \delta.
    If x is irrational we get |0-a|=|a|=|f(a)|.

    From the triangle inequality we know that ||f(x)|-|f(a)|| \leq |f(x)-f(a)| \leq |f(a)| (the last step was attained from above).

    \Rightarrow 0< |f(x)|-|f(a)|<||f(x)|-|f(a)|| \leq |f(x)-f(a)| \leq |f(a)|

    \Rightarrow 0< |f(x)|-|f(a)| \leq |f(a)|

    \Rightarrow 0<|f(a)|<|f(x)| \leq 2|f(a)| (adding f(a) to all the sides and making it greater than 0).

    but 0<\frac{|f(a)|}{2}<|f(a)| so putting this in the inequality gives:

    \Rightarrow 0<\frac{|f(a)|}{2}<|f(a)|<|f(x)| \leq 2|f(a)|

    and isolating the bits we need gives the required result 0<\frac{|f(a)|}{2}<|f(x)|
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. continuity
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 20th 2011, 01:36 PM
  2. y=x^3 (continuity)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 16th 2010, 10:50 PM
  3. continuity
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 9th 2010, 04:23 PM
  4. Continuity
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 7th 2009, 09:10 PM
  5. Continuity, Uniform Continuity
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 1st 2009, 08:36 PM

Search Tags


/mathhelpforum @mathhelpforum