Results 1 to 2 of 2

Math Help - Countable Subset

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    3

    Countable Subset

    I have the following to prove:

    Let f : R --> R have the property that lim x-->a− f(x) = f(a). Show that there is a countable subset A contained in R (A c R) such that f is continuous on R \ A.

    Where R is the set of Real Numbers


    This is what I have come up with so far, I would like to know if its any good, if not where my mistakes are. My homework is for tomorrow!
    Thank you sooo much!

    First we know that the function may be discontinuous at a, but the
    limit tells us that the left side (interval:]-∞,a[ ) of the function is
    continuous. Therefore there must be a maximal finite interval with
    upper bound smaller than a, where the function is continuous. Let this
    interval be denoted by a lower bound b; b element of Real numbers and
    b>-∞. Say we start at [b,c] with (b – c) be the smallest subinterval
    in f, we know this interval is continuous. We do this continuously for
    all different endpoints, with upper bound c<a and b>-∞ . So by
    induction assume [b,c] is continuous on f, then [b+1,c+1] is continuous
    on f as long as c+1<a and b+1> Therefore we can say there are a
    finite number of discontinuities in any finite interval in R. Let any
    of these subintervals (we can even let it be the union of all
    subintervals) be denoted by S. We know that S is continuous, and S c R, because S has endpoints which are not infinity, S is a countable set. We let R\S=A; A CR,
    A is also a countable set. A may be discontinuous, because it will
    contain the point a. If we look at this in the contrary way, R\A=S ,
    and we have proven that S is continuous on all its range. Therefore
    there IS a countable subset A, such that f is continuous on R\A=S.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Jimena View Post
    I have the following to prove:

    Let f : R --> R have the property that lim x-->a− f(x) = f(a). Show that there is a countable subset A contained in R (A c R) such that f is continuous on R \ A.

    Where R is the set of Real Numbers

    This is what I have come up with so far, I would like to know if its any good, if not where my mistakes are. My homework is for tomorrow!
    Thank you sooo much!

    First we know that the function may be discontinuous at a, but the
    limit tells us that the left side (interval:]-∞,a[ ) of the function is
    continuous. Therefore there must be a maximal finite interval with
    upper bound smaller than a, where the function is continuous. Let this
    interval be denoted by a lower bound b; b element of Real numbers and
    b>-∞. Say we start at [b,c] with (b – c) be the smallest subinterval
    in f, we know this interval is continuous. We do this continuously for
    all different endpoints, with upper bound c<a and b>-∞ . So by
    induction assume [b,c] is continuous on f, then [b+1,c+1] is continuous
    on f as long as c+1<a and b+1> Therefore we can say there are a
    finite number of discontinuities in any finite interval in R. Let any
    of these subintervals (we can even let it be the union of all
    subintervals) be denoted by S. We know that S is continuous, and S c R, because S has endpoints which are not infinity, S is a countable set. We let R\S=A; A CR,
    A is also a countable set. A may be discontinuous, because it will
    contain the point a. If we look at this in the contrary way, R\A=S ,
    and we have proven that S is continuous on all its range. Therefore
    there IS a countable subset A, such that f is continuous on R\A=S.
    You may be going about this the wrong way. Particularly because I think the above argument assumes that an interval in \mathbb{R} is countable. In fact (0,1)\simeq\mathbb{R}.

    If that is not what you meant let me know and I will come back and give a hint. Otherwise, maybe reconsider your argument.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ORD subset of L; |ORD| uncountable; |L| countable --?
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: November 15th 2010, 06:45 AM
  2. Prove every subset of N is either finite or countable
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: January 1st 2009, 02:57 AM
  3. Subset of A countable Set is countable
    Posted in the Advanced Math Topics Forum
    Replies: 7
    Last Post: November 2nd 2008, 01:47 PM
  4. Replies: 11
    Last Post: October 11th 2008, 07:49 PM
  5. Replies: 4
    Last Post: October 11th 2008, 02:42 PM

Search Tags


/mathhelpforum @mathhelpforum