1. ## 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.

2. Originally Posted by Jimena
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 $\displaystyle \mathbb{R}$ is countable. In fact $\displaystyle (0,1)\simeq\mathbb{R}$.

If that is not what you meant let me know and I will come back and give a hint. Otherwise, maybe reconsider your argument.