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**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.*