Blumberg’s theorem
October 8th, 2005 by WaltHere’s a bizarre theorem I’d never heard of before: Blumberg’s theorem. It states that for any real function, there is a dense subset of the reals on which the function becomes continuous.
Here’s a bizarre theorem I’d never heard of before: Blumberg’s theorem. It states that for any real function, there is a dense subset of the reals on which the function becomes continuous.
October 13th, 2005 at 5:43 am
I wonder if this theorem provides a post-hoc justification for the working practices of physicists and engineers in the 19th century. They assumed, for example, the “obvious” claim that the infinite limit of a sequence of continuous functions, if it existed, would be continuous. Cauchy and others proved this claim false with counter-examples. But, since every such limit function would have a continuous sub-function defined on a dense subset, it made no difference in practice.
October 15th, 2005 at 3:57 am
Peter : saying something is continuous on a dense subset is really not saying much. A dense subset can be of zero measure, like Q. A function can be constant on Q and very wild on R-Q. Even if this function would be continuous in restriction to Q, you could certainly not say that it is “almost continuous”. For this kind of statements you have to use measure theory or Baire categories, not just density.
October 18th, 2005 at 12:25 pm
Re Fabien’s comment: A slight digression - it is actually impossible for a function to be continuous EXACTLY on Q and discontinuous on all of R-Q. I forget the proof, but it has something to do with G-delta sets (I think) on the real line. (Q is G-delta whereas R-Q is not).
October 19th, 2005 at 9:23 am
Hi karthik. You’re right, in fact the set of points of discontinuity of a real function must be an F-sigma set, that is a coutable union of closed sets.
October 19th, 2005 at 9:34 am
…and R-Q in not an F_sigma, whereas Q is. (you swapped G-delta for F-sigma)