Re: Functions continuous only at the rationals and only at the integers

nicegirl_...@xxxxxxxxx wrote:

Yes, great idea. Your idea works in any metric space
that contains a proper dense subset, right?

The function dist(E,x) will still be continuous, but
we'll need a real-valued function that is everywhere
discontinuous. You can get one by using the
characteristic function of a set such that
both the set and its complement is dense
("proper dense" isn't enough; forget metric
spaces for a moment -- in the reals, note that
all real numbers except the number 2 is a proper
dense set), but I don't know if it's possible
to have an everywhere discontinuous real-valued
function without such a set existing. [That is,
I don't know if the property for a metric space
to have an everywhere discontinuous real-valued
function on it is less stringent than the property of
having a subset such that both it and its complement
is dense. It probably is less stringent, and more
than likely someone will say so, and why, soon after
this post appears.]

Now that I'm thinking about it, I think this
construction can be used to show that any F_sigma
set of the reals (a set that can be expressed as
a countable union of closed sets) is the discontinuity
set for some function f:R --> R. Write the F_sigma
set as a union of closed sets E_n, define f_n to be
discontinuous exactly on E_n, and then form the
sum n=1 to infinity of (1/2^n)*f_n. The 1/2^n factors
are to ensure not just convergence, but uniform
convergence (Weierstrass M-test), so that the
continuity stuff will work out. There's a little
bit of additional work needed, because I suppose
we need to make sure that a discontinuity point
of one of the f_n's doesn't get canceled out by
the discontinuities that other f_n's might have
at the same point (for example, the characteristic
function of the rationals plus the characteristic
function of the irrationals, which is a sum of two
nowhere continuous functions, is about as nice as
possible -- it's the constant function '1').
There's a more precise pointwise version of
the uniform convergence of functions theorem
that might be all we need, but I don't know it
off the top of my head. But this is getting past
anything you need to worry over at this point
However, if anyone is interested, the first
page of one of the earlier chapters of Oxtoby's
book "Measure and Category" has this argument,
or something pretty close to it.

I need to leave now, and I don't think I'll
be back on the internet until tomorrow morning.
(It's 5:05 P.M. local time for me now.)

Dave L. Renfro

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