From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: G deltas
Date: Sat, 16 Sep 2000 17:42:37 -0500
Newsgroups: sci.math
Summary: [missing]

On Sat, 16 Sep 2000 hero_no_1@my-deja.com wrote:

> It is a fact that the set of continuity points of a function
> defined on a metric space is a G-delta , is the converse true
> ,that given any G-delta A we can find a function whose countinuity
> points is A.

It's a well-known fact that the set of continuity points of a function
f:R --> R is a G-delta, and conversely, every G-delta subset of R is
the set of continuity points of a function f:R --> R. But I gather
from  your mention of "metric space" that you are after something much
more general. Here's what I know about that.

Theorem. If f:X --> Y, where X is any topological space but Y is a
metric space, then the set of continuity points of f is a G-delta.

Proof. Let D_n be the union of all open sets U of X such that f(U) has
diameter < 1/n in Y; the set of continuity points of f is just the
intersection of all the sets D_n.

Note that we need metrizability for the *codomain*, not the domain.
The "fact" you stated, about functions defined on a metric space, with
nothing assumed about the codomain, is incorrect. Let Y = R union {p},
where p is not in R, and give Y the funny topology where the open sets
are Y itself and all subsets not containing the point p. Let A be an
arbitrary subset of R. Define f:R --> Y by setting f(x) = p if x is in
A, and f(x) = x otherwise; then the set of continuity points of f is
equal to A.

The converse you asked about is false for metric spaces in general:
e.g., if X has an isolated point x, than X\{x} is a clopen set but is
not the set of continuity points of any function defined on X. Here is
a fairly general converse:

Theorem. If X is a topological space containing two disjoint dense
subsets S and T, then every G-delta set in X is the set of continuity
points of some function f:X --> R.

Proof. Let A be a G-delta subset of X; say A is the intersection of
the open sets A_1, A_2, ..., A_n, ... . If x is in A, let f(x) = 0. If
x is not in A, let n(x) be the least n for which x is not in A_n, and
let f(x) = 1/n(x) if x is in S, f(x) = -1/n(x) if x is not in S.

Is this a homework problem?

-- 
"Any clod can have the facts, but having opinions is an art."--McCabe
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: G deltas
Date: Sat, 16 Sep 2000 18:22:34 -0500
Newsgroups: sci.math

On Sat, 16 Sep 2000, Fred Galvin wrote:

> Theorem. If X is a topological space containing two disjoint dense
> subsets S and T, then every G-delta set in X is the set of continuity
> points of some function f:X --> R.

By the way, assuming the axiom of choice, every metric space without
isolated points satisfies the hypothesis of the theorem.

-- 
"Any clod can have the facts, but having opinions is an art."--McCabe
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: G deltas
Date: Sat, 16 Sep 2000 19:09:39 -0500
Newsgroups: sci.math

On Sat, 16 Sep 2000, Fred Galvin wrote:

> On Sat, 16 Sep 2000, Fred Galvin wrote:
> 
> > Theorem. If X is a topological space containing two disjoint dense
> > subsets S and T, then every G-delta set in X is the set of continuity
> > points of some function f:X --> R.
> 
> By the way, assuming the axiom of choice, every metric space without
> isolated points satisfies the hypothesis of the theorem.

For "metric space without isolated points" read "topological space in
which every point has a countable neighborhood base and no one-point
set is open".

-- 
"Any clod can have the facts, but having opinions is an art."--McCabe
==============================================================================

From: jean_philippe.boucheron@cybercable.fr (j-p boucheron)
Subject: Re: G deltas
Date: Sun, 17 Sep 2000 00:57:12 +0200
Newsgroups: sci.math

<hero_no_1@my-deja.com> wrote:

> It is a fact that the set of continuity points of a function defined on
> a metric space is a G-delta , is the converse true ,that given any
> G-delta A we can find a function whose countinuity points is A.

In any metric space, I don't know, but it is true if A is a subset of R.

Put R\A=B=F_0 U F_1 U ... U F_n U ...
where each F_n is closed and the sequence (F_n)_n is increasing.

There is a f_n:R->R with :
* 0=<f_n=<1 ;
* f_n is continuous at each x \in R\F_n ;
* the oscillation of f_n at each x \in F_n is 1.

To obtain such a function, consider Q_n=Q inter F_n, and define
f_n=characteristic function of F_n\Q_n.


Define f=\sum_n (1/n!) f_n.

Then f is continuous at each point of A.
Moreover, the oscillation of f at each point of F_n\F_{n+1} is at least
(1/n!) - \sum_{k>n} (1/k!) > 0.

Then f is discontinuous at each point of B.


-- 
jpb
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: G deltas
Date: Sun, 17 Sep 2000 01:24:59 -0500
Newsgroups: sci.math

On Sat, 16 Sep 2000, Fred Galvin wrote:

> Theorem. If X is a topological space containing two disjoint dense
> subsets S and T, then every G-delta set in X is the set of continuity
> points of some function f:X --> R.

Here is a better answer to the question: in which spaces X is every
G-delta set the set of continuity points of some function f:X --> R?
An obvious necessary condition (since the empty set is a G-delta) is
the existence of a nowhere-continuous function g:X --> R, and that is
also sufficient. (The theorem quoted above uses the stronger
assumption that there is a nowhere-continuous function g:X --> {0,1}.)

THEOREM. Let X be a topological space. If there is a
nowhere-continuous function g:X --> R, then every G-delta set in X is
the set of continuity points of some function f:X --> R.

PROOF. The function h(x) = 3/2 + (arctan g(x))/pi is nowhere
continuous, and takes values in the open interval (1,2). Let A be the
intersection of a decreasing sequence of open sets A_n (n = 1,2,...).
If x is in A, let f(x) = 0. If x is not in A, let n(x) be the least n
for which x is not in A_n, and let f(x) = h(x)/2^{n(x)}.

Note that, if x is in A_n, then 0 <= f(x) <= 1/2^n. Since A_n is open,
this shows that f is continuous at each point of A. Now consider a
point x not in A; say n(x) = n. If x is in the closure of A_n, then f
is discontinuous at x because f(x) > 1/2^n. On the other hand, if x is
not in the closure of A_n, then x has a neighborhood on which f
coincides with the nowhere-continuous function h/2^n.

EXAMPLE. Let X be the set of natural numbers, topologized so that the
collection of all nonempty open sets is a free ultrafilter. Then X
does not contain two disjoint dense sets, but there is a
nowhere-continuous function f:X --> R, namely, the function f(x) = x.

QUESTION. Can you find a less pathological (e.g., Hausdorff) example?

-- 
"Any clod can have the facts, but having opinions is an art."--McCabe