From: Fred Galvin 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 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 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 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=n} (1/k!) > 0. Then f is discontinuous at each point of B. -- jpb ============================================================================== From: Fred Galvin 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