The points at which a f is continuous is a G-delta

[SiddharthM]

Now that I have finished with the 7 chapters of rudin i plan to study Part 1 of royden's real analysis. Here is a problem regarding borel sets

Let f be a real valued function defined for all reals. Prove that the set of points at which f is continuous can be written as a countable intersection of open sets.

set y=f(x)

Consider the family of open sets, N_{1/n} (y) for every real y, call it V_n (y). Taking preimages of these sets will yield some open sets and some not. Let O_n be the union of the collection of preimages that are open, then O_n is open. Put
O=intersection of O_n.

If f is continuous at x the preimage of every V_n(f(x)) is open so it is contained in O_n which means that x is in O. Now if x is in O then x is in O_n for every n. Let e>0, pick a N so that 1/N<e/2. Then x is inside the preimage of some V_N (z). Since this the preimage of this V_N (z) is open, call it N(z), there exists a delta>0 so that if d(x,y)<delta then y is in N(z). Also
|f(x)-f(y)| <\= |f(x)-f(z)|+|f(y)-f(z)|<e

we have shown O is the set of points where f is continuous.

Can someone tell me if I'm missing something, i just feel like i am.

cheers!

 

[rudinreader]

Consider the family of open sets, N_{1/n} (y) for every real y, call it V_n (y).
...
If f is continuous at x the preimage of every V_n(f(x)) is open so it is contained in O_n which means that x is in O.

I had to stop reading at this point because f continuous at x does not imply $$f^{-1}(N_e(f(x)))$$ is open. For instance, take f(x) = 0 for |x| <= 1, and f(x) = 2 for x > 1. You get $$f^{-1}((-1,1)) = [-1,1]$$. You are ensured the preimages are open only when f is continuous everywhere.

 

[SiddharthM]

Sorry the assertion was completely false but it can be reformulated so it becomes a local property. If f is continuous at x for every epsilon neighborhood around f(x) (called V_n (x)) there exists a delta neighborhood (called D_n (x)) so that f(D_n(x)) is contained in V_n(x).

Consider the family of V_n(x) for all x in R, when such a D_n (x) exists put it in O_n. So O_n is the union of a family of open sets and is therefore open.
Let O_n be the union of D_n's when they exist as such.

Again O=intersection of O_n

If f is continuous at x then there will always be a neighborhood D_n (x) so that f(D_n(x)) is contained in V_n (x) so that x is in O. Now let e>0 and x in O, there exists a N so that 1/N<e/2. Then x is in some D_n(z) which is a neighborhood of z and is open, so there exists a neighborhood of x contained inside D_n(z). Now for every y in D_n(z) we know that
|f(z)-f(y)|<e/2 so that
|f(x)-f(y)|<|f(x)-f(z)|+|f(z)+f(y)|<e
as long as y is in the neighborhood of x which is contained inside D_n(z).

 

[rudinreader]

When f is continuous at x, you associate an open $$D_n(x) \subset B(x,1/n)$$ so that $$f(D_n(x)) \subset B(f(x),1/n)$$. Then $$O_n$$ is the union of all $$D_n(x)$$, such that f is continuous at x. Clearly, $$O = \cap O_n$$ contains all the continuity points (because each continuity point by definition is in a $$D_n \subset O_n$$), and also O is a countable intersection of open sets.

You still need to show O does not contain a discontinuity point. If a discontinuity point p is in O. Then p is in some $$D_n(p_n) \subset B(p_n,1/n)$$ for all n (where $$f(D_n(p_n)) \subset B(f(p_n),1/n)$$). Then fix e > 0 such that for every d > 0, there is a q such that d(p,q) < d and d(f(p),f(q)) > e. Then for 1/N < e/2, you have p in D_N(p_N), and thus some q in D_N(p_N) such that d(f(p),f(q)) > e, but also $$d(f(p),f(q)) \leq d(p,p_N) + d(p_N,q) < e/2 + e/2$$, a contradiction.