G-delta set theorem

[dimitri151]

Does this proof have any holes? I'm not sure about passing the lim through intersection in the second to last line.

[micromass]

You take the limit of sets? How do you define this?

I'm a bit suspicious of the proof. Especially since you never mention the function f. And you dont seem to use the fact that f is continuous. I may have overlooked it tho...

[dimitri151]

Yeah, that's the hole. I can use the same argument to show the set of points at which f is discontinuous is a g-delta set. But I know that in that case it's a f-sigma set. I have to use the continuity property some how. Thanks!

p.s. You can take the limit of a set. Happens all the time.

[mathwonk]

as i recall, the proof that the set of points of continuity is a whatever, uses the concept of variation. I.e. given e>0 consider all x such that on some nbhd N of x, |f(y)-f(z)| < e for all y,z in N.

I think this set is open and the intersection of all these sets for e = 1/n is the set of points of continuity of f. what do you think?

[dimitri151]

I agree completely. That's exactly the argument I'm using, or trying to use. I form -1/n, +1/n interval around each point of continuity and take the limit. But it strikes me as half circular though I can't quite put my finger on where. But why can't you use the same argument for the set of points which are discontinuous. Let D be the set of points at which f is discontinuous , let N be an open ball or nbhd around each point in D then take the limit as N goes to zero around each point in D and voila all that remains is D, so the points at which f is discontinuous is a G delta set too. There's a hole.

[micromass]

You'll have to use continuity of f to show that

\bigcap_{n\in \mathbb{N}}{I_n}=\mathcal{C}

for the rest, your proof seems fine.

And yes, I am aware that you can take a limit of sets, but the problem is that I've seen various definitions of this. I just wanted to know which one you used...

[Landau]

p.s. You can take the limit of a set. Happens all the time.To take a limit you need a topology. There is no standard topology on 'the set of all sets'. So I suggest explaining to the reader what your definition of 'limit of a set' is.

[dimitri151]

Heaven forbid we have to specify the topology every time we talk about a theorem. It's explicit that our set is the real numbers, R^1. You probably need to get out of the house if you're wondering whether or not it's a topology induced by a the euclidean metric or something else. Also remember the "set of all sets" [of X] is a topology of X.
For the definition of limits of sequences I think it's fairly standard (on a set X with subsets a_n, and topology not specified) lim sup a_n =inf sup {a_n}, lim inf a_n=sup inf {a_n}, and lim a_n exists only when lim sup a_n=lim inf a_n. I also used a_n=(f(n),g(n)), then lim (as n->infinity) a_n=(lim f(n), lim g(n)). You also look to see if a sequence of sets that's a union or intersection of some sets is monotone increasing (decreasing) so then lim (as n->infinity) = a_n or a_1 whatever the case may be, (informally-you can get rid of the union/intersection symbol). N'est pas?

[micromass]

Ah, so for a sequence of sets, you put

\limsup_{n}{A_n}=\bigcap_{n=1}^{+\infty}{\bigcup_{ k=n}^{+\infty}{A_k}}~\text{and}~\liminf_{n}{A_n}=\ bigcup_{n=1}^{+\infty}{\bigcap_{k=n}^{+\infty}{A_k }}

and the limit of these sets exists if the limsup equals the liminf. This is indeed a much used defintion in measure theory and probability. But when talking about analysis, I would always specify it, since there are other possible definitions, such as

\lim_{n}{A_n}=\{\lim_n{a_n}~\vert~\forall n: a_n\in A_n \}

Or the limit of some sets can also be the limit points of the filter generated by this sets,...

[dimitri151]

Pretty good with the laytex. That last limit looks like the limit of a_n equals the closure of A_n as n->infinity. It's an entirely different sequence of sets. Or at least you've added an additional restriction.

[micromass]

Well to bad this is one of the diseases of mathematics. Sometimes, everybody uses other definitions. And, like here, it's best to mention which definition is used...

[dimitri151]

I suppose it can only help you have greater clarity in the long run. It's just a little tedious.