Hurkyl said:
It's hard to help if we don't know what definition of continuity you want to use, what proof you're looking at, and where you're having trouble with it.
Are you using the "
f is continuous iff the inverse image of an open set is open" definition? You know that if this condition holds for neighborhoods in the range, it's true for all open sets in the range, right? There are only two kinds of neighborhoods in
R: open intervals containing 0 and open intervals not containing 0.
Ok, here it goes.
As said, we want to prove that f : x --> 1/x is continuous at every point of
R\{0}.
A real function defined on an open interval I = <a, b> is said to be continuous at a point c from I if for
every \epsilon > 0 there exists
at least one \delta > 0 such that for every x from I, for which |x-c|<\delta holds, |f(x)-f(c)|<\epsilon holds, too. Further on, a function is continuous on the interval I if it is continuous at
every point c from I.
Now, let \epsilon>0 be given. We have to find some \delta>0 such that, for x\neq 0, we have |x-c|<\delta \Rightarrow |f(x)-f(c)| = |\frac{x-c}{x\cdot c}|<\epsilon.
Further on, the proof says than now we consider all real numbers x such that |x-c|<\frac{1}{2}|c|. Now, this relation does not hold for x = 0, so we assured that x cannot be 0. Further on, it is equivalent to c-\frac{1}{2}|c|<x<c+\frac{1}{2}|c|, so it is obvious that for every c\neq 0, any x lies in an interval which does not contain 0, hence x cannot be 0. The first thing that troubles me: could we also consider all the reals x such that |x-c|<\frac{1}{9}|c|, for example? Is that formulation chosen just because it is 'easy' to calculate with 2? (Sorry if it's a dumb question, but I want to understant everything correctly.)
Further on, the proof says that |x-c|<\frac{1}{2}|c| implies |x|>\frac{1}{2}|c|, which is obvious, and, as a further consequence, we have |x\cdot c|>\frac{1}{2}|c|\cdot |c|=\frac{1}{2}|c|^2 (1). Now, the key part of the proof suggests that all above implies:
|x-c|<\frac{1}{2}|c|\Rightarrow |f(x)-f(c)|=|\frac{x-c}{x\cdot c}|\leq \frac{2|x-c|}{|c|^2}. (2)
I'm not really sure where that comes from. Obviously the point was to write the
implied term in terms of |x-c| somehow to be able to set up a relation between \delta and \epsilon. But where did that come from? Is it from the fact that (1) implies \frac{2|x\cdot c|}{|c|^2}> 1, so when we multiply |\frac{x-c}{x\cdot c}| with something greater than 1, it must become greater or equal to |\frac{x-c}{x\cdot c}| (equal in the case x=c)?
The rest of the proof is clearer to me, but this parts bothers me somehow, and I feel it is important to (finally) start with completely understanding these proofs, otherwise, I feel I won't ever be able to cope with analysis at a satisfying level. Sorry if I'm boring, but I'd appreciate some confirmation/help/criticism on the presented part of the proof. Thanks in advance.
