Re: Analysis problem and continuity

In article <y99xk.9449$cn7.4708@xxxxxxxxxxxxxxxxxxxx>,
"Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote:

"Carl R." <solrac140@xxxxxxxxxxx> wrote in message
news:fd66f253-0435-4d81-92b8-9a03e85b7a4e@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


Let f: R - > C be a map from the set of real numbers to the comple
numbers.
Define phi(x,delta):= sup { | f(s) - f(t) | : s, t in (x-delta, x
+delta)}
Let r(x):= inf { phi(x,delta) : delta >0 }


Prove that

a) f is continuous in x iff r(x) = 0.

So here's my work.

Asumme f is continuous in x then for all e>0 we can find delta >0 such
that
if |y-x|< delta then |f(y) - f(x) | < epsilon.
But if |y-x|< delta then y,s are in (x-delta,x+delta) , thus
phi(x,delta): = sup { epsilon } = epsilon.
Thus r(x) := inf { epsilon : epsilon > 0} = 0 because epsilon>0.


This is the general idea but the proof is pretty weak and not formulated in
the "general style".

Let phi(x, d) = sup{|f(s) - f(t)|} for s and t in (x - d, x + d)

phi has already been defined; no need to include it in the proof.
Also, "for s and t ..." should be included in the { }.

suppose f is continuous,

|s - t| < d ==> |f(s) - f(t)| < e

No. Let e > 0 first. Then there exists d ...

then 0 <= phi(x,d) < e (this is actually a bit vague... it is true exactly
because sup(f(x)) takes on its maximum... there is a theorem and it happens
because f(x) is continuous)

Actually it's false. A continuous function on a closed bounded
interval assumes its maximum. But (x - d, x + d) is open. The most you
can say here is phi(x,d) <= e. Which of course is good enough.

hence 0 <= r(x) < phi(x,d) < e

since e can be chosen arbitrarily we have r(x) = 0.


This is a bit better, but not necessarily the best. It's logic is much clear
than what you have said. Your's is a bit more "intuitive" but it also shows
and it is not the standard method of proving things.

i.e,., your last 2 statements are not common. It is almost exactly what I
said but you skip the steps:


phi(x,delta): = sup { epsilon } = epsilon.

why does phi(x,delta) = sup{epsilon}? Surely it can only be <= epsilon

Thus r(x) := inf { epsilon : epsilon > 0} = 0 because epsilon>0.

the same here... why does r(x) = inf{epsilon} = 0?

You must always go back to the definitions!!!!!!!!

inf(X) <= X <= sup(X)!!!



(it is because epsilon can be chosen arbitrarily small, which you can see
how I wrote it.


Where I'm stuck is the part that if r(x) = 0 then f is continuous in
x. Can you please help?


always start with your definitions:

inf{phi(x,d)} = 0

implies that phi(x,d) can be made arbitrarily small.

e.g., there exists a delta that makes it smaller than we wish

i.e.,

phi(x,d) < e for some d

but now phi(x,d) = sup{|f(s) - f(t)|}

but this is easy

|f(s) - f(t)| <= phi(x,d) < e

so we can choose |s - t| < d

which will make r(x) < e

(definition of infimum)

which is just

|f(s) - f(t)| <= phi(x,d) < e

which shows that f(x) is continuous.



I'll leave you to put it all together.
.

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