Re: Analysis problem and continuity

In article
<50fc6221-a286-40f3-8f24-377730e4df1e@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Carl R." <solrac140@xxxxxxxxxxx> wrote:

On 8 sep, 21:58, The World Wide Wade <aderamey.a...@xxxxxxxxxxx>
wrote:
In article
<d0c02c70-bfb5-45d5-86fe-1590390a8...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
 "Carl R." <solrac...@xxxxxxxxxxx> wrote:





On 8 sep, 07:50, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx> wrote:
"Carl R." <solrac...@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)

suppose f is continuous,

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

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)

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.- Ocultar texto de la cita -

- Mostrar texto de la cita -

Thanks very much John, I understood your proof, very clear!
Can you please give a hint to show that the set of discontinuity
points of f is a G_delta set?

That will be difficult, because it's false. Reread the problem.- Ocultar
texto de la cita -

- Mostrar texto de la cita -

The set of continuity points, that is, sorry.

Show that for any e > 0, {x : r(x) < e} is open. Then show {x : r(x) =
0} is the intersection of countably many of such open sets.
.

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