Analyticity on a closed set.

Hi:

Let C be the field of complex numbers and let D(z0,m) denote the open

disk with center z0 and radius m. Let U be an open subset of C,

closure of D(a,R) contained in U and f:U->C. Suppose further that

f is analytic on the closure of D(a,R). Proposition: for no r > R is f

analytic on D(a,r). Can this be true? Thank you for reading.

Analycity on a closed set.

I'm very sorry I didn't explain myself clearly. No. I'm only told f is analytic

on the closure of the disk. However, first I should have wondered if there

is such D(a,r) in S (I forgot to establish that D(a,r) must be contained in

S, but reading your proof I see you understood it that way). This is a

purely topological question and your proof answers it in the affirmative.

So the question remains. Is there a D(a,r) in S such that r > R and f

analytic on D(a,r)? Best regards and thank you for your time.

Enrique

Analycity on a closed set.

Perfectly well. I have now revised the definition of analytic function as

given in the book I am studying and see it has a note saying: if T is an

arbitrary subset of the complex (not necessarily open), the terminology

"f is analytic on T" is used to mean that f is analytic on some open set

containing T. This would have saved me effort if only I had read more

carefully. By the way, it's like being in heaven to have such a nice people

at one's disposal to help in math problems.

P.S.: I could have used the note as is or I could have taken T to be

a point in the complex and used your proof. By the way again: whom

could I contact to get help on using the text editor. Thank you very

much for your kindness.