/sci/ - Science & Math » Thread #14274944

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Anonymous Wed 02 Mar 18:30:29 2022 No.14274944 View Reply Original Report

Quoted By:

What if the limit point is outside D or on its border? Can you give me an example of this situation where the theorem doesn't work?

Anonymous

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Anonymous Wed 02 Mar 2022 18:31:24 No.14274946 Report

Quoted By:

bump

Anonymous

Anonymous Thu 03 Mar 2022 03:49:16 No.14275897 Report

Quoted By:

Take D to be the closed unit disc and $f(z)=sin(z)$ .

It's a standard result that the zeros of $f$ are precisely the integer multiples of $\pi$ . That is, the only zero of $f$ in D is $0. <br /><br />A single point set contains no limit points, and so <script type="math/tex"> f$ has no limit points in D. This is what we expect because $f$ is obviously not identically zero on D.

Anonymous

Anonymous Thu 03 Mar 2022 03:51:29 No.14275902 Report

Quoted By:

apologies for the typos. Should read

'a single point set contains no limit points, and so'

Anonymous

Anonymous Thu 03 Mar 2022 05:30:32 No.14276024 Report

Quoted By:

The problem is that $f$ and $g$ need not be continuous (or even exist) on the closure of $D$ . Just set $f \equiv g$ outside $D$ .

The proof of this theorem relies upon $D$ being both open and connected (and hence path connected).

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