MHB Theorem 1.8: Sets or Domains in the Complex Plane - Palka Ch.2

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 2: The Rudiments of Plane Topology ...

I need help with an aspect of Theorem 1.8 ...

Theorem 1.8 (preceded by its "proof") reads as follows:

https://www.physicsforums.com/attachments/7337In the above text from Palka Ch.2 we read the following:

"Let $$A$$ be a set in the complex plane ... ... "Now it seems that from what Palka has written in the quoted text above, that $$A$$ cannot be an arbitrary set ... anyway not a scattered set of points in the complex plane ... is that correct?

It seems that $$A$$ must be a connected region or domain in the complex plane ... is that right?

[ ... ... Note that Palka does not use the term "connected region" or "region" but does refer (without definition as far as I can tell, to "plane set" ... ]

Can someone please clarify the above concerns ...

Peter===============================================================================It may help readers of the above post to have access to Palka's basic notation and terminology regarding plane topology ... so I am proving the same ... as follows:View attachment 7338
View attachment 7339
View attachment 7340

Hope that helps ... ...

Peter

 

[Opalg]

Now it seems that from what Palka has written in the quoted text above, that $$A$$ cannot be an arbitrary set ... anyway not a scattered set of points in the complex plane ... is that correct?

It seems that $$A$$ must be a connected region or domain in the complex plane ... is that right?

No, Theorem 1.8 applies to an arbitrary subset $A$ of the complex plane.

I think that maybe you are misled by thinking that an isolated point $x$ in $A$ cannot be the limit of a sequence in $A$. But in fact it is the limit of such a sequence: you can just use the constant sequence in which every element is $x$ itself.

 

[Math Amateur]

No, Theorem 1.8 applies to an arbitrary subset $A$ of the complex plane.

I think that maybe you are misled by thinking that an isolated point $x$ in $A$ cannot be the limit of a sequence in $A$. But in fact it is the limit of such a sequence: you can just use the constant sequence in which every element is $x$ itself.

oh! indeed ... yes ... thanks Opalg ...

... so you are saying that where $$A$$ is a set of isolated points scatted across the complex plane that we choose $$z_n = z_0$$ for our point belonging to $$A \cap \Delta ( z_0, \frac{1}{n} ) $$ ... ... and do so again and again for $$n+1, n+2, \ ...$$ in order to manufacture the required sequence $$ \left\langle z_1 \right\rangle $$ ... ... Is that correct?

Peter***EDIT***

... BUT ... problem ! ... Theorem 1.8 would then imply that the point $$z_0$$ belongs to the closure of the set $$A$$ ... but surely $$z_0$$ does not belong to the closure of $$A$$ ... ? ... does it?

Oh ... maybe $$z_0$$ DOES belong to the closure of the set $$A$$ ... because for every $$r \gt 0$$, the open disk $$\Delta ( z_0, r )$$ contains $$z_0$$ and hence has a non-empty intersection with $$A$$ ... and also, of course, with $$\mathbb{C} \sim A$$ ... ...Can someone please clarify the above for me ...

Peter

 

[Opalg]

Oh ... maybe $$z_0$$ DOES belong to the closure of the set $$A$$ ... because for every $$r \gt 0$$, the open disk $$\Delta ( z_0, r )$$ contains $$z_0$$ and hence has a non-empty intersection with $$A$$

Yes, that is correct. The closure of a set $A$ always includes the whole of $A$.