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

It seems that \(\displaystyle A\) 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 \(\displaystyle A\) cannot be an arbitrary set ... anyway not a scattered set of points in the complex plane ... is that correct?

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

Peter***EDIT***

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

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

Peter

 

[Opalg]

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

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