Perfect Sets in R^k are uncountable - Issue/problem 2

[Math Amateur]

I am reading Walter Rudin's book, Principles of Mathematical Analysis.

Currently I am studying Chapter 2:"Basic Topology".

I have a second issue/problem with the proof of Theorem 2.43 concerning the uncountability of perfect sets in \(\displaystyle R^k\).

Rudin, Theorem 2.43 reads as follows:View attachment 3805In the above proof we read:

"Suppose \(\displaystyle V_n\) has been constructed so that \(\displaystyle V_n \cap P\) is not empty. Since every point of \(\displaystyle P\) is a limit point of P, there is a neighbourhood \(\displaystyle V_{n+1}\) such that

(i) \(\displaystyle \overline{V}_{n+1} \subset V_n
\)

(ii) \(\displaystyle x_n \notin \overline{V}_{n+1}
\)

(iii) \(\displaystyle V_{n+1} \cap P\) is not empty

... ... "I do not understand how the fact that every point of \(\displaystyle P\) is a limit point of \(\displaystyle P\) allows us to claim that there is a \(\displaystyle V_{n+1}\) such that the above 3 conditions hold ... indeed the whole thing is a bit mysterious, since there is no given process for selecting the points \(\displaystyle x_1, x_2, x_3\), ... and so \(\displaystyle x_{n+1}\) may be a considerable distance from \(\displaystyle x_n\), making it difficult for \(\displaystyle \overline{V}_{n+1} \subset V_n \) to hold ...

Can someone please explain how the above fact follows ... ...

Peter

 

[Fallen Angel]

Hi Peter,

The points $x_{1},x_{2},...$ are not selected, Rudin is supposing that the set is countable, so they are all the points.

Being a limit point means that for any neighborhood of $x_{n}$ there are at least one point different from it. In particular, there exist another point that we will call $x_{n+1}$ in $V_{n}$, now we can take a neighborhood of this point satisfying this three conditions, for example doing the following:

As soon as $x_{n+1}\in V_{n}$(open) there exists $\delta>0$ such that $B(x_{n+1},2\delta)\subset V_{n}$
Let $\epsilon=\dfrac{|x_{n}-x_{n+1}|}{2}$, now the ball centered in $x_{n+1}$ with radius $min \{\delta ,\epsilon\}$ clearly satisfies all conditions.

 

[Math Amateur]

Hi Peter,

The points $x_{1},x_{2},...$ are not selected, Rudin is supposing that the set is countable, so they are all the points.

Being a limit point means that for any neighborhood of $x_{n}$ there are at least one point different from it. In particular, there exist another point that we will call $x_{n+1}$ in $V_{n}$, now we can take a neighborhood of this point satisfying this three conditions, for example doing the following:

As soon as $x_{n+1}\in V_{n}$(open) there exists $\delta>0$ such that $B(x_{n+1},2\delta)\subset V_{n}$
Let $\epsilon=\dfrac{|x_{n}-x_{n+1}|}{2}$, now the ball centered in $x_{n+1}$ with radius $min \{\delta ,\epsilon\}$ clearly satisfies all conditions.

Hi Fallen Angel ... thanks for your help ...

Peter

 

[Math Amateur]

Hi Fallen Angel ... thanks for your help ...

Peter

Hi all,

I understand that the text: Principles of Mathematical Analysis by Walter Rudin is highly regarded, but at times I feel the proofs are stated in less than effective and clear ways pedagogically speaking anyway ... . Indeed In particular I had considerable difficulty in following the proof of Theorem 2.43 that every non-empty perfect set in \(\displaystyle \mathbb{R}^k\) is uncountable - as indicated in the above post ...

Given this difficulty I went to the text, Understanding Analysis by Stephen Abbott and found a proof of the above theorem in \(\displaystyle \mathbb{R}\). The proof is essentially the same as the proof in \(\displaystyle \mathbb{R}^k\) and Abbott's text makes the proof strategy much clearer.

I am producing Abbott's proof for the interests of MHB members:

View attachment 3821
View attachment 3822
https://www.physicsforums.com/attachments/3823

I must say that on this Theorem at least, Abbott gives a VERY clear explanation.Peter

 

[Euge]

Hi all,

I understand that the text: Principles of Mathematical Analysis by Walter Rudin is highly regarded, but at times I feel the proofs are stated in less than effective and clear ways pedagogically speaking anyway ... . Indeed In particular I had considerable difficulty in following the proof of Theorem 2.43 that every non-empty perfect set in \(\displaystyle \mathbb{R}^k\) is uncountable - as indicated in the above post ...

Given this difficulty I went to the text, Understanding Analysis by Stephen Abbott and found a proof of the above theorem in \(\displaystyle \mathbb{R}\). The proof is essentially the same as the proof in \(\displaystyle \mathbb{R}^k\) and Abbott's text makes the proof strategy much clearer.

I am producing Abbott's proof for the interests of MHB members:

View attachment 3821
View attachment 3822
https://www.physicsforums.com/attachments/3823

I must say that on this Theorem at least, Abbott gives a VERY clear explanation.Peter

This was one of the reasons why I didn't recommend Rudin's book as a main text for self-study (in response to one your threads in the chat room). Rudin's style is concise and at times very terse. More than a decade ago, when I was a college student, the text I used for analysis was a Dover book whose level is a bit higher than Rudin's, but suitable for independent study. I'm glad that Abbott's book is working for you.

 

[Math Amateur]

This was one of the reasons why I didn't recommend Rudin's book as a main text for self-study (in response to one your threads in the chat room). Rudin's style is concise and at times very terse. More than a decade ago, when I was a college student, the text I used for analysis was a Dover book whose level is a bit higher than Rudin's, but suitable for independent study. I'm glad that Abbott's book is working for you.

Thanks for those comments Euge ... ...

Yes, your reasoning regarding Rudin was very accurate ...

Always glad to know your views and your advice ...

Thanks again ...

Peter