Perfect Sets in R^k are uncountable

[Math Amateur]

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

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

I am concerned that I do not fully understand the proof of Theorem 2.43 concerning the uncountability of perfect sets in \(\displaystyle R^k\).

Rudin, Theorem 2.43 reads as follows:
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In the above proof, Rudin writes:

"Let \(\displaystyle V_1\) be any neighbourhood of \(\displaystyle x_1\). If \(\displaystyle V_1\) consists of all \(\displaystyle y \in R^k\) such that \(\displaystyle | y - x_1 | \lt r\), the closure \(\displaystyle \overline{V_1}\) of \(\displaystyle V_1\) is the set of all \(\displaystyle y \in R^k\) such that \(\displaystyle | y - x_1 | \le r\)."

Now, I am assuming that the above two sentences that I have quoted from Rudin's proof are a recipe or formula to be followed in constructing \(\displaystyle V_2, V_3, V_4\), and so on ... ... is that right?

I will assume that is the case and proceed ...

Rudin, then writes:

"Suppose \(\displaystyle V_n\) has been constructed, so that \(\displaystyle V_n \cap P\) is not empty ... ... "My question is as follows:

Why does Rudin explicitly mention that he requires \(\displaystyle V_n\) to be constructed so that \(\displaystyle V_n \cap P\) is not empty?

Surely if \(\displaystyle V_n\) is constructed in just the same way as \(\displaystyle V_1\) then \(\displaystyle V_n\) is a neighbourhood of \(\displaystyle x_n\) ... ... and therefore we are assured that \(\displaystyle V_n \cap P\) is not empty ... ... aren't we? ... ... and so there is no need to mention that \(\displaystyle V_n\) needs to be constructed in a way to assure this ...

Can someone please clarify this issue ...

Hope someone can help ...

Peter

 

[Fallen Angel]

Hi Peter,

Rudin is constructing some specific neighborhood for those $x_{i}$. (Usually called $B(x_{i},r_{i})$, balls of center $x_{i}$ and radius $r_{i}$).

What he says next is , suppose $V_{n}$ is constructed in this way, then $V_{n}\cap P$ is non empty

 

[Math Amateur]

Hi Peter,

Rudin is constructing some specific neighborhood for those $x_{i}$. (Usually called $B(x_{i},r_{i})$, balls of center $x_{i}$ and radius $r_{i}$).

What he says next is , suppose $V_{n}$ is constructed in this way, then $V_{n}\cap P$ is non empty

Thanks for the help, Fallen Angel ...

But ... I must say I still have a problem, I think, ... because Rudin does not say:

"If \(\displaystyle V_n\) has been so constructed then \(\displaystyle V_n \cap P\) is not empty"He says:

"Suppose \(\displaystyle V_n\) has been constructed so that \(\displaystyle V_n \cap P\) is not empty"

which seems to me that \(\displaystyle V_n\) has to be constructed in a specific way to assure that \(\displaystyle V_n \cap P\) is not empty.

What do you think?

Peter

***NOTE***

Although I differ in my interpretation of Rudin's sentence ... I suspect that you are correct given the 'facts' of the mathematics ...

 

[Fallen Angel]

Hi Peter,

Reading again my post I think it wasn't clear.

By definition, all this are limit points, so for any neighborhood $V_{n}$ of $x_{n}$ we have $V_{n}\cap P\neq \emptyset$.

Now in $\Bbb{R}^{k}$ the balls form a basis for the usual topology, which means that every open set can be written as a countable union of balls.

A set $U\subseteq\Bbb{R}^{k}$ being open means that for every $u\in U$ exists a radius $r_{u}$ such that $u\in B(u,r_{u})\subseteq U$.

Then in the proof we can consider balls centered at the points $x_{1},x_{2},\ldots$ as neighborhoods.

 

[blue_raver22]

Hello Peter,

Thank you for your question and for sharing your concerns about the proof of Theorem 2.43 in Rudin's book. I will try to provide a clear explanation to help clarify the issue.

Firstly, it is important to understand the definition of a perfect set in R^k. A perfect set is a closed set that is also infinite and has no isolated points. In other words, every point in the set is a limit point of the set itself. This means that for any point x in the perfect set P, there exists a sequence of distinct points in P that converges to x.

Now, in the proof of Theorem 2.43, Rudin is trying to show that any perfect set in R^k is uncountable. To do this, he constructs a sequence of nested neighborhoods V_1, V_2, V_3, ... such that V_n \cap P is not empty for all n. This is important because it ensures that for any point x in P, there exists a sequence of points in P that converges to x. This is necessary in order to show that P is infinite and has no isolated points.

To answer your question, yes, V_n is constructed in the same way as V_1, but it is necessary to explicitly mention that V_n \cap P is not empty because this is a crucial step in the proof. Without this condition, we cannot guarantee that P is infinite and has no isolated points, which are essential properties of a perfect set.

I hope this helps to clarify the issue. If you have any further questions, please don't hesitate to ask. Good luck with your studies!

Best,