Perfect Sets in R^k are uncountable

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In summary: The set $U$ of all these neighborhoods forms a continous subset of $\Bbb{R}^{k}$.This means that $U$ has a finite intersection with $V_{n}$, since for every $u\in U$ there exists a ball centered at $u$ with radius $r_{u}$, such that $u\in V_{n}$. So by the definition of $U$ being open, $U\cap V_{n}$ does not exist."
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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:
View attachment 3806

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
 
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  • #2
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
 
  • #3
Fallen Angel said:
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 ...
 
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  • #4
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.
 
  • #5
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,
 

FAQ: Perfect Sets in R^k are uncountable

1. What is a perfect set in R^k?

A perfect set in R^k is a subset of the real numbers that is closed, uncountable, and has no isolated points. In other words, every point in the set is a limit point, and there is no way to list or count all of the points in the set.

2. How do you prove that perfect sets in R^k are uncountable?

The proof for this statement involves using the Cantor diagonalization method. It begins by assuming that a perfect set in R^k is countable, and then constructing a new point that is not in the set, thus creating a contradiction. This shows that the original assumption was false, and therefore, perfect sets in R^k must be uncountable.

3. Can perfect sets in R^k be finite?

No, perfect sets in R^k are always uncountable, which means they must have an infinite number of elements. This is because if a perfect set were to be finite, it would have a maximum element, which would contradict the definition of a perfect set having no isolated points.

4. Are all uncountable sets in R^k perfect?

No, not all uncountable sets in R^k are perfect. For example, the set of irrational numbers is uncountable, but it does have isolated points, such as the square root of 2. Perfect sets in R^k are a specific type of uncountable set that must meet certain criteria, including being closed and having no isolated points.

5. Why are perfect sets in R^k important in mathematics?

Perfect sets in R^k have important applications in various areas of mathematics, including topology, analysis, and measure theory. They provide examples of uncountable sets that have interesting and useful properties, and they are also closely related to other important concepts such as Cantor sets and fractals.

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