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 $$R^k$$.

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

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

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 $$V_2, V_3, V_4$$, and so on ... ... is that right?

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

Rudin, then writes:

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

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

Surely if $$V_n$$ is constructed in just the same way as $$V_1$$ then $$V_n$$ is a neighbourhood of $$x_n$$ ... ... and therefore we are assured that $$V_n \cap P$$ is not empty ... ... aren't we? ... ... and so there is no need to mention that $$V_n$$ 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 $$V_n$$ has been so constructed then $$V_n \cap P$$ is not empty"He says:

"Suppose $$V_n$$ has been constructed so that $$V_n \cap P$$ is not empty"

which seems to me that $$V_n$$ has to be constructed in a specific way to assure that $$V_n \cap P$$ 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.