This is a question regarding exercise 27 in chapter 2 from baby Rudin:

How does one prove that P is perfect?

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- Nov 19th 2010, 07:08 AMjefferson_lcBaby Rudin - ch 2, ex 27
This is a question regarding exercise 27 in chapter 2 from baby Rudin:

How does one prove that P is perfect? - Nov 19th 2010, 07:28 AMTheEmptySet
- Nov 19th 2010, 11:52 AMHallsofIvy
That depends on two things: (1) What is the definition of P? (2) What is the definition of "perfect"?

- Nov 19th 2010, 12:25 PMChris L T521
Here's the problem from Rudin's "Principles of Mathematical Analysis":

Quote:

Originally Posted by**Chapter 2, Exercise 27**

- Nov 21st 2010, 12:24 PMDrexel28
It should be uncountable, otherwise it's trivially true that $\displaystyle P$ is perfect since it' empty!

**Lemma:**Let $\displaystyle X$ be a second countable metric (or topological) space with countable basis $\displaystyle \mathcal{B}$ and $\displaystyle E\subseteq X$. Then, if $\displaystyle \Omega=\{U_\alpha\}_{\alpha\in\mathcal{A}}$ is an open cover for $\displaystyle E$ then $\displaystyle \Omega$ admits a countable subcover.

**Proof:**For each $\displaystyle x\in E$ there exists a neighborhood $\displaystyle N_x$ containing it. But, since $\displaystyle \mathcal{B}$ is a basis we may find some basic open set $\displaystyle B_x\in\mathcal{B}$ for which $\displaystyle x\in B_x\subseteq N_x$. Evidently then $\displaystyle \left\{B_x:x\in E\right\}$ is an open cover for $\displaystyle E$, and since it's a subset of $\displaystyle \mathcal{B}$ it must be countable. Thus, for each element $\displaystyle B_{x_0}$ of $\displaystyle \left\{B_x:x\in E\right\}$ choosing some $\displaystyle N_{x_0}\in\Omega$ for which $\displaystyle B_{x_0}\subseteq N_{x_0}$ produces a countable subcover for $\displaystyle \Omega$ as desired. $\displaystyle \blacksquare$

So, we now claim that $\displaystyle P$, as described in the problem, is perfect. To sync our notations call $\displaystyle D(P)$ (the derived set) the set of all limit points of $\displaystyle P$. Thus, we aim to show that $\displaystyle D(P)=P$.

So, to first see that $\displaystyle D(P)\subseteq P$ let $\displaystyle x\in D(P)$. Then, for each neighborhood $\displaystyle U$ of $\displaystyle x$ we have that there is some $\displaystyle y\in D(P)\cap U$. But, since $\displaystyle U$ is a neighborhood of $\displaystyle y$ we see that $\displaystyle U\cap E$ is uncountable. Thus, since $\displaystyle U$ was arbitrary it follows that $\displaystyle x\in P$.

Conversely, suppose that $\displaystyle x\notin D(P)$ then, there exists a neighborhood $\displaystyle U$ of $\displaystyle x$ for which every element $\displaystyle y$ of $\displaystyle U-\{x\}$ has a neighborhood $\displaystyle N_y$ for which $\displaystyle \#\left(N_y\cap E\right)\leqslant\aleph_0$ clearly then $\displaystyle \left\{N_y\right\}_{y\in U-\{x\}}$ is an open cover for $\displaystyle U-\{x\}$. Thus, by our lemma it must admit a countable subcover $\displaystyle \left\{N_{y_n}\}_{n\in\mathbb{N}}$. Note though that $\displaystyle \displaystyle \#\left(\left(U-\{x\}\cap E\right)\right)=\#\left(\bigcup_{n\in\mathbb{N}}N_ {y_n}\cap E\right)=\#\left(\bigcup_{n\in\mathbb{N}}\left(N_{ y_n}\cap E\right)\right)\leqslant \aleph_0$ (this last part since by construction $\displaystyle N_{y_n}\cap E$ is countable and the countable union of countable sets is countable). and thus $\displaystyle \#\left(U\cap E\right)\leqslant\aleph_0$ and thus $\displaystyle x\notin P$. Thus, $\displaystyle \mathbb{R}^n-D(P)\subseteq \mathbb{R}^n-P\Leftrightarrow P\subseteq D(P)$.

Thus, putting these two together gives $\displaystyle D(P)=P$.

So, now for the second part of the problem let $\displaystyle \mathcal{B}$ be the countable basis for $\displaystyle \mathbb{R}^n$ (the set of all open balls with rational radii at points of $\displaystyle \mathbb{Q}^n$) and let $\displaystyle \mathcal{W}=\left\{B\in\mathcal{B}:\#\left(B\cap E\right)\leqslant\aleph_0\right\}$. We claim that $\displaystyle \displaystyle \bigcup_{B\in\mathcal{W}}B=\mathbb{R}-P$. To see this let $\displaystyle \displaystyle x\in\bigcup_{B\in\mathcal{W}}B$, then $\displaystyle x\in B_0$ for some $\displaystyle B_0\in\mathcal{W}$, but evidently then $\displaystyle B_0$ is a neighborhood of $\displaystyle x$ for which $\displaystyle \#\left(B_0\cap E\right)\leqslant\aleph_0$ and so $\displaystyle x\notin P$. Conversely, let $\displaystyle x\notin P$, then there is some neighborhood $\displaystyle U$ of $\displaystyle x$ for which $\displaystyle \#\left(U\cap E\right)\leqslant\aleph_0$, and since $\displaystyle \mathcal{B}$ is a basis we man find some $\displaystyle B_0\in\mathcal{B}$ for which $\displaystyle x\in B_0\subseteq U$ and so evidently $\displaystyle \#\left(B_0\cap E\right)\leqslant\aleph_0$ and so $\displaystyle B_0\in\mathcal{W}$ and thus $\displaystyle \displaystyle x\in B_0\subseteq\bigcup_{B\in\mathcal{W}}B$. It follows that $\displaystyle \displaystyle \mathbb{R}^n-P=\bigcup_{B\in\mathcal{W}}B$ as desired. Note though that $\displaystyle \displaystyle \#\left(E\cap \left(\mathbb{R}^n-P\right)\right)=#\left(E\cap\bigcup_{B\in\mathcal{ W}}B\right)=\#\left(\bigcup_{B\in\mathcal{W}}\left (B\cap E\right)\right)\leqslant \aleph_0$ as desired (this past part gotten since by construction $\displaystyle \#\left(B\cap E\right)\leqslant\aleph_0$ for each $\displaystyle B\in\mathcal{W}$ and since $\displaystyle \mathcal{W}\subseteq\mathcal{B}$ we have that $\displaystyle \mathcal{W}$ is countable).