# Baby Rudin - ch 2, ex 27

• Nov 19th 2010, 07:08 AM
jefferson_lc
Baby 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 AM
TheEmptySet
Quote:

Originally Posted by jefferson_lc
This is a question regarding exercise 27 in chapter 2 from baby Rudin:

How does one prove that P is perfect?

It would be helpful if you posted the full question. Most people don't have a stack of old textbooks with them at all times.
• Nov 19th 2010, 11:52 AM
HallsofIvy
That depends on two things: (1) What is the definition of P? (2) What is the definition of "perfect"?
• Nov 19th 2010, 12:25 PM
Chris L T521
Here's the problem from Rudin's "Principles of Mathematical Analysis":

Quote:

Originally Posted by Chapter 2, Exercise 27
Define a point $p$ in a metric space $X$ to be a condensation point of a set $E\subset X$ if every neighborhood of $p$ contains uncountably many points of $E$.

Suppose $E\subset \mathbb{R}^k$, $E$ is countable, and let $P$ be the set of all condensation points of $E$. Prove that $P$ is perfect and that at most countably many points of $E$ are not in $P$. In other words, show that $P^c\cap E$ is at most countable. Hint: Let $\{V_n\}$ be a countable base for $\mathbb{R}^k$, let $W$ be the union of those $V_n$ for which $E\cap V_n$ is at most countable, and prove $P=W^c$.

• Nov 21st 2010, 12:24 PM
Drexel28
Quote:

Originally Posted by Chris L T521
Here's the problem from Rudin's "Principles of Mathematical Analysis":

It should be uncountable, otherwise it's trivially true that $P$ is perfect since it' empty!

Lemma: Let $X$ be a second countable metric (or topological) space with countable basis $\mathcal{B}$ and $E\subseteq X$. Then, if $\Omega=\{U_\alpha\}_{\alpha\in\mathcal{A}}$ is an open cover for $E$ then $\Omega$ admits a countable subcover.
Proof: For each $x\in E$ there exists a neighborhood $N_x$ containing it. But, since $\mathcal{B}$ is a basis we may find some basic open set $B_x\in\mathcal{B}$ for which $x\in B_x\subseteq N_x$. Evidently then $\left\{B_x:x\in E\right\}$ is an open cover for $E$, and since it's a subset of $\mathcal{B}$ it must be countable. Thus, for each element $B_{x_0}$ of $\left\{B_x:x\in E\right\}$ choosing some $N_{x_0}\in\Omega$ for which $B_{x_0}\subseteq N_{x_0}$ produces a countable subcover for $\Omega$ as desired. $\blacksquare$

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

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

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

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

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