If you find any mistakes, please make a comment! Thank you.

Solution to Principles of Mathematical Analysis Chapter 2 Part C

Chapter 2 Basic Topology

Exercise 21

(By ghostofgarborg)

(a) We note that when $\epsilon > 0$ \[ \| \mathbf p (t \pm \epsilon) – \mathbf p(t) \| = \epsilon \| \mathbf b – \mathbf a \| \]and if $ | t – t^\prime | < \epsilon / \| \mathbf b – \mathbf a \| $, then $ \| \mathbf p(t) – \mathbf p(t^\prime) \| < \epsilon $. Therefore, if $\mathbf p(t) = \mathbf y$, then \[ N_{\epsilon / \| \mathbf b – \mathbf a\|}(t) \subset \mathbf p^{-1}(N_\epsilon(\mathbf y)) \]Consequently, the inverse image under $\mathbf p$ of an open set is open, and the inverse image of a closed set is closed. Clearly $A_0$ and $B_0$ are disjoint and non-empty, so this allows us to conclude that \[ \bar A_0 \cap B_0 \subseteq \mathbf p^{-1}(\bar A) \cap \mathbf p^{-1}(B) = \mathbf p^{-1}(\bar A \cap B) = \emptyset \] And similarly when interchanging $A$ and $B$. This shows that $A_0$ and $B_0$ are separated.

(b) Restrict $\mathbf p$ to $[0,1]$, and define $A_0$ and $B_0$ correspondingly. The result in (a) still holds. Since $[0,1]$ is connected, we must have $ \mathbf [0,1] \supsetneq A_0 \cup B_0$. Choose $t_0 \in [0,1] \setminus A_0 \cup B_0$. Then $t_0 \neq 0$ and $t_0 \neq 1$, and $\mathbf p(t_0) \not \in A \cup B$.

(c) Let $S \subset \mathbb R^k$. The above result shows that if $\mathbf p(t) \in S$ for all $\mathbf a, \mathbf b \in S$ and $t \in [0, 1]$, then $S$ is connected. This is precisely the condition that $S$ is convex.

Exercise 22

(By ghostofgarborg) Assume for contradiction that $\mathbb R^k = \bigcup^\infty F_n$ where each $F_n$ is closed and has non-empty interior. Note that $\mathbb Q^k$ is countable. Choose $\epsilon > 0$. If $(x_1, \cdots, x_n) \in \mathbb R^k$, pick rational $r_i$ such that $|r_i – x_i| < \epsilon/ \sqrt k$. Then \[ \|(x_1, \cdots, x_k) – (r_1, \cdots, r_k) \| < \left( \sum_1^k \epsilon^2 / k \right)^{\frac 1 2} = \epsilon \]Consequently, any neighborhood around any point of $\mathbb R^k$ contains a point of $\mathbb Q^k$, and $\overline{\mathbb Q^k} = \mathbb R^k$, which implies separability.

Exercise 23

(By ghostofgarborg) Let $B$ be a countable, dense subset of $X$. Note that the collection
$$ \mathcal B = \{ N_{\frac 1 n}(x) : x \in B, n \in \mathbb N \}$$ is countable. We claim that it is a base. Let $x \in U$, $U$ open. Then $ N_\epsilon(x) \subset U$ for some $\epsilon$. Pick $k$ such that $ \frac 1 k < \frac \epsilon 2$. By density, we can find $y \in B$ such that $d(x,y) < \frac 1 k$. Then $x \in N_{\frac 1 k}(y)$, which is in $\mathcal B$, and $N_{\frac 1 k}(y) \subset N_\epsilon(x) \subset U$. This finishes the proof.

Exercise 24

(By ghostofgarborg) We can assume $X$ is infinite. Pick $\delta > 0$, pick an $x_1 \in X$, and inductively pick $x_{n+1}$ such that $d(x_i, x_{n+1}) > \delta$ for all $ 1 \leq i < n + 1$ if possible. This process must terminate. Otherwise, $\{ x_i \}_{i=1}^\infty $ would have a limit point $x$, and $N_{\frac \delta 2}(x)$ would contain two distinct points of the sequence, a contradiction.

$X$ can therefore be covered with a finite number of neighborhoods of radius $\delta$ around a finite number of points $x_1, \cdots, x_k$. Let $\mathcal B$ be the collection of all such neighborhoods for $\delta = \frac 1 n$, $n \in \mathbb N$. This is a countable collection. Given a neighborhood $N_\epsilon(x)$ of $x \in X$, pick $k$ such that $\frac 1 k < \frac \epsilon 2$. Then $x$ is covered by some neighborhood $N \in \mathcal B$ of radius $ \frac 1 k$, and $N \subset N_\epsilon(x)$. This shows that $\mathcal B$ is a countable base.

Exercise 25

(By ghostofgarborg) A finite number of neighborhoods of radius $\delta$ cover $K$. By the reasoning in ex. 24, $K$ has a countable base.

To see that this implies that $K$ is separable, let $\{ U_n \}$ be a countable base, and choose $x_n \in U_n$, so that we obtain a countable subset $S = \{x_n\}_{n=1}^\infty$. Any open set in $K$ contains $U_n$, and therefore an element of $S$. The $S$ is therefore dense in $K$.

Exercise 26

(By ghostofgarborg) We know that $X$ is separable by ex. 24, and that it has a countable base by ex. 23. Since any open set is a union of base sets, we can reduce any open cover to a countable subcover. Let $\mathcal G = \{ G_n \}$ be such a subcover. Assume $\mathcal G$ has no finite subcover, so that $ F_n = (G_1 \cup \cdots \cup G_n)^c$ is non-empty for all $n$. Pick $x_n \in F_n$. They constitute an infinite subset, and therefore have a limit point $x$. This point is in $G_k$ for some $k$. But then $F_k^c$ is an open set around $x$ that contains only a finite number of the $x_n$, contradicting $x$ being a limit point. Any open cover of $X$ must therefore have a finite subcover, so $X$ is compact.

Exercise 27

(By ghostofgarborg) Let $\{ V_n \}$ be a countable base for $\mathbb R^k$. A point $x$ is a condensation point of $E$ if and only if every $V_n$ that contains $x$ contains an uncountable number of points of $E$. Consequently, $P^c$ is the union of those $V_n$ for which $V_n \cap E$ is at most countable. $P^c \cap E$ is then at most countable.

Exercise 28

(By ghostofgarborg) We observe that the proof in ex. 27 goes through for any separable metric space. Assume that $E$ is closed. Since $P$ consists of limit points of $E$, $P \cap E = P$. We can therefore write $E = P \cup (P^c \cap E)$, which is a union of a perfect set and a set which is at most countable.

Exercise 29

(By ghostofgarborg) Let $V$ be the open set in question. We know that the collection $\mathcal U$ consisting of all segments of the form $(r – \alpha, r + \beta)$, $r,\alpha \in \mathbb Q$, $\alpha, \beta \in \mathbb Q^+$, is a countable basis for $\mathbb R$.

Let $\mathcal V$ be the subcollection of sets of $\mathcal U$ that are contained in $V$. Let $I_x$ be the union of all sets $U \in \mathcal V$ such that $U$ intersects a segment in $ \mathcal V$ that contains $x$. It is clear that $I_x$ is a segment, $I_x \subset V$, and that if $y \in V$ then either $I_x = I_y$ or they are disjoint.

The collection $ \{ I_x \}_{x \in V} $ covers $V$, and since $\mathbb R$ is separable, it can be reduced to a countable subcover $\mathcal I$. It is then clear that $\mathcal I$ has the desired properties.

Exercise 30

(By ghostofgarborg) Assume for contradiction that $\mathbb R^k = \bigcup^\infty F_n$ where each $F_n$ is closed and has non-empty interior. Let $N_0$ be a ball of finite radius around a point $x_1 \in F_1$, so that $\bar N_0$ is compact. Assume $N_{i-1}$ is open and does not contain any points of $F_1, \cdots F_{i-1}$. This set must contain a point $x_i$ not in $F_i$, as it otherwise would belong to the interior of $F_i$. Furthermore, $x_i$ must be contained in a neighborhood $N_i \subseteq N_{i-1}$ that does not intersect $F_i$, as $x_i$ otherwise would be a limit point of $F_i$ and therefore belong to $F_i$. We can choose $N_i$ such that $\bar N_i \subseteq N_{i-1}$, and we observe that it does not contain any points of $F_1, \cdots, F_i$.

Now observe that each $\bar N_i$ is compact, and that $ \bar N_{i+1} \subseteq \bar N_{i} $, so that by the corollary to thm. 2.36, $I = \bigcap_i \bar N_i$ is non-empty. However, by construction, if $ x \in I$, then $x \not \in F_i$ for any $i$. But this implies that $x \not \in \bigcup F_i = \mathbb R^k$, a contradiction.

Baby Rudin 数学分析原理完整第二章习题解答


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu