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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


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