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## Solution to Principles of Mathematical Analysis Chapter 2 Part B

### Chapter 2 Basic Topology

#### Exercise 11

(By ghostofgarborg) $d_2$ and $d_5$ are metrics, the others are not. We note that $d_3(1,-1) = 0$, and therefore violates condition (a). $d_4(0,1) \neq d_4(1,0)$, and violates condition (b). We further note that $d_1(1, -1) = 4 > 2 = d_1(1,0) + d_1(0,-1)$ which violates (c).

We now note the following:

Let $f : \mathbb R^{\geq 0} \to \mathbb R^{\geq 0}$ be a strictly increasing function such that $f(0) = 0$, which is subadditive, i.e.: $f(a + b) \leq f(a) + f(b)$and let $d$ be a metric. Then $f \circ d$ is a metric. That $f \circ d$ satisfies condition (a) follows from the injectivity of $f$,
and from the fact that $f(0) = 0$. That it satisfies $(b)$ follows because $d$ does. Lastly, if
$d(a,c) \leq d(a,b) + d(b,c)$then by virtue of being increasing and subadditive:
$f \circ d(a,c) \leq f(d(a,b) + d(b,c)) \leq f \circ d(a,b) + f \circ d(b,c)$which establishes (c).

To show that $d_2$ and $d_5$ are metrics, it therefore suffices to show that $\sqrt{x}$ and $\frac{x}{1+x}$ satisfy the criteria on $f$ above.

We know from chapter 1 that $\sqrt{x}$ is strictly increasing, and that $\sqrt{0} = 0$. Subadditivity follows by noting that $a+b \leq a + b + 2\sqrt{ab} = (\sqrt a + \sqrt b)^2$ Therefore, $d_2$ is a metric.

The function $\frac{x}{1+x} = 1 – \frac{1}{1+x}$ is clearly strictly increasing, and $0/(1+0) = 0$.
Moreover, $\frac{a+b}{1+a+b} = \frac{a}{1+a+b} + \frac{b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ Consequently, $d_5$ is a metric.

#### Exercise 12

(By ghostofgarborg) Let $\mathcal U$ be an open cover for $K$. Then there exists a $U \in \mathcal U$ for which $0 \in U$. By openness, there exists an $\epsilon > 0$ such that $N_\epsilon(0) \subset U$. By the archimedean property, there exists an $N$ such that $N\epsilon > 1$, i.e. such that $\frac 1 n < \epsilon$ for $n > N$. Therefore, $N_\epsilon(0)$ contains all but a finite number of elements of $K$. We can now pick a set in $\mathcal U$ for each of the remaining points, and obtain a finite cover.

#### Exercise 13

(By ghostofgarborg) Note that $\left| \frac 1 n – \frac {1}{n-1} \right| = \frac 1 {n(n-1)} > \frac 1 {n^2}$, so that the sets $S_n = \{ \frac 1 n + \frac 1 m : m \geq n^2 \} \cup \{ \frac 1 n \}$ are disjoint. Consider the set $S = \bigcup_n S_n \cup \{ 0 \}$. The set is clearly bounded, so we need to show that it is closed, and that the limit points are countable.

We claim that the limit points are precisely $L = \{ \frac 1 n : n \in \mathbb N \} \cup {0}$. Since $L$ is contained in $S$ and is countable, that will suffice to prove our claim. It is easy to see that these points are limit points. To see that there are no other, observe that any point not in $[0,2]$ or the previously mentioned set is either strictly greater than the lub or strictly smaller than the glb of all but at most one of the disjoint sets $S_j$. It would consequently have to be a limit point of $S_j$, but all such limit points are contained in $L$. Lastly, we can easily see that $0$ is a limit point, but $2$ is not.

#### Exercise 14

(By ghostofgarborg) Let $U_n = (\frac 1 n, 1 – \frac 1 n)$. This collection evidently covers $(0,1)$, but has no finite subcover.

#### Exercise 15

(By ghostofgarborg) Possible examples from $\mathbb R$:

Closed: Let $S_n = [n, \infty)$. Then $\bigcap_n S_n$ is empty.

Bounded: Let $S_n = (0, \frac 1 n)$. Then $\bigcap S_n$ is empty.

#### Exercise 16

(By ghostofgarborg) Note that $E = [\sqrt 2, \sqrt 3] \cap \mathbb Q$. It follows from thm. 2.23 and thm. 2.30 that $E$ is closed. Boundedness is clear.

Let $a_n$ be a sequence in $\mathbb Q$ that converges from above to $\sqrt 2$ (as described in chapter 1). Let $A_n = (a_n, \sqrt 3) \cap \mathbb Q$. This is an open cover of $E$, but clearly has no finite subcover. $E$ is consequently not compact.

$E$ is open by thm 2.30, since $E = (\sqrt 2, \sqrt 3) \cap \mathbb Q$.

#### Exercise 17

(By ghostofgarborg) Let $A$ be the set of sequences on the digits $0$ and $1$. We get an injection $A \to E$ by associating with a sequence $a_n$ the number whose $n$th decimal is $4$ if $a_n = 1$ and $7$ otherwise. By thm. 2.14, this makes $E$ uncountable.

$E \subset [0.4, 0.8]$, and is therefore not dense in $[0,1]$.

Since $E$ is bounded, compactness is equivalent to closedness. Assume $x \not \in E$, and let $x_n$ be the $n$th digit in its decimal expansion. For some $N$, $x_N$ is not equal to $4$ or $7$. Choose $\Delta$ such that the $N$th decimal digit of all $x$ in a $\Delta$-neighborhood of $x$ is $x_n$. (Note that this can be done.) This is a neighborhood of $x$ that does not intersect $E$. Therefore $E^c$ is open and $E$ is closed, hence compact.

Let $x \in E$, choose $\epsilon > 0$, and choose $n$ such that $10^{-n} < \epsilon$. The number obtained by changing the $n$th digit of $x$ from $4$ to $7$ or vice versa, is contained in an $\epsilon$-neighborhood of $x$. This means that $x$ is a limit point, so $E$ is perfect.

#### Exercise 18

(By ghostofgarborg) Yes. Here is one example: Choose two irrational numbers, e.g. $\alpha = \sqrt 2$ and $\beta = \sqrt 3$, and let $\{ a_n \}$ be an enumeration of the rational numbers in $E_1 = [\alpha, \beta]$.

Let $r_1$ be the first element of $\{ a_n \}$ that is in $E_{1}$, and pick irrational numbers $\alpha_1 < r_1$, $\beta_1 > r_1$ such that $\alpha < \alpha_1 < r_1 < \beta_1 < \beta$. By removing the interval $(\alpha_1, \beta_1)$, we end up with a union of closed intervals $E_2 = [\alpha, \alpha_1] \cup [\beta_1, \beta]$ By repeating this process for each closed interval in this union and continuing this process in the same vein as the construction of the Cantor set, we end up with a sequence of compact sets $E_1 \supset E_2 \supset \cdots$Now consider $E = \bigcup_{n=1}^\infty E_n$. The reasoning on p. 41 carries through with small modifications, so that we can deduce that $E$ is perfect. By construction, there are no rational numbers in $E$.

#### Exercise 19

(By ghostofgarborg)

(a) Assume $A$ and $B$ are closed and disjoint $\bar A \cap B = A \cap B = \emptyset$ $A \cap \bar B = A \cap B = \emptyset$so that they are separated.

(b) $A^c$ is closed, so by thm 2.27 (c), $\bar B \subset A^c$, and $A \cap \bar B = \emptyset$. Similarly by interchanging $A$ and $B$.

(c) $A$ and $B$ are open and disjoint sets. They are separated by (b).

(d) Fix a point $x$ of a connected metric space $M$, and assume for contradiction that the set of other points is non-empty and at most countable. Then $D = \{ d(x, y) \}_{y \neq x}$ is countable, and we can pick $r \in \mathbb R \setminus D$ such that $r > 0$ and such that there is a $y$ with $d(x,y) > r$. By the choice of $r$, $M = \{ y \in M : d(x,y) < r \} \cup \{ y \in M : d(x,y) > r \}$which by (c) would imply that $M$ is disconnected, a contradiction.

#### Exercise 20

(By Dan kyp44 Whitman) We claim first that closures of connected sets are always connected. To show this consider any connected set $E$ and consider nonempty sets $A$ and $B$ where $\bar E = A \cup B$. Now, let $\breve{E} = \bar E – E$, $C = A – \breve{E}$, and $D = B – \breve{E}$.

First we show that that $E = C \cup D$. So consider any $x \in E$. Then $x \in E \cup E’ = \bar E$, from which it follows that $x \in A$ or $x \in B$ since $\bar E = A \cup B$. Since $x \in E$ it follows that $x \not\in \breve{E}$. Then $x \in A – \breve{E} = C$ or $x \in B – \breve{E} = D$ so that $x \in C \cup D$. Thus $E \subseteq C \cup D$. Now consider any $x \in C \cup D$. If $x \in C$ then $x \in A$ and $x \not\in \breve{E}$. It follows that $x \not\in \bar E$ or $x \in E$, which is logically equivalent to $x \in \bar E \to x \in E$. But since $x \in A$, $x \in \bar E = A \cup B$ so that indeed $x \in E$. A similar argument shows that $x \in E$ if $x \in D$. Therefore $C \cup D \subseteq E$, hence $E = C \cup D$ as claimed.

Now since $E$ is connected and $E = C \cup D$, by the definition of connectedness $\bar C \cap D \neq \emptyset$ and $C \cap \bar D \neq \emptyset$. Thus there is an $x \in \bar C \cap D$ so that $x \in \bar C$ and $x \in D$. Since $x \in D$, $x \in B$ by definition, and since $x \in \bar C$, $x \in C \cup C’$. If $x \in C$ then $x \in A$ by definition so that $x \in A \cup A’ = \bar A$. On the other hand if $x \in C’$ then $x$ is a limit point of $C$. In this case consider any neighborhood $N_r(x)$. Then there is a $y \in N_r(x)$ where $y \neq x$ and $y \in C$. So since $C = A – \breve{E}$, $y \in A$. From this it follows that $x$ is also a limit point of $A$ since $N_r(x)$ was arbitrary and $y \neq x$. Thus $x \in A’$ so that $x \in A \cup A’ = \bar A$. Hence in either case $x \in \bar A$ so since also $x \in B$, $x \in \bar A \cap B$ so that $\bar A \cap B \neq \emptyset$. A similar argument shows that $A \cap \bar B \neq \emptyset$, thereby showing that $\bar E$ is connected.

However there is an example of a set in $\mathbb R^2$, that is connected but whose interior is not. Such an example can be constructed from the following simple sets: $A = N_1((-2,0))$$B = N_1((2,0))$$C = \{(x,0) : x \in (-2, 2)\}$Then $E = A \cup B \cup C$ is clearly connected and yet its interior is not since the interior of $C$ in $\mathbb R^2$ is the empty set, thereby separating the interiors of $A$ and $B$.

(By ghostofgarborg) Closures of connected sets are connected: Let $E$ be connected, and assume $\bar E = A \sqcup B$, both non-empty, such that $\bar A \cap B = A \cap \bar B = \emptyset$. Since $E$ is connected, $\overline{(A \cap E)} \cap (B \cap E) \subset \bar A \cap B = \emptyset$$(A \cap E) \cap \overline{(B \cap E)} \subset A \cap \bar B = \emptyset$force $A \cap E = \emptyset$ or $B \cap E = \emptyset$, so that $A \subset E^\prime \setminus E$ or $B \subset E^\prime \setminus E$. Without loss of generality, assume it is the former, so that $E \subset B$. Then $A \cap \bar B \supset A \cap \bar E = A$, a contradiction.

Interiors of connected sets need not be connected: Let $E = \{ (x,y) \in \mathbb R^2 : x \geq 0, y \geq 0 \} \cup \{ (x,y) \in \mathbb R^2 : x \leq 0, y \leq 0 \}$Given two points $\mathbf x_1 \neq \mathbf x_2 \in E$, the map $f : [0,1] \to E$ $f(t) = \begin{cases} \mathbf x_1 – 2t \mathbf x_1 & 0 \leq t \leq \frac 1 2 \\ 2t\mathbf x_2 – \mathbf x_2 & \frac 1 2 < t \leq 1 \end{cases}$ is a continuous path between them, showing connectedness. However, $E^\circ = \{ (x,y) \in \mathbb R^2 : x > 0, y > 0 \} \cup \{ (x,y) \in \mathbb R^2 : x < 0, y < 0 \}$which is the union of two open, disjoint sets and therefore disconnected.

Baby Rudin 数学分析原理完整第二章习题解答

#### Linearity

1. For problem 20, the proof for the “closure” part can be much easier. Aim for the contraposition. If $\overline{E}=A\cup B$ for separated $A$ and $B$, then $E=E\cap\overline{E}=E\cap(A\cup B)=(E\cap A)\cup(E\cap B)$. The remaining work is to show that $(E\cap A)$ and $(E\cap B)$ are separate. $\overline{(E\cap A)}\cap(E\cap B)\subset\overline{A}\cap B=\varnothing$, $(E\cap A)\cap\overline{(E\cap B)}\subset A\cap\overline{B}=\varnothing$.