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

### Chapter 2 Basic Topology

#### Exercise 1

(By ghostofgarborg) Let $S$ be any set. The empty set $\emptyset$ is a subset of $S$ iff for all $x \in \emptyset$, $x \in S$. This is vacuously true, so we are done.

#### Exercise 2

(By ghostofgarborg) Let $\mathbb A$ be the set of algebraic numbers. Let $p$ be a polynomial over $\mathbb C$. Using the division algorithm for polynomials over a field, we observe that $(z – \alpha) \mid p(x)$ iff $\alpha$ is a root of $p$. We can deduce that a polynomial of degree $n$ has at most $n$ roots.

Let $P_{n}$ be the set of polynomials $p$ in $\mathbb Z[x]$ for which the coefficients $a_i$ satisfy $$\sum_{i=1}^m |a_i| = n – \deg p .$$ This is a finite set, and $\bigcup P_n = \mathbb Z[x]$. Let $V_{n}$ be the corresponding set of all roots of the polynomials in $P_{n}$. By the above observation, this set is finite. Consequently, $\mathbb A = \bigcup_n V_n$ is at most countable. Since $\mathbb Z \subset \mathbb A$, it is also at least countable.

#### Exercise 3

(By ghostofgarborg) This must be the case, as the set of real numbers otherwise would be countable.

#### Exercise 4

(By ghostofgarborg) No. We know the set of rational numbers is countable. Since
$$\mathbb R = \mathbb Q \cup (\mathbb R \setminus \mathbb Q)$$ and $\mathbb R$ is uncountable, we must have that $(\mathbb R \setminus \mathbb Q)$ is uncountable by theorem 2.12.

#### Exercise 5

(By ghostofgarborg) Take e.g. the set
$\{ \frac 1 n : n \in \mathbb N \} \cup \{ 1 + \frac 1 n : n \in \mathbb N \} \cup \{ 2 + \frac 1 n : n \in \mathbb N \}.$

#### Exercise 6

(By ghostofgarborg) Let $x$ be a limit point of $E^\prime$. Any neighborhood $N$ of $x$ contains an $x^\prime \in E^\prime$ for which $x^\prime \neq x$.

By theorem 2.22, $N \setminus \{x\} = N \cap \{x\}^c$ is open, and therefore contains a neighborhood $N^\prime$ of $x^\prime$. $N^\prime$ intersects $E$ since $x^\prime \in E^\prime$, and consequently so does $N$. This implies that $x \in E^\prime$, and since this holds for any limit point, $E^\prime$ is closed.

Since $E \subseteq \bar E$, a limit point of $E$ is a limit point of $\bar E$. If $x$ is a limit point of $\bar E = E \cup E^\prime$, $N_{\frac 1 n}(x)$, must intersect either $E$ for an infinite number of values of $n \in \mathbf N$, or it must intersect $E^\prime$ for an infinite number of values of $n$. (Otherwise, there would be an $N$ for which $N_{\frac 1 n}(x)$ did not intersect either whenever $n > N$, a contradiction.)

Consequently, since $N_\epsilon(x) \subseteq N_{\epsilon^\prime}(x)$ whenever $\epsilon < \epsilon^\prime$, all neighborhoods of $x$ intersect either $E$ or $E^\prime$. In both cases, $x \in E^\prime$. This finishes the proof.

(By Matt frito Lundy) Let $x \in (E’)^c$. Then $x \notin E’$, so $x$ is not a limit point of $E$. Hence there exists a neighborhood $N$ of $x$ that contains no points of $E$, except possibly $x$ itself.

If $N$ contained a limit point $y$ of $E$, then there would exist a neighborhood $N_y$ of $y$ that contains a point of $E$ other than $y$ such that $N_y \subset N$. This would contradict the fact that $x$ is a limit point of $E$, so $N$ must not contain any limits points of $E$. So $N \subset (E’)^c$, and $(E’)^c$ is open, thus $E’$ is closed.

If $x$ is a limit point of $E$, then $E \subset \bar E$ implies that $x$ is also a limit point of $\bar E$.If $x$ is a limit point of $\bar E$ but not a limit point of $E$, then there is a neighborhood $N$ of $x$ such that $N$ contains no points of $E$ other than possibly $x$ itself. But $x$ is a limit point of $\bar E$ means that there is a $q \in N$ such that $q \neq x$ and $q \in \bar E$. If $q \in E$, we have arrived at a contradiction.If $q \in E’$, then $q$ is a limit point of $E$, and there is some neighborhood of $N_q$ of $q$ such that $N_q \subset N$ that contains a point of $E$, which is also a contradiction. Therefore, $E$ and $\bar E$ share the same limit points.

$E$ and $E’$ need not share the same limit points. Consider any segment $E = (a,b)$ in $R^1$.

(By analambanomenos) Let $x\notin E’$. Then there is a neighborhood $N$ of $x$ which contains no elements of $E$ (other than possibly $x$). Hence every element of $y\neq x$ of $N$ has has a small neighborhood contained in $N$ containing no points of $E$, and so $y\notin E’$ also. Hence $E’^c$ is open, and so $E’$ is closed.

Since $E\subset\bar E$, $E’\subset\bar{E}’$. Since $\bar E$ is closed by Theorem 2.27(a), $x\in\bar{E}’$ is an element of $\bar{E}=E\cup E’$. Suppose $x\notin E’$. Then it has a neighborhood with no elements of $E$ other than $x$ itself, and since $E’$ is closed, $x$ has a neighborhood which contains no elements of $E’$. This contradicts the assumption that $x\in\bar{E}’=(E\cup E’)’$, so $x\in E’$.

$E$ and $E’$ do not always have the same limit points. Take the answer to Exercise 5, for example. $E’$ is a set of 3 points, so $E”=\emptyset$.

#### Exercise 7

(By ghostofgarborg)

(a) Assume $1 \leq i \leq n$. Since $A_i \subseteq B_n$, a limit point of $A_i$ is a limit point of $B_n$. Let $x$ be a limit point of $B_n$, and consider the neighborhoods $N_k = N_{\frac 1 k}(x)$ for $k \in \mathbf N$. There must be a $j$ for which there is no $K$ such that $k > K$ implies that $N_k$ does not intersect $A_j$, as the negation of that statement would imply that $x$ is not a limit point of $B_n$. Since $N_\ell \subset N_k$ whenever $\ell > k$, this implies that $N_k$ intersects $A_j$ for all $k \in \mathbb N$. Given any neighborhood $N_\epsilon(x)$ of $x$, we can find a $k$ such that $N_k \subseteq N_\epsilon(x)$, so all neighborhoods of $x$ intersect $A_j$, and $x \in \bar A_j$.

(b) It is clear that a limit point of $A_i$ is a limit point of $B$, so that the inclusion holds. To see that the inclusion is proper, let $A_i = (\frac 1 i, 1)$. Then $\bar A_i = [\frac 1 i, 1]$, so that $\bigcup_i \bar A_i = (0,1]$. However, $B = \bigcup_i (\frac 1 i, 1) = (0,1)$, so that $\bar B = [0,1]$, and $\bar B \supsetneq \bigcup \bar A_i$.

(By analambanomenos) Recall Theorem 1.27(c): $\bar E\subset F$ for every closed set $F\subset X$ such that $E\subset F$.

(a) Since $\cup_1^n\bar{A}_i$ is a closed set that contains $B_n$, we have $\bar{B}_n\subset\cup_1^n\bar{A}_i$. Conversely, since $\bar{B}_n$ is a closed set that contains $A_i$, $i=1,\ldots,n$, we have $\bar{A}_i\subset\bar{B}_n$, $i=1,\ldots,n$, so $\cup_1^n\bar{A}_i\subset\bar{B}_n$. Hence $\bar{B}_n=\cup_1^n\bar{A}_i$.

(b) Since $\bar B$ is a closed set containing each $A_i$, we have $\bar{A}_i\subset\bar{B}$ for each $i$, so $\cup_1^\infty\bar{A}_i\subset\bar B$.

#### Exercise 8

(By ghostofgarborg)

Open sets: Yes. Any point $x$ in an open set $E$ is contained in a neighborhood $N_\epsilon(x) \subset E$. Any point $y$ such that $d(y,x) < \epsilon$ is contained in $E$. It is clear that any neighborhood of $x$ contains such a point $y$.

Closed sets: No. A single point set $E = \{x\}$ is closed, but is not a limit point, since no neighborhood of $x$ contains a point $y \in E$ such that $y \neq x$.

#### Exercise 9

(By ghostofgarborg)

By definition, if $x \in E^\circ$, then $x$ in an open subset $U \subset E$. Let $U \subset E$ be open. Then any point of $U$ has a neighborhood inside $E$, so that $U \subset E^\circ$. This establishes that $E^\circ = \bigcup_{\substack{U \subset E \\ U \text{ open}}} U.$(a) $E^\circ$ is a union of open sets, and is therefore open. (Note: This is also true if the union is empty.)

(b) If $E = E^\circ$, then we know from (a) that $E$ is open. For the converse, begin by noting that $E^\circ \subseteq E$, since it is a union of subsets of $E$. If $E$ is open, then $E$ is in the union, so that $E \subseteq E^\circ$ as well, from which equality follows.

(c) This is clear, as $G$ is included in the union.

(d) By theorem 2.27 (a) and (e) $(E^\circ)^c = \bigcap_{\substack{U \subset E \\ U \text{ open}}} U^c = \bigcap_{\substack{V \supset E^c \\ V \text{ closed}}} V = \overline{E^c}.$

(e) No. Let $E = \mathbb Q$ in $\mathbb R$. Then $E^\circ = \emptyset$, but $(\bar E)^\circ = \mathbb R^\circ = \mathbb R$.

(f) No. Let $E = \mathbb Q$ in $\mathbb R$. Then $\bar E = \mathbb R$, but $\overline{ E^\circ } = \bar \emptyset = \emptyset$.

#### Exercise 10

(By ghostofgarborg) We see that $d(x,x) = 0$, and $d(x,y) \neq 0$ whenever $x \neq y$. We also have $d(x,y) = d(y,x)$. Lastly, consider $d(x,z) \leq d(x,y) + d(y,z)$The inequality is obvious if $x = z$, and otherwise either $y \neq x$ or $y \neq z$, so that the inequality still holds. The function $d$ is thus a metric.

Note that every point set is open, since $\{x\} = N_{1/2}(x)$. Consequently, every subset of $X$ is open, and therefore every subset of $X$ is closed. Let $S$ be a subset of $X$, and consider the open cover $\{x\}_{x \in S}$. This cover has a finite subcover iff $S$ is finite. Any compact set must therefore be finite, and conversely, it is easy to see that any finite set is compact. It is thus clear that the compact sets are precisely those that are finite.

Baby Rudin 数学分析原理完整第二章习题解答