Calculus TA

Answer (1) Let $X$ be a metric space, $A \subset X$ be a complete subset, and $p \in X$ be a limit point of $A$. We want to show that $p \in A$. There exists a sequence $a_n$ such that $a_n \rightarrow p$. Note that $a_n$ is a Cauchy sequence, and therefore $a_n \rightarrow a \in A$. Then $p=a \in A$. (2) Consider $A=\mathbb{Q} \cap [0,2]$ in a metric space $\mathbb{Q}$. Note that $A$ is closed in $\mathbb{Q}$. Consider a sequence $q_n \in A$ converging to $\sqrt{2}$. Then $q_n$ is a Cauchy sequence. However, $q_n$ does not converge in $A$, which implies $A$ is not complete.

Problem Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact. Answer By Problem 23 and 24, $X$ has a countable base. Consider an arbitrary open cover $\{G_\alpha\}_{\alpha \in A}$ of $X$. For any $\alpha \in A$ and $x \in G_\alpha$, $x$ is contained in some base $V_{\alpha,x} \subset G_\alpha$. Since the base element $V_{\alpha,x}$ is contained in countable family, $G_\alpha = \bigcup_{x \in G_\alpha}V_{\alpha,x}$ has a countable subcover, denoted by $\{G_n\}_{n=1}^\infty$. Suppose there is no finite subcollection of $\{G_n\}$ covers $X$. Then $F_n:=(G_1 \cup \cdots \cup G_n)^c$ is nonempty for each $n$. Let $E$ be a set which contains a point from each $F_n$. If $E$ is a finite set, then $E$ is covered by finite subcover of $\{G_n\}$, say $G_{n_1},\cdots,G_{n_N}$. However, it contradicts to the fact that $E$ has to contain point in $(G_1 \cup \cdots \cup G_{n_\ast})^c$ where $n_\ast = \max(n_1,\cdots,n_N)$. Hence, $E$ is infinite and ha...

Problem Regard $\mathbb{Q}$, the set of all rational numbers, as a metric space, with $d(p,q)=|p-q|$. Let $E$ be the set of all $p \in \mathbb{Q}$ such that $2 < p^2 < 3$. Show that $E$ is closed and bounded in $\mathbb{Q}$, but that $E$ is not compact. Is $E$ open in $\mathbb{Q}$? Answer (1) Closed: $E=\mathbb{Q} \cap ((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3})) = \mathbb{Q} \cap ([-\sqrt{3},-\sqrt{2}] \cup [\sqrt{2},\sqrt{3}])$. Define $F:=[-\sqrt{3},-\sqrt{2}] \cup [\sqrt{2},\sqrt{3}]$  $\Rightarrow$  $F^c$ is open in $\mathbb{R}$  $\Rightarrow$  $\mathbb{Q} \cap F^c$ is open in $\mathbb{Q}$ by Theorem 2.30  $\Rightarrow$  $\mathbb{Q} \cap F$ is closed in $\mathbb{Q}$. (2) Bounded: $d(p,0) < \sqrt{3}$ for all $p \in E$, which means $E$ is bounded in $\mathbb{Q}$. (3) Not compact: $E$ is not compact in $\mathbb{R}$, and hence is not compact in $\mathbb{Q}$ by Theorem 2.33. (4) Open: Since $((-\sqrt{3},-\sqrt{2}) \cup (\sqrt{2},\sqrt{3}))$ is ...

Problem Let $K \subset \mathbb{R}^1$ consist of $0$ and the numbers $1/n$, for $n=1,2,3,\cdots$. Prove that $K$ is compact directly from the definition (without using Heine-Borel theorem). Answer Let $\{U_\alpha\}_\alpha \in A$ be a open cover of $K$. Then there exists $\alpha_0 \in A$ such that $0 \in U_{\alpha_0}$. Since $U_{\alpha_0}$ is open, there exists $r>0$ such that $N_r(0) \subset U_{\alpha_0}$. If we choose $N \in \mathbb{N}$ such that $1/N < r$, then $1/n \in N_r(0) \subset U_{\alpha_0}$ for $n \geq N$. For $n=1,\cdots,N-1$, there exist $\alpha_n \in A$ such that $1/n \in U_{\alpha_n}$. Then $K$ can be covered by $U_{\alpha_0},U_{\alpha_1},\cdots,U_{\alpha_{N-1}}$.

Problem Let $X$ be an infinite set. For $p \in X$ and $q \in X$, define $$ d(p,q) = \begin{cases} 1 \;\; (\text{if $p \neq q$}) \\ 0 \;\; (\text{if $p = q$}). \end{cases} $$ Prove that this is a metric. Which subsets of the resulting metric space are open? Which are closed? Which are compact? Answer (1) We first show that $d$ is a metric. (a) By the definition, $d(p,q)>0$ if $p \neq q$ and $d(p,p)=0$. (b) If $p \neq q$ then $d(p,q)=1=d(q,p)$. If $p = q$ then $d(p,q) = 0 = d(q,p)$. (c) If $p \neq q$ then $d(p,q) = 1 \leq d(p,r) + d(r,q)$ because one of $d(p,r)$ and $d(r,q)$ is 1. If $p = q$ then $d(p,q) = 0 \leq d(p,r) + d(r,q)$ for any $r \in X$. (2) For any $A \subset X$, $N_{1/2}(p) = \{ p \} \subset A$ for all $p \in A$, which means every subset of $X$ is open. (3) Every subset of $X$ has an open complement, and therefore is closed. (4) Finite set is obviously compact. For infinite set $A \subset X$, an open cover $\{N_{1/2}(x)\}_{x \in A}$ of $A$ does not have a finite subcover,...

Problem Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable. Answer For $n \in \mathbb{N}$, choose sequentially $x_i \in X$ for $i=1,2,\cdots$ satisfying $d(x_i,x_j) \geq 1/n$ for $j=1,\cdots,i-1$. Then, this process must stop after a finite number of steps (If this process does not stop, then it contradicts to the fact that every infinite subset has a limit point). Define $S_n$ be the set of such points. Note that $S:=\bigcup_{n=1}^\infty S_n$ is countable. It remains to show that $S$ is dense in $X$. Fix $x \in X$ and $\epsilon>0$. We can choose $n \in \mathbb{N}$ such that $1/n<\epsilon$. Then $x \in N_{1/n}(x_\ast)$ for some $x_\ast \in S_n$, which implies $x_\ast \in N_\epsilon(x)$.

Problem A metric space is called separable if it contains a countable dense subset. Show that $\mathbb{R}^k$ is separable. Answer Consider $\mathbb{Q}^k \subset \mathbb{R}^k$, which is a countable set. Fix $x=(x_1,\cdots,x_k) \in \mathbb{R}^k$ and $\epsilon > 0$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists $y_i \in \mathbb{Q}$ such that $y_i \in N_{\epsilon/k}(x_i)$ for each $i=1,\cdots,k$. Then, $y=(y_1,\cdots,y_k) \in N_\epsilon(x)$, which implies $\mathbb{Q}^k$ is dense in $\mathbb{R}^k$. 

Problem Is every point of every open set $E \subset \mathbb{R}^2$ a limit point of $E$? Answer the same question for closed sets in $\mathbb{R}^2$. Answer (1) Consider an open set $E \subset \mathbb{R}^2$. For $x \in E$, there exists $\epsilon>0$ such that $N_\epsilon(x) \subset E$. For any arbitrary $\delta >0$, $y = x + (0,\min(\delta/2,\epsilon/2))$ is contained in $N_\delta(x)$ and $y \neq x$, which means $x$ is a limit point of $E$. (2) Consider a closed set $E = \{(1,1)\} \subset \mathbb{R}^2$. $(1,1)$ is not a limit point of $E$.

Problem Let $A_1,A_2,A_3,\cdots$ be subsets of a metric space. (a) If $B_n=\bigcup_{i=1}^n A_i$, prove that $\overline{B_n} = \bigcup_{i=1}^n \overline{A_i}$, for $n=1,2,3,\cdots$. (b) If $B=\bigcup_{i=1}^\infty A_i$, prove that $\overline{B} \supset \bigcup_{i=1}^\infty \overline{A_i}$. Answer (a) ($\subset$) Let $x \in \overline{B_n} = B_n \cup (B_n)'$. If $x \in B_n$ then $x \in \bigcup_{i=1}^n A_i \subset \bigcup_{i=1}^n \overline{A_i}$. If $x \in (B_n)'$, to prove by contradiction, suppose $x \notin \overline{A_i}$ for all $i=1,\cdots,n$. Then there exists $\epsilon_i >0$ such that $N_{\epsilon_i}(x) \cap A_i = \emptyset$. For $\epsilon=\min(\epsilon_1,\cdots,\epsilon_n)$, $N_{\epsilon}(x) \cap B_n = \emptyset$, which is contradiction. ($\supset$) Let $x \in \overline{A_i}$ for some $i=1,\cdots,n$. If $x \in A_i$ then $x \in B_n$. If $x \in (A_i)'$ then for $\epsilon >0$, there exists $y \in A_i$ such that $y \in N_\epsilon(x)$ and $y \neq x$. Since $y$ is also c...

Problem Prove that every open set in $\mathbb{R}^1$ is the union of an at most countable collection of disjoint segments. Answer Let $K_l:=\{[m,m+1/2^l)\;|\; m \in \mathbb{Z}/2^l\}$. Then $K_l$ is a collection of coutable disjoint segments with length $1/2^l$. For open set $G \subset \mathbb{R}^1$, define $$S_0:=\{Q \in K_0\;|\; Q \subset G\}.$$ $$S_n:=\{Q \in K_n\;|\; Q \subset G, \text{$Q \not\subset Q'$ for some $Q' \in \bigcup_{i=1}^{n-1}K_i$}  \}.$$ $$S:=\bigcup_{n=1}^\infty S_n.$$ Then $S$ is at most countable collection of disjoint segments and $\bigcup_{Q \in S}Q \subset G$. For $x \in G$, there exists $N \in \mathbb{N}$ such that $N_{1/2^N}(x) \subset G$. Then, we can find $m \in \mathbb{Z}/2^N$ such that $m \leq x$ and $m \in N_{1/2^N}(x)$. This means $x \in Q$ for some $Q \in \bigcup_{n=1}^N S_n$. Hence $G \subset \bigcup_{Q \in S}Q$.

Problem Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $E$ and $\overline{E}$ have the same limit points. (Recall that $\overline{E}=E \cup E'$.) Do $E$ and $E'$ always have the same limit points? Answer In Rudin's book, the definition of "limit point" is A point $p$ is a limit point of the set $E$ if every neighborbood of $p$ contains a point $q \neq p$ such that $q \in E$. (1) $E'$ is closed. Let $x$ be a limit point of $E'$. For $\epsilon >0$, there exists $y \in E'$ such that $y \in N_\epsilon(x)$ and $x \neq y$. Since $y \in E'$, there exists $z \in E$ such that $z \in N_{\epsilon - |x-y|}(y)$ and $z \neq y$. Note that $z \neq x$ and $z \in N_\epsilon(x)$. This implies $x \in E'$. (2) $E$ and $\overline{E}$ have the same limit points. Obviously, $E' \subset \overline{E}'$. Conversely, $\overline{E}' = E' \cup (E')' \subset E'$ by (1) and problem 7. (3) For ...

Answer (1) Not connected Note that $[0,1] \cap \mathbb{Q} = [0,\sqrt{2}/2) \cup (\sqrt{2}/2,1]$ and $[0,\sqrt{2}/2),(\sqrt{2}/2,1]$ are separated. (2) Not compact Consider the open cover $\{(-1,\sqrt{2}/2-1/n) \cup (\sqrt{2}/2+1/n,2)\}_{n=2}^\infty$. For any finite subcover $\{(-1,\sqrt{2}/2-1/n_k) \cup (\sqrt{2}/2+1/n_k,2)\}_{k=1}^N$, there exists $q \in N_{1/n_\ast}(\sqrt{2}/2) \cap \mathbb{Q}$ such that $q \notin \bigcup_{k=1}^N(-1,\sqrt{2}/2-1/n_k) \cup (\sqrt{2}/2+1/n_k,2)$ where $n_\ast>\max(n_1,\cdots,n_N)$.

Answer In Rudin's book, the definition of "connected" is A set $E \subset X$ is said to be connected if $E$ is not a union of two nonempty separated sets. Two subsets $A$ and $B$ of a metric space $X$ are said to be separated if both $A \cap \overline{B}$ and $\overline{A} \cap B$ are empty (1) Connected Suppose $(0,1)$ is not connected. Then there are separated sets $A$ and $B$ such that $(0,1)=A \cup B$. Let $a \in A, b \in B$ and assume $a<b$ without loss of generality. Since $C:=[a,b] \cap \overline{A}$ is closed, $x:=\sup C \in C \subset \overline{A}$. Note that $x \neq b$ because $\overline{A} \cap B = \emptyset$. Then, $(x,b] \subset B$  $\Rightarrow$  $[x,b] \subset \overline{B}$  $\Rightarrow$  $x \in \overline{B}$. This implies $x \notin A,B$, but $a \leq x \leq b$  $\Rightarrow$  $x \in (0,1)=A \cup B$, which is contradiction. (2) Not compact Consider the open cover $\{(0,1-1/n)\}_{n=2}^\infty$. For any finite subcover $(0,1-1/n_1),\cdots,(0,1...

Problem Prove the following properties (for subset of $\mathbb{R}^n$). (a) $\text{int}(\text{int}(A)) = \text{int}(A)$. (b) $\text{int}(A \cup B) \supset \text{int}(A) \cup \text{int}(B)$ (c) $\text{int}(A \cap B) = \text{int}(A) \cap \text{int}(B)$. Answer (a) For $x \in \text{int}(\text{int}(A))$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset \text{int}(A)$, which means $x \in \text{int}(A)$. For $x \in \text{int}(A)$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset A$. For $y \in D(x,\epsilon)$, $D(y,\epsilon-|x-y|) \subset D(x,\epsilon) \subset A$, which implies $D(x,\epsilon) \subset \text{int}(A)$. Hence, $x \in \text{int}(\text{int}(A))$. (b) For $x \in \text{int}(A)$, there exists $\epsilon >0$ such that $D(x,\epsilon) \subset A \subset A \cup B$  $\Rightarrow$  $x \in \text{int}(A \cup B)$  $\Rightarrow$  $\text{int}(A) \subset \text{int}(A \cup B)$. Similarly, $\text{int}(B) \subset \text{int}(A \cup B)$. (c) For $x \in \t...

Problem Discuss whether the following sets are open or closed: (a) (1,2) in $\mathbb{R}$ (b) [2,3] in $\mathbb{R}$ (c) $\bigcap_{n=1}^\infty [-1,1/n)$ in $\mathbb{R}$ (d) $\mathbb{R}^n$ in $\mathbb{R}^n$ (e) A hyperplane in $\mathbb{R}^n$ (f) $\{r \in (0,1) \;|\; \text{$r$ is rational} \}$ in $\mathbb{R}$ (g) $\{(x,y) \in \mathbb{R}^2 \;|\; 0<x \leq 1\}$ in $\mathbb{R}^2$ (h) $\{x \in \mathbb{R}^n \;|\; \|x\|=1\}$ in $\mathbb{R}^n$. Answer (a) Open: For $x \in (1,2)$, we can choose $\epsilon = \min(x-1,2-x)$ so that $D(x,\epsilon) \subset (1,2)$. (b) Closed: For $x \in [2,3]^c$, we can choose $\epsilon = \min(2-x,x-3)$ so that $D(x,\epsilon) \subset [2,3]^c$. (c) Closed: We first prove that $[-1,0] = A:=\bigcap_{n=1}^\infty [-1,1/n)$. Since $[-1,0] \subset [-1,1/n)$ for all $n \in \mathbb{N}$, $[-1,0] \subset A$. To prove $A \subset [-1,0]$, we assume that $x \in A$. Then obviously, $x \geq -1$. If $x > 0$, then there exists $n \in \mathbb{N}$ such that $x > 1/n$, which means ...