Calculus TA

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 Give an example of an open cover of the segment $(0,1)$ which has no finite subcover. Answer Consider a family of sets $\{(1/n,1)\}_{n=2}^\infty$. For any $x \in (0,1)$, there exists $N \in \mathbb{N}$ such that $1/N < x$, which means $x \in (1/N,1)$. This means that $\{(1/n,1)\}_{n=2}^\infty$ covers $(0,1)$. Consider an arbitrary finite subcover $\{(1/n_1,1),\cdots,(1/n_K)\}$ of $\{(1/n,1)\}_{n=2}^\infty$. Then $\bigcup_{k=1}^K(1/n_k,1)=(1/n_\ast,1)$ where $n_\ast=\max(n_1,\cdots,n_K)$, which cannot cover $(0,1)$.

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 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 $G \subset \bigcup_{Q \in S}Q$.

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 Suppose $k \geq 3,\mathbf{x},\mathbf{y} \in \mathbb{R}^k, |\mathbf{x} - \mathbf{y}|=d>0$, and $r>0$. Prove: (a) If $2r > d$, there are infinitely many $\mathbf{z} \in \mathbb{R}^k$ such that $$ |\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}| = r.$$ (b) If $2r = d$, there is exactly one such $\mathbf{z}$. (c) If $2r < d$, there is no such $\mathbf{z}$. Answer (a) Consider $\mathbf{w} \in \mathbb{R}^k$ such that $|\mathbf{w}|^2=r^2-d^2/4$ and $\mathbf{w} \cdot (\mathbf{x} - \mathbf{y}) = 0$. Then, the components of $\mathbf{w}=(w_1,\cdots,w_k)$ satisfy the following equations. $$w_1(x_1 - y_1) + \cdots + w_k(x_k - y_k) = 0,$$ $$w_1^2 + \cdots + w_k^2 = r^2 - d^2/4.$$ For $k \geq 3$ and $2r > d$, there are infinitely many $w_1,\cdots,w_k$ satisfying the above. For every $\mathbf{w}$ satisfying the above, $\mathbf{z} = \frac{\mathbf{x} + \mathbf{y}}{2} + \mathbf{w}$ satisfies $$ |\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}| = r.$$ Hence, there there are...

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 ...

Problem Is the set of all irrational real numbers countable? Answer No. Suppose that the set of all irrational real numbers is countable. Then the set of real number is countable since it is a union of two countable sets, which is contradiction.

Answer Define a function $f:(0,1) \rightarrow (0,1]$ as $$f(x) = \begin{cases} 2x, & \text{$x = 1/2^n$ for $n=1,2,\cdots$} \\ x, & \text{otherwise} \end{cases}$$ (1) Injective: suppose $f(x_1) = f(x_2)$. If $f(x_1)=f(x_2)=1/2^n$ for some $n \in \{0,1,2,\cdots\}$, then $x_1=1/2^{n+1}=x_2$. Otherwise, $x_1=f(x_1)=f(x_2)=x_2$. (2) Surjective: suppose $y \in (0,1]$. If $y = 1/2^n$ for some $n \in \{0,1,2,\cdots\}$, then $y = f(1/2^{n+1})$. Otherwise, $y = f(y)$.

Problem Show that $d=\inf(S)$ iff $d$ is a lower bound for $S$ and for any $\epsilon > 0$ there is an $x \in S$, such that $d \geq x - \epsilon$. Answer ($\Rightarrow$) Let $d = \inf(S)$ then obviously $d$ is a lower bound of $S$. For $\epsilon >0$, $d+\epsilon$ is cannot be a lower bound of $S$. Therefore, there exists $x \in S$ such that $x < d + \epsilon$  $\Rightarrow$  $d \geq x - \epsilon$. ($\Leftarrow$) Let $d$ be a lower bound satisfying the given condition. Suppose $c$ is another lower bound of $S$ and $c > d$. Then for $\epsilon:=(c - d)/2$, there exists $x \in S$ such that $d \geq x - \epsilon > x - 2\epsilon = x - (c - d)$  $\Rightarrow$  $c > x$. This contradicts the fact that $c$ is a lower bound.

Problem For sets $A,B \subset \mathbb{R}$, let $A+B=\{x+y | \text{$x \in A$ and  $y \in B$}\}$. Show that $\sup(A+B) = \sup(A) + \sup(B)$. Make a similar statement for inf's. Answer Let $a = \sup(A), b= \sup(B)$. Then for $z \in A+B$, $z = x+y$ for some $x \in A, y \in B$ and $z = x+y \leq a+b$, which means $a+b$ is an upper bound of $A+B$. For $\epsilon > 0$, there exists $x_\epsilon \in A, y_\epsilon \in B$ such that $x_\epsilon > a - \epsilon/2, y_\epsilon > b - \epsilon/2$  $\Rightarrow$  $x_\epsilon + y_\epsilon > a+b - \epsilon$. This implies $a+b = \sup(A+B)$. For inf's we can also prove the following in similar way. $$ \inf(A+B) = \inf(A) + \inf(B)$$

Problem Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $$\inf A = -\sup(-A).$$ Answer Let $\alpha = \inf A$. For all $x \in A$, $\alpha \geq x$  $\Longrightarrow$  $-\alpha \geq -x$. Therefore, $-\alpha$ is upper bound of $-A$. To prove $-\alpha$ is the supremum of $-A$, suppose that there exists another upper bound $\beta$ of $-A$. It remains to show that $\beta \geq -\alpha$. For all $x \in A$, $-x \leq \beta$  $\Longrightarrow$ $x \geq -\beta$, which means $-\beta$ is lower bound of $A$. Since $\alpha=\inf A$, $\alpha \geq -\beta$ $\Longrightarrow$ $\beta \geq -\alpha$.

Problem Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $D$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$. Answer Since $E$ is nonempty, there exists $x \in E$. Then $\alpha \leq x \leq \beta$.

Problem If $r$ is rational ($r \neq 0$) and $x$ is irrational, prove that $r+x$ and $rx$ are irrational. Answer If $r+x$ is rational then $-r+r+x=x$ is rational, which is contradiction. If $rx$ is rational then $(1/r)rx=x$ is rational, which is contradiction.

Problem Fix $b>1$, $y>0$, and prove that there is a unique real $x$ such that $b^x=y$, by completing the following outline. (This $x$ is called the logarithm of $y$ to the base $b$.) (a) For any positive integer $n$, $b^n - 1 \geq n(b-1)$. (b) Hence $b-1 \geq n(b^{1/n}-1)$. (c) If $t>1$ and $n>(b-1)/(t-1)$, then $b^{1/n}<t$. (d) If $w$ is such that $b^w < y$, then $b^{w+(1/n)} < y$ for sufficiently large $n$; to see this, apply part (c) with $t=y \cdot b^{-w}$. (e) If $b^w > y$, then $b^{w-(1/n)} > y$ for sufficiently large $n$. (f) Let $A$ be the set of all $w$ such that $b^w < y$, and show that $x = \sup A$ satisfies $b^x = y$. (g) Prove that this $x$ is unique. Answer (a) Note that $$b^n - 1 =(b-1)(1+b+b^2 + \cdots + b^{n-1}).$$ Since $b>1$, $$1+b+b^2+\cdots+b^{n-1} \geq n.$$ Hence $$b^n - 1 = (b-1)(1+b+b^2 + \cdots + b^{n-1}) \geq n(b-1).$$ (b) It is enough to show that $b^{1/n} > 1$. If $b^{1/n} \leq 1$, then $b = (b^{1/n})^n \leq 1^n = 1$, whic...

Problem Prove Proposition 1.15. The axioms for multiplication imply the following statements. (a) If $x \neq 0$ and $xy = xz$ then $y = z$. (b) If $x \neq 0$ and $xy = x$ then $y = 1$. (c) If $x \neq 0$ and $xy = 1$ then $y = 1/x$. (d) If $x \neq 0$ then $1/(1/x) = x$. Answer (a) $xy = xz$   $\overset{(M5)}{\Longrightarrow}$   $(1/x)(xy) = (1/x)(xz)$   $\overset{(M3)}{\Longrightarrow}$   $((1/x)x)y = ((1/x)x)z$   $\overset{(M2)}{\Longrightarrow}$   $(x(1/x))y = (x(1/x))z$   $\overset{(M5)}{\Longrightarrow}$   $1y = 1z$   $\overset{(M4)}{\Longrightarrow}$   $y = z$. (b) $xy = x$   $\overset{(M4)}{\Longrightarrow}$   $xy = 1x$   $\overset{(M2)}{\Longrightarrow}$   $xy = x1$   $\overset{(a)}{\Longrightarrow}$   $y = 1$. (c) $xy = 1$   $\overset{(M5)}{\Longrightarrow}$   $xy = x(1/x)$   $\overset{(a)}{\L...

Problem Prove that there is no rational number whose square is 12. Answer Suppose there exists a rational number $p$ that satisfies $p^2=12$. Since $p$ is a rational number, there are some integers $m,n$ that are not both multiples of 3 and $p=m/n$. Then, we have $12n^2=m^2$. This shows that $m$ is a multiple of 3. Then $n$ also has to be a multiple of 3 and this is a contradiction.