Calculus TA

Problem Call a mapping of $X$ into $Y$ open if $f(V)$ is an open set in $Y$ whenver $V$ is an open set in $X$. Prove that every continuous open mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$ is monotonic. Answer Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is an open mapping but not a monotonic function. Then, without loss of generality, there exists $a,b,c \in \mathbb{R}$ such that $a<b<c$, $f(a) < f(b)$, and $f(c) < f(b)$. Note that the interval $[a,c] \subset \mathbb{R}$ is compact, so there exists $p \in [a,c]$ such that $f(p) = \sup f([a,c])$. $p$ is an interior point of $[a,c]$ since $f$ does not attain a maximum at $a$ or $c$. Choose $\epsilon > 0$ such that $N_\epsilon(p) \subset (a,c)$. Then $f(N_\epsilon(p))$ is open because $f$ is an open mapping. For $f(p) \in f(N_\epsilon(p))$, we can choose $\epsilon_1>0$ such that $N_{\epsilon_1}(f(p)) \subset f(N_\epsilon(p))$. This means $f(q) > f(p)$ for some $q \in N_\epsilon(p) \subset [a,c]$, which is contradic...

Problem Show that the requirement in the definition of uniform continuity can rephrased as follows, in terms of diameters of sets: To every $\epsilon > 0$ there exists a $\delta > 0$ such that $\text{diam}f(E)<\epsilon$ for all $E \subset X$ with $\text{diam}E < \delta$. Definition 4.18 Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. We say that $f$ is uniformly continuous on $X$ if for every $\epsilon >0$ there exists $\delta >0$ such that $d_Y(f(p),f(q))<\epsilon$ for all $p$ and $q$ in $X$ for which $d_X(p,q) < \delta$. Answer Let us denote Definition 4.18 and the new definition as Definition A and Definition B. We will show that Definition A implies Definition B and vice versa. A $\Rightarrow$ B: For $\epsilon>0$, there exists $\delta_A>0$ such that $d_Y(f(p),f(q))<\epsilon/2$ for all $p$ and $q$ in $X$ for which $d_X(p,q) < \delta_A$. Let $\delta:=\delta_A$ and consider $E \subset X$ satisfying $\text{diam}E < \de...

Problem Let $f$ be a continuous real function on $\mathbb{R}^1$., of which it is known that $f'(x)$ exists for all $x \neq 0$ and that $f'(x) \rightarrow 3$ as $x \rightarrow 0$. Does it follow that $f'(0)$ exists? Answer $f'(0) = \lim_{h \rightarrow 0}\frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0}\frac{\frac{d}{dh}(f(0+h)-f(0))}{\frac{d}{dh}h}=\lim_{h \rightarrow 0} f'(h) = 3$. 

Problem If $$ C_0 + \frac{C_1}{2} + \cdots + \frac{C_{n-1}}{n} + \frac{C_n}{n+1} =0, $$ where $C_0,\cdots,C_n$ are real constants, prove that the equation $$ C_0 + C_1x + \cdots + C_{n-1}x^{n-1} + C_nx^n = 0 $$ has at least one real root between 0 and 1. Answer Define a function $p:[0,1] \rightarrow \mathbb{R}$ as $$ p(x) = C_0x + \frac{C_1}{2}x^2 + \cdots + \frac{C_n}{n+1}x^{n+1}. $$ Then, $p$ is differentiable on $[0,1]$ and $p(0) = p(1) = 0$. By the mean value theorem, there exists $x \in (0,1)$ such that $$ C_0 + C_1x + \cdots + C_{n-1}x^{n-1} + C_nx^n = p'(x) = \frac{p(1) - p(0)}{1-0} = 0.$$

Problem If $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Assume there is no $x \in \mathbb{R}$ such that $f(x)=0=f'(x)$. Show that $S=\{x\;|\;0\leq x \leq 1,f(x)=0\}$ is finite. Answer Note that $S=f^{-1}(0)$ is closed since $f$ is continuous and $\{0\}$ is closed. Also, $S$ is bounded since $S \subset [0,1]$. These imply $S$ is compact. If $S$ is infinite, there is a sequence $\{x_n\} \subset S$ and its subsequence $x_{n_k} \rightarrow x \in S$. Then $$ f'(x) = \lim_{k \rightarrow \infty} \frac{f(x) - f(x_{n_k})}{x-x_{n_k}} =0. $$ Hence, $S$ has to be finite.

Problem Let $f:(0,1) \rightarrow \mathbb{R}$ be uniformly continuous. Must $f$ be bounded? Answer Since $f$ is a uniformly continuous, there exists a $\delta >0$ such that $|x-y|<\delta$  $\Rightarrow$  $|f(x)-f(y)|<1$. Suppose $f$ is not bounded. Then, we can choose a sequence $x_n \in (0,1)$ such that $f(x_{n+1}) > f(x_n) + 1$. Since $x_n$ is a sequence in the compact set $[0,1]$, there exists a convergent subsequence $x_{n_k}$. Since $x_{n_k}$ is a Cauchy sequence, there exists $K \in \mathbb{N}$ such that $|x_{n_k} - x_{n_l}|<\delta$ for all $k,l \geq K$. However, $|f(x_{n_K}) - f(x_{n_{K+1}})|>1$, which is contradiction.

Problem Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x \in I$. Answer In Rudin's book, the intermediate value theorem states that Let $f$ be a continuous real function on the interval $[a,b]$. If $f(a) < f(b)$ and if $c$ is a number such that $f(a) < c < f(b)$, then there exists a point $x \in (a,b)$ such that $f(x)=c$. Suppose $f(x) \neq x$ for all $x \in I$. Define a function $g(x) = f(x) -x$. Then $g$ is continuous and $g(0) = f(0) >0$, $g(1)=f(1) - 1 <0$. By the intermediate value theorem, there exists a point $c \in (0,1)$ such that $g(c)=0$  $\Rightarrow$  $g(c) = c$, which is a contradiction.

Problem If $s_1 = \sqrt{2}$, and $$ s_{n+1}=\sqrt{2 + \sqrt{s_n}}\;\;\;(n=1,2,3,\cdots),$$ prove that $\{s_n\}$ converges, and that $s_n < 2$ for $n=1,2,3,\cdots$. Answer Note that $s_1 < 2$. If $s_n < 2$ then $$ s_{n+1} = \sqrt{2 + \sqrt{s_n}} < \sqrt{2+\sqrt{2}} < 2. $$ By mathematical induction, $s_n < 2$ for $n=1,2,3,\cdots$. Now we claim that $s_n$ is increasing, and hence convergent. Note that $s_1 = \sqrt{2} < \sqrt{2 + \sqrt{2}}$. If $s_{n-1} < s_n$ then $$ s_{n+1} = \sqrt{2+\sqrt{s_n}} > \sqrt{2 + \sqrt{s_{n-1}}} = s_n$$. By mathematical induction, $s_n$ is increasing.

Problem If $f$ is defined on $E$, the graph of $f$ is the set of points $(x,f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact. Answer $\Rightarrow$: Consider a map $\phi(x):=(x,f(x))$. Since $\phi$ is continuous and $E$ is compact, the graph $\phi(E)$ is compact. $\Leftarrow$: Suppose $f$ is not continuous on $E$. Then for any sequence $x_n \rightarrow x$, $f(x_n) \not\rightarrow f(x)$. Since $\phi(E)$ is compact, the sequence $(x_n,f(x_n)) \in \phi(E)$ has a convergent subsequence $(x_{n_k},f(x_{n_k})) \rightarrow (p,q) \in \phi(E)$. Since $x_n \rightarrow x$, $x_{n_k} \rightarrow x = p$. Furthermore, $q=f(p)$ because $(p,q)$ is on the graph. Hence, $f(x_{n_k}) \rightarrow f(x)$, which is a contradiction.

Problem For any two real sequences $\{a_n\}$, $\{b_n\}$, prove that $$ \limsup_{n \rightarrow \infty} (a_n + b_n) \leq \limsup_{n \rightarrow \infty} a_n + \limsup_{n \rightarrow \infty} b_n $$ Answer Note that there exists $n_k$ such that $a_{n_k} + b_{n_k} \rightarrow \limsup_{n \rightarrow \infty} (a_n + b_n)$. $\limsup_{n \rightarrow \infty} (a_n + b_n) = \lim_{k \rightarrow \infty} (a_{n_k} + b_{n_k}) \leq \lim_{k \rightarrow \infty} (\sup_{m \geq k}a_{n_m} + \sup_{m \geq k}b_{n_m}) = \lim_{k \rightarrow \infty}\sup_{m \geq k}a_{n_m} + \lim_{k \rightarrow \infty}\sup_{m \geq k}b_{n_m} = \limsup_{n \rightarrow \infty} a_n + \limsup_{n \rightarrow \infty} b_n$.

Problem Let $x_n$ be a Cauchy sequence in $\mathbb{R}$ and let $A_n=\sup\{x_n,x_{n+1},\cdots\}$ and $B_n=\inf\{x_n,x_{n+1},\cdots\}$. Prove $A_n$ converges to the same limit as $B_n$, which in turn is the same as the limit of $x_n$. Answer Since $\mathbb{R}$ is complete, $x_n \rightarrow x \in \mathbb{R}$.  For $\epsilon>0$, there exists $N$ such that $|x_n - x|< \epsilon$ for all $n \geq N$. Since $A_n=\sup\{x_n,x_{n+1},\cdots\}$, $A_n \geq x_n >x - \epsilon$ for all $n \geq N$. On the other hand, note that $x_n < x+\epsilon$ for all $n \geq N$ $\Rightarrow$  $A_n \leq x + \epsilon$ for all $n \geq N$  $\Rightarrow$  $|A_n - x| \leq \epsilon$ for all $n \geq N$. This means $A_n \rightarrow x$. Since $B_n=\inf\{x_n,x_{n+1},\cdots\}$, $B_n \leq x_n < x + \epsilon$ for all $n \geq N$. On the other hand, note that $x_n > x-\epsilon$ for all $n \geq N$ $\Rightarrow$  $B_n \leq x - \epsilon$ for all $n \geq N$  $\Rightarrow$  $|B_n - x| \l...

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 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 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...