Feb 23

Here is a solution for Problem 17.

Suppose it is contractive. Then there is a $$c\in(0,1)$$ such that for all $$n\in\N$$, $1>c>\left|\frac{a_{n+2}-a_{n+1}}{a_{n+1}-a_n}\right|=\frac{n}{n+2}\to1.$ The Sandwich Theorem (3.9) implies $$c=1$$. This contradiction forces the conclusion that $$a_n$$ cannot be contractive.

Here is a solution for Problem 18.

By Theorem 3.22, $$a_n$$ converges to some $$A\in\R$$. Theorem 3.14 implies $$A=L$$.

Problem 20. (Due Mar 2.) Let $$x_n$$ be a sequence with range $$\{0,1,2,3,4,5,6,7,8,9\}$$. Prove that $$\sum_{n=1}^\infty x_n10^{-n}$$ converges and its sum is in the interval $$[0,1]$$.

Problem 21. (Due Mar 2.) Prove that $$6.17272727272\cdots$$ is a rational number.

Feb 21

Here is a solution for Problem 16.

To see $$a_n$$ is bounded, note that for every $$n\in\N$$, $\frac{1}{2} = \frac{n}{2n} \lt \sum_{k=n+1}^{2n}1/k=a_n \lt \frac{n}{n+1} \lt 1.$ Then \begin{align*} a_{n+1}-a_n &= \sum_{k=n+2}^{2(n+1)}\frac{1}{k}-\sum_{k=n+1}^{2n}\frac{1}{k}\\ &= \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1}\\ &= \frac{1}{4n^2+6n+2}\\ &>0 \end{align*} shows $$a_n$$ is increasing. Theorem 3.11 implies $$a_n$$ converges. Theorem 3.22 now implies $$a_n$$ is a Cauchy sequence.

Here is a solution for Extra Credit 3.

By using upper and lower rectangles on the graph of $$y=1/x$$, it can be seen that $a_n+\frac{1}{n} > \int_{n+1}^{2n}\frac{dx}{x} > a_n.$ (See the figure below for the case when $$n=6$$.) Calculate the integral and rearrange this inequality. $\frac{1}{n} > \ln\left( \frac{2n}{n+1} \right)-a_n > 0.$ The Sandwich Theorem (3.9) implies $$a_n\to\ln2$$.

Feb 19

Problem 19. (Due Feb 26.) If $$a_n$$ is a convergent sequence and $$b_n$$ is a sequence such that $$|a_m-a_n|\ge|b_m-b_n|$$ for all $$m,n\in\N$$, then $$b_n$$ converges.

Here is a solution for Problem 15.

Since $$a_n$$ is bounded, there is an interval $$J=[\alpha,\beta]$$ such that $$\{n:a_n\in J\}=\N$$. Let $$\e>0$$ choose $$N\in\N$$ such that $\frac{\beta-\alpha}{N}\lt\e.$ Partition $$J$$ by $$\alpha=x_0\lt x_1\lt\cdots\lt x_N=\beta$$, where $$x_i-x_{i-1}=(\beta-\alpha)/N$$ for $$1\le i\le N$$. If $$S_i=\{n:x_{i-1}\le a_n\le x_i\}$$ for $$1\le i\le N$$, then $$\N=\bigcup_{i=1}^N S_i$$. At least one of the $$S_i$$, say $$S_{i_0}$$ must be infinite, else $$\N$$ is a finite union of finite sets. Let $$I$$ be any interval of length $$\e$$ containing $$[x_{i_0-1},x_{i_0}]$$.

If $$a_n=n$$, then no interval of finite length can contain an infinite number of terms from the sequence, so the boundedness of the sequence is necessary.

Feb 17

Here is a solution for Problem 14.

First, it will be shown $$a_n\gt1$$for all $$n\in\N$$.

It is clear that $$a_1=3\gt1$$. Suppose $$a_n>1$$ for some $$n$$. Then $a_{n+1}=2-\frac{1}{a_n}>2-1=1,$ and it follows by induction that $$a_n>1$$ for all $$n\in\N$$.

Next, it will be shown that $$a_n$$ is strictly decreasing.

Clearly $$a_1=3>5/3=a_2$$. Suppose $$a_n\lt a_{n-1}$$ for some $$n>1$$. Then \begin{align*} a_{n+1}-a_n &= 2-\frac{1}{a_n}-a_n\\ &= 2-\frac{1}{a_n}-\left( 2-\frac{1}{a_{n-1}} \right)\\ &=\frac{1}{a_{n-1}}-\frac{1}{a_n}\\ &= \frac{a_n-a_{n-1}}{a_{n-1}a_n}\\ &\lt 0 \end{align*} In the last step, the induction hypothesis was invoked, along with the facts that $$a_{n-1}>a_n>1>0$$. It follows by induction that $$a_n$$ is a strictly decreasing sequence bounded below by $$1$$.

Theorem 3.11 implies $$a_n\to\ell$$ for some $$\ell\ge1$$. Using this \begin{align*} \lim_{n\to\infty}a_{n+1}&=\lim_{n\to\infty}\left( 2-\frac{1}{a_n} \right)\\ \ell&=2-\frac{1}{\ell}\\ \ell^2-2\ell+1&=0\\ (\ell-1)^2&=0\\ \ell&=1 \end{align*}

Feb 16

Problem 17. (Due Feb 23.) $$a_n=1/n$$ is a Cauchy sequence which is not contractive.

Problem 18. (Due Feb 23.) If $$a_n$$ is a Cauchy sequence and $$b_n$$ is a subsequence of $$a_n$$ such that $$b_n\to L$$, then $$a_n\to L$$.

Feb 14 💕

Problem 16. (Due Feb 21.) Prove that $$a_n=\sum_{k=n+1}^{2n}\frac{1}{k}$$ is a Cauchy sequence.

Extra Credit 3. (Due Feb 21.) Find $$\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$$. (You will probably have to use some calculus.)

Here is a solution for Extra Credit 2.

Note first that $$(2n)!\ge\prod_{k=n}^{2n}k> n^{n+1}$$. Therefore $\sqrt[2n]{(2n)!}>\sqrt[2n]{n^{n+1}}\ge\sqrt{n}.$ Next, since $$(2n+1)!\ge\prod_{k=n}^{2n+1}k>\prod_{k=n}^{2n}k> n^{n+1}$$, $\sqrt[2n+1]{(2n+1)!}>\sqrt[2n+1]{n^{n+1}}\ge\sqrt{n}.$ Therefore, $$a_n\ge\sqrt{n}\to\infty$$, and the proof is done.

For a nice application of this limit, see the article by Charles C. Mumma, $$n!$$ and The Root Test, Amer. Math. Monthly, 93(7):561, August-September 1986.

Feb 13

Here is a solution for Problem 10.

Let $$p=q/r$$, where $$q,r\in\Z$$. \begin{align*} q+x\in\Q &\iff \exists m,n\in\Z\left( \frac{q}{r}+x=\frac{m}{n} \right)\\ &\iff \exists m,n\in\Z\left( x=\frac{m}{n}-\frac{q}{r}=\frac{mr-nq}{nr} \right)\\ &\iff x\in\Q \end{align*}

Here is a solution for Problem 11.

First note that if $$A=\emptyset$$, then $$B=\R$$ and $$\lub A=\glb B=-\infty$$. Also, if $$A$$ is not bounded above, then $$B=\emptyset$$ and $$\lub A=\glb B=\infty$$.

So, suppose $$A\ne\emptyset\ne B$$ and $$A$$ is bounded above. Let $$\alpha=\lub A$$ and $$\beta=\glb B$$. Since $$\alpha$$ is an upper bound for $$A$$, it follows that $$\alpha\in B$$. This, in turn, implies $$\beta\le\alpha$$.

Suppose $$\beta\lt\alpha$$ and $$\e=(\alpha-\beta)/2$$. According to Theorem 2.19, there is a $$\gamma\in(\alpha-\e,\alpha]\cap A$$ and a $$\delta\in[\beta,\beta+\e)\cap B$$. So, $$\delta\lt\gamma$$, where $$\delta$$ is an upper bound for $$A$$ and $$\gamma\in A$$. This is clearly impossible, and we must conclude $$\alpha=\beta$$, as desired.

Here is a solution for Problem 12.

Clearly, $$\sigma(1)\ge1$$. Suppose $$\sigma(n)\ge n$$ for some $$n\in\N$$. Then, $$\sigma(n+1)>\sigma(n)\ge n$$, implies $$\sigma(n+1)\ge n+1$$. By induction, it follows that $$\sigma(n)\ge n,\ \forall n\in\N$$.

Here is a solution for Problem 13.

Let $$\e>0$$ and $$N\in\N$$ such that $$N>11/9\e$$. If $$n\ge N$$, then \begin{align*} \left|a_n-2\right| &= \left|\frac{4n-1}{3n+2}-\frac{4}{3}\right|\\ &= \frac{11}{3(3n+2)}\\ &\le \frac{11}{3(3N+2)}\\ &\lt \frac{11}{3\left(3\frac{11}{9\e}+2\right)}\\ &= \frac{11}{\frac{11}{\e}+6}\\ &\lt \frac{11}{\frac{11}{\e}}\\ &= \e. \end{align*} This shows Definition 3.3 is satisfied, so $$a_n\to4/3$$.

Feb 12

Problem 15. (Due Feb 19.) Let $$a_n$$ be a bounded sequence. Prove that given any $$\e>0$$, there is an interval $$I$$ with length $$\e$$ such that $$\{n:a_n\in I\}$$ is infinite. Is it necessary that $$a_n$$ be bounded?

Here is a solution for Problem 9.

Let $$I=(a,b).$$

First, assume $$a>0$$. According to Corollary 2.23(b), there is an $$n\in\N$$ such that $$1/n\lt b-a$$. Corollary 2.23(a) shows $$\{k\in\N:k>an\}\ne\emptyset$$. Let $$m=\min\{k\in\N:k>an\}$$, so $$m/n>a$$. In order to arrive at a contradiction, suppose$$m/n\ge b$$. Then $\frac{m-1}{n}=\frac{m}{n}-\frac{1}{n}\ge b-\frac{1}{n}>b-(b-a)=a$ is a contradiction of the choice of $$m$$ and it must be that $$m/n\lt b$$.

Therefore $$m/n\in(a,b)$$.

If $$a\lt 0$$, then use the Archimedean property to find an $$N\in\N$$ such that $$N>-a$$. From above, there is a $$q\in(a+N,b+N)\cap\Q$$. Clearly, $$q-N\in(a,b)\cap\Q$$ because $$\Q$$ is closed under addition.

Feb 9

Problem 14. (Due Feb 16.) Let $$a_1=3$$ and $$a_{n+1}=2-1/a_n$$ for $$n\in\N$$. Analyze the sequence.

Feb 7

Extra Credit 2. (Due Valentine’s Day.) Determine the limit of $$a_{n}=\sqrt[n]{n!}$$. (Hint: If $$n$$ is even, then show that $$n!>(n/2)^{n/2}$$.)

Feb 5

Here is a solution for Problem 8.

Since $$\alpha$$ is an upper bound for $$S$$, $$(\alpha,\infty)\cap S=\emptyset$$. Because $$\alpha\in S$$, it follows that for all $$\e>0$$, $$\alpha\in(\alpha-\e,\alpha]\cap S$$, so $$(\alpha-\e,\alpha]\cap S\ne\emptyset$$. Theorem 2.19 implies $$\alpha=\lub{S}$$.

Here is a solution for Problem 7.

By assumption, $$a\ge0$$. Assume $$a>0$$. The given condition gives $$a>a\ge0$$, which is impossible. This contradiction forces the conclusion $$a=0$$.

Feb 1

Problem 12. (Due Feb 12.) If $$\sigma:\N\to\N$$ is strictly increasing, then $$\sigma(n)\ge n$$ for all $$n\in\N$$.

Problem 13. (Due Feb 12.) Let the sequence $$\displaystyle a_n=\frac{4n-1}{3n+2}$$. Use the definition of convergence for a sequence to show $$a_n$$ converges.

Here is a solution for Problem 6.

Case 1: Suppose $$a$$ and $$b$$ are negative. Then $$-a$$ and $$-b$$ are positive and $$(-a)(-b)=ab=c$$. The two-out-of-three rule implies $$c\gt0$$.
Case 2: Suppose $$a$$ and $$c$$ are negative. Corollary 2.8 shows $$a^{-1}\lt0$$. Since $$ab=c\iff b=ca^{-1}$$ and case 1 implies $$b>0$$.
Case 3: If $$b$$ and $$c$$ are negative, this is the same as case 2.

Jan 31

Problem 10. (Due Feb 9.) Let $$q\in\mathbb{Q}$$ and $$x\in\R$$ Prove $q+x\in\Q \iff x\in\Q.$

Problem 11. (Due Feb 9.) If $$A\subset\R$$ and $$B=\{x:x\text{ is an upper bound for }A\}$$, then $$\lub(A)=\glb(B)$$.

Jan 30

Here is a solution for Problem 5.

($$\Rightarrow$$) Suppose$$|x|\le y$$. By Theorem 2.11(a), $$|x|\ge0$$ for all $$x$$, so Theorem 2.5(b) shows $$y\ge0$$, and consequently $$-y\le0$$ by Axiom 7(b). If $$x\ge0$$, then Theorem 2.11(b) implies $$\label{ineqp5.1} -y\le0\le x\le y. \tag{3}$$ If $$x\lt0$$, then $$-x=|x|>0$$. Multiplying the assumed inequality $$-x=|x|\le y$$ by $$-1$$ gives $$\label{ineqp5.2} -y\le x\lt0\le y. \tag{4}$$ Combining (3) and (4) gives $$-y\le x\le y$$, as desired.

($$\Leftarrow$$) Assume $$-y\le x\le y$$. If $$x\ge0$$, then $$|x|=x$$, and it is obvious $$|x|\le y$$. If $$x\lt0$$, then multiplying the inequality $$-y\le x$$ by $$-1$$ gives $$-x=|x|\le y$$, as desired. Therefore, in all cases, $$|x|\le y$$.

Jan 29

Problem 9. (Due Feb 7.) If $$I$$ is an interval, then $$\Q\cap I\ne\emptyset$$.

Jan 28

Here is a solution for Problem 4.

By Axiom 4, $$1\ne0$$. Apply Theorem 2.4.

Here is a solution for Problem 3.

Since $$A\subset A\cup B$$, it's clear that \begin{align}\label{p3.1} \aleph_0\le\card{A}\le\card{A\cup B}.\tag{1} \end{align} By assumption, there are two bijections $$\phi:A\to\N$$ and $$\psi:B\to\N$$. Define $$\zeta:A\cup B\to\N$$ by \begin{align*} \zeta(x) = \begin{cases} 2\phi(x), & x\in (A\setminus B) \cup (A\cap B)\\ 2\psi(x)-1, & x\in B\setminus A \end{cases} \end{align*} It is straightforward to show $$\zeta$$ is injective. This implies $\card{A\cup B}\le\aleph_0.\tag{2}$ Combining (1) and (2) shows $$\card{A\cup B}=\aleph_0$$.

Jan 26

Problem 7. (Due Feb 5.) If $$\F$$ is an ordered field and $$a\in\F$$ such that $$0\le a\lt \e$$ for every $$\e\gt0$$, then $$a=0$$.

Problem 8. (Due Feb 2.) If $$\alpha$$ is an upper bound for $$S$$ and $$\alpha\in S$$, then $$\alpha=\lub S$$.

Jan 24

Problem 6. (Due Jan 31.) Let $$\F$$ be an ordered field and $$a,b,c\in\F$$. If $$ab=c$$ and two of $$a$$, $$b$$ and $$c$$ are negative, then the third is positive.

Jan 22

Problem 5. (Due Jan 29.) Prove Theorem 2.11(d) from the notes.

Jan 21

Here is a solution for Extra Credit 1.

One way to do this is to figure out the first few sets in the sequences $$A_n$$ and $$B_n$$. \begin{align*} &B_1=\Z\setminus f(\N)=\Z\setminus\N=\{-n:n\in\omega\}=\{0,-1,-2,-3,\dots\}\\ % &A_1=g(B_1)=\{g(-n):n\in\omega\}=\{1+3n:n\in\omega\}=\{1,4,7,10,\dots\}\\ % &B_2=A_1\\ % &A_2=g(B_2) = \{g(1+3n):n\in\omega\} = \{2+9n:n\in\omega\} = \{2,11,20,29,\dots\}\\ % &B_3=A_2\\ % &A_3=g(B_3) =\{g(2+9n):n\in\omega\} =\{5+27n:n\in\omega\} =\{5,32,59,86,\dots\}\\ % &B_4=A_3\\ % &A_4 =g(B_4) =\{g(5+27n):n\in\omega\} =\{14+81n:n\in\omega\} =\{14,95,176,\dots\}\\ \end{align*} Since all elements of $$A_n$$ for $$n>3$$ exceed $$6$$, it follows that $$7\in\tilde{A}$$ and $$6\in A\setminus\tilde{A}$$. Therefore, $$h(7)=g^{-1}(7)=-2$$ and $$h(6)=f(6)=6$$.

It's an interesting puzzle to determine the form of $$A_k$$ for an arbitrary $$k$$. It turns out that $A_1=\{1+3n:n\in\omega\}$ and for $$k>1$$, $A_k=\left\{1+\sum_{i=1}^{k-1}3^{i-1}+3^kn:n\in\omega\right\} =\left\{3^k\left(n+\frac{1}{2}\right)+\frac{1}{2}:n\in\omega\right\}.$

I've not been able to find a satisfying formula giving the elements of $$\tilde{A}$$.

Here is a solution for Problem 2.

It must be shown that $$g\circ f$$ is surjective and injective.

Let $$z\in C$$. Since $$g$$ is surjective, there is a $$y\in B$$ such that $$g(y)=z$$ and since $$f$$ is surjective, there is an $$x\in A$$ such that $$f(x)=y$$. Since, $$g\circ f(x)=g(y)=z$$ and $$z$$ is an arbitrary element of $$C$$, it follows that $$g\circ f$$ is surjective.

If $$x_1,x_2\in A$$ with $$x_1\ne x_2$$, then $$f(x_1)\ne f(x_2)$$ because $$f$$ is injective. In the same way, since $$g$$ is injective, $$g(f(x_1))\ne g(f(x_2))$$ and it follows that $$g\circ f$$ is injective.

Jan 19

Chapter 1 was finished and the first seven axioms of $$\R$$ were introduced.

Problem 3. (Due Jan 26) If $$A$$ and $$B$$ are sets such that $$\card{A}=\card{B}=\aleph_0$$, then $$\card{A\cup B}=\aleph_0$$.

Problem 4. (Due Jan 26) In an ordered field $$\F$$ show $$0\lt1$$.

Jan 17

To solve Problem 1, let $$S=\{a,\{a\}\}$$

for some $$a$$. Then $S\cap\powerset{S}=\{a,\{a\}\}\cap\{\emptyset,\{a\},\{\{a\}\}\{a,\{a\}\}\}=\{\{a\}\}\ne\emptyset.$
Jan 12

Problem 2. (Due Jan 19.) If $$f:A\to B$$ and $$g:B\to C$$ are bijections, then so is $$g\circ f:A\to C$$.

Extra Credit 1. (Due Jan 19.) Using the notation from the proof of the Schröder-Bernstein Theorem, let $$A=\mathbb{N}$$, $$B=\mathbb{Z}$$, $$f(n)=n$$ and $g(n)=\begin{cases}1-3n,&n\le0\\3n-1,&n>0\end{cases}.$ Calculate $$h(6)$$ and $$h(7)$$.

Jan 10

The Schröder-Bernstein theorem was proved.

Jan 8

Problem 1. (Due Jan 17.) Is there a set $$S$$ such that $$S\cap\powerset{S}\ne\emptyset$$?

Read §1.1–1.5.

Jan 3

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