Here is a solution for Problem 38.
Let \(\ell(x)\) be the line determined by \((a,f(a))\) and \((b,f(b))\) and \(F(x)=f(x)-\ell(x)\). Then \(F\in C([a,b]\cap D((a,b))\) and \(F(a)=F(x_0)=F(b)=0\). According to Rolle's Theorem there are \(c_1\in(a,x_0)\) and \(c_2\in(x_0,b)\) such that \(F'(c_1)=F'(c_2)=0\). Since \(f'\in D((a,b))\), it must be that \(F'\in D((a,b))\) and \(F'\in C([c_1,c_2])\). Another application of Rolle's Theorem gives a \(c\in(c_1,c_2)\) such that \[ 0=F''(c)=f''(c)+\ell''(c)=f''(c), \] because \(\ell\) is a linear function.
Here is a solution for Problem 36.
Suppose \(|f'(x)|\lt B\) for all \(x\) and \(\e\gt 0\). Since \(x_n\) is a Cauchy sequence, there is an \(N\in\N\) such that \[ n,m\ge N\implies|x_n-x_m|\lt\frac{\e}{B}. \] If \(n,m\ge N\), then the Mean Value Theorem implies the existence of \(c\) between \(x_n\) and \(x_m\) satisfying \[ |f(x_n)-f(x_m)|=|f'(c)(x_n-x_m)|\lt B|x_n-x_m|\lt B\frac{\e}{B}=\e. \] Therefore, \(f(x_n)\) is a Cauchy sequence and must converge.
Here is a solution for Problem 37.
The line tangent to \(f\) at the point \((a,f(a))\) is \(\ell(x)=f(a)+f'(a)(x-a)\). If \(x\in I\), Taylor’s Theorem (7.18) implies there is a \(c\) between \(a\) and \(x\) such that \[ f(x)-\ell(x)=\frac{f''(c)}{2}(x-a)^2\ge0. \]
Problem 38. (Due April 23.) Suppose \(f\) is continuous on \([a,b]\) and \(f''\) exists on \((a,b)\). If there is an \(x_0\in(a,b)\) such that the line segment between \((a,f(a))\) and \((b,f(b))\) contains the point \((x_0,f(x_0))\), then there is a \(c\in(a,b)\) such that \(f''(c)=0\).
Here is a solution for Problem 35.
Suppose \(f(n)=0\) for some \(n\in\omega\) Let \(\alpha\in(n,n+1)\) and \(\beta\in[n,\alpha]\) such that \(|f(\beta)|=\max\{|f(x)|:n\le x\le\alpha\}\). According to the Mean Value Theorem there is a \(c\in(n,\beta)\) such that \[ |f(\beta)| = |f(\beta)-f(n)| = |f'(c)(\beta-n)| \le |f(c)|(\beta-n) \le |f(\beta)|(\beta-n). \] Since \(\beta-n\lt1\), the only way this can be true is if \(f(\beta)=0\). This shows \(f(x)=0\) for \(n\le x\le\alpha\). Because \(\alpha\) is an arbitrary element of \((n,n+1)\), we conclude \(f(x)=0\) on \([n,n+1]\).
By assumption \(f(0)=0\). The argument given above shows \(f(x)=0\) when \(0\le x\le1\). A straightforward induction argument implies \(f(x)=0\) whenever \(x\ge0\).
Problem 37. (Due April 20.) Suppose that \(I\) is an open interval and that \(f''(x)\ge0\) for all \(x\in I\). If \(a\in I\), then show that the part of the graph of \(f\) on \(I\) is never below the tangent line to the graph at \((a,f(a))\).
Here is a solution for Problem 34.
Since \(f'(a)\) exists, \begin{align*} f'(a) &= \lim_{h\to0}\frac{f(a+h)-f(a)}{h}\\ &= \lim_{h\downarrow0}\frac{1}{2}\left(\frac{f(a+h)-f(a)}{h}+\frac{f(a)-f(a-h)}{h}\right)\\ &= \lim_{h\to0}\frac{f(a+h)-f(a-h)}{2h}\\ &= f^s(x). \end{align*} On the other hand, if \(f(x)=|x|\), then \[ \lim_{h\to0}\frac{f(0+h)-f(0-h)}{2h}=\lim_{h\to0}\frac{|h|-|-h|}{2h}=0 \] so \(f^s(0)=0\), but we know \(f'(0)\) does not exist.
Problem 36. (Due April 18.) Suppose that \(f:(a,b)\to\R\) is differentiable and \(f'\) is bounded. If \(x_n\) is a sequence from \((a,b)\) such that \(x_n\to a\), then \(f(x_n)\) converges.
Here is a solution for Problem 33.
If \(f\) is not bounded, then for each \(n\in\N\) there is an \(x_n\) such that \(|f(x_n)|>n\). The Bolzano-Weierstrass theorem gives a Cauchy subsequence \(x_{n_k}\to L\in[a,b]\). But, \(|f(x_{n_k}|>n_k\) shows \(f\) has no limit at \(L\). Therefore, \(f\) must be bounded.
To see it is not necessarily true when the interval is open, consider \(f(x)=1/x\) on the interval \((0,1)\). For each \(x\in(0,1)\), let \(\d_x=\min\{x/2,(1-x)/2\}\). Then \(f\) is bounded by \(2/x\) on \(\B{x}{\d_x}\), but is not bounded on \((0,1)\).
Problem 35. (Due April 16.) Let \(f\) be continuous on \([0,\infty)\) and differentiable on \((0,\infty)\). If \(f(0)=0\) and \(|f'(x)|\le|f(x)|\) for all \(x\gt0\), then \(f(x)=0\) for all \(x\ge0\).
Here is a solution for Extra Credit 6.
First, notice that \(f(x)=f(x+0)=f(x)+f(0),\ \forall x\in\R\implies f(0)=0\). Using this, \[ 0=f(0)=f(x+(-x))=f(x)+f(-x)\implies f(-x)=-f(x). \] Let \(x_0\in\R\) and \(x_n\) be a sequence converging to \(x_0\). Then \(x_0-x_n\to0\) and the additivity of \(f\) and continuity of \(f\) at \(0\) imply \[ \lim_{n\to\infty}|f(x_0)-f(x_n)| = \lim_{n\to\infty}|f(x_0-x_n)|=f(0)=0 \implies \lim_{n\to\infty}f(x_n)=f(x). \] Corollary 6.12 implies \(x_0\in C(f)\) and, because \(x_0\) is arbitrary, \(C(f)=\R\).
The equation \(f(x+y)=f(x)+f(y)\) is called Cauchy’s functional equation. It turns out that pretty much any “niceness” condition placed on \(f\) forces \(f\) to have the form \(f(x)=cx\). Examples of “niceness” conditions include differentiability or continuity at a single point, or even local boundedness.
Problem 34. (Due Friday, April 13.) If \(f\) is defined on an open set containing \(x_0\), the symmetric derivative of \(f\) at \(x_0\) is defined as \[ f^s(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0-h)}{2h}. \] Prove that if \(f'(x)\) exists, then so does \(f^s(x)\). Is the converse true?
Here is a solution for Problem 31.
Let \(\e>0\) and \(\d=\e/\alpha\). If \(|x-y|\lt\d\), then \[|f(x)-f(y)|\lt\alpha|x-y|\lt\alpha \e/\alpha=\e\] shows \(x\in C(f)\) for all \(x\in\R\). (In fact, this shows \(f\) is uniformly continuous.)
Here is a solution for Problem 32.
By Definition 6.9, there is a \(\d>0\) such that whenever \(x\in G=\B{a}{\d}\), then \(f(x)\in(0,2f(a))\subset(0,\infty)\).
Problem 33. (Due April 11.) If \(f:[a,b]\to\R\) has a limit at every point, then \(f\) is bounded. Is this true for \(f:(a,b)\to\R\)?
Check your scores on Blackboard to make sure our numbers are the same.
Here is a solution for Problem 30.
Let \(\e\in(0,L)\). Since \(\lim_{x\to a}f(x)=L\), we can choose \(\d>0\) so when \(0\lt|x-a|\lt\d\), then \[ |f(x)-L|\lt\e \iff L-\e\lt f(x)\lt L+\e \implies f(x)\gt0 \] because \(0\lt\e\lt L\).
Extra Credit 6. (Due April 9.) If \(f:\R\to\R\) satisfies \(f(x+y)=f(x)+f(y)\) for all \(x\) and \(y\) and \(0\in C(f)\), then \(f\) is continuous.
Problem 31. (Due April 6.) If \(f:\R\to\R\) and there is an \(\alpha>0\) such that \(|f(x)-f(y)|\le\alpha|x-y|\) for all \(x,y\in\R\), then show that \(f\) is continuous.
Problem 32. (Due April 6.) If \(f:\R\to\R\) and \(a\in C(f)\) with \(f(a)>0\), then there is a neighborhood \(G\) of \(a\) such that \(f(G)\subset(0,\infty)\).
Here is a solution for Problem 29.
Let \[ f(x)=\frac{|x-a|}{x-a} \text{ and } g(x)=-f(x) \] so that \(f(x)+g(x)=0\) and \(f(x)g(x)=-1\) when \(x\ne a\).
Problem 30. (Due April 2.) If \(\displaystyle \lim_{x\to a}f(x)=L>0\), then there is a \(\d\gt0\) such that \(f(x)\gt0\) when \(0\lt|x-a|\lt\d\).
Here is a solution for Problem 28.
Let \(x\) be a limit point of \(S'\). According to Theorem 5.9, it is enough to show \(x\in S'\). Theorem 5.6 gives a sequence \(x_n\) from \(S'\setminus\{x\}\) such that \(x_n\to x\). For each \(n\in\N\), let \(\d_n=|x_n-x|>0\). From Definition 5.5, \(y_n\in\B{x_n}{\d_n}\cap S\) can be chosen. Then \[ |y_n-x|=|y_n-x_n+x_n-x|\le|y_n-x_n|+|x_n-x|\lt2\d_n\to0 \] shows \(y_n\to x\) and Theorem 5.6 implies \(x\in S'\).
Here is a solution for Extra Credit 5.
Suppose \(f\) is not bounded above. Then for each \(n\in\N\) there is an \(x_n\in[a,b]\) such that \(f(x_n)>n\). The Bolzano-Weierstrass theorem gives a convergent subsequence \(x_{n_k}\to x_0\in[a,b]\). By assumption, there is a \(\d_{x_0}>0\) such that \(f\) is bounded on \(\B{x_0}{\d_{x_0}}\). There is an \(N\in\N\) such that \(\{f(x_{n_k}):k\ge N\}\subset\B{x_0}{\d_{x_0}}\). Since \(f(x_{n_k})>n_k\), the set \(\{f(x_{n_k}):k\ge N\}\) is unbounded, leading to a contradiction. Therefore, \(f\) cannot be unbounded above. A similar argument holds, if \(f\) is unbounded below.
Problem 29. (Due Mar 30.) Give examples of functions \(f\) and \(g\) such that neither function has a limit at some number \(a\), but \(f+g\) does. Do the same for \(fg\).
Here is a solution for Problem 26.
\[ \bigcap_{n\in\N}\left(-\frac{1}{n},1\right)=[0,1) \]
Here is a solution for Problem 27.
For \(S\) an uncountable subset of \R, set \[ S_n=S\cap[-n,n],\ \forall n\in\N. \] Then \(S=\bigcup_{n\in\N}S_n\) and there must be an \(n_0\in\N\) such that \(S_{n_0}\) is infinite, else Exercise 1.25 shows \(S\) is countable. Since \(S_{n_0}\) is bounded and infinite, Theorem 5.8 implies there is an \(x\in S_{n_0}'\). Theorem 5.6 gives a sequence \(x_n\in S_{n_0}\) converging to \(x\). Since \(S_{n_0}\subset S\), Theorem 5.6 also shows \(x\in S'\).
Problem 28. (Due Mar 26.) If \(S\subset\R\), then \(S'\) is closed.
Extra Credit 5. (Due Mar 26.) Let \(f:[a,b]\to\R\) be a function such that for every \(x\in[a,b]\) there is a \(\d_x>0\) such that \(f\) is bounded on \((x-\d_x,x+\d_x)\). Prove \(f\) is bounded.
Here is a solution for Problem 25.
From DeMorgan’s Theorem, \[ G\setminus F=F^c\cap G \text{ and } F\setminus G=F\cap G^c. \] The first of these is open because \(F^c\) is open and Theorem 5.2(b) shows the intersection of two open sets is open. The second is closed because \(G^c\) is closed and Corollary 5.3(a) shows the intersection of closed sets is closed.
Here is a solution for Problem 23.
Let \(a_n=\cos\frac{n\pi }{3}\) and \(b_n=\sin\frac{\pi }{n}\).
First, note that \[ a_1+a_2+a_3+a_4+a_5+a_6 = \frac{1}{2}-\frac{1}{2}-1-\frac{1}{2}+\frac{1}{2}+1=0. \] From this it is readily seen that \(s_n=\sum_{k=1}^na_k\) is periodic with period six. This implies \(s_n\) is a bounded sequence. (It can be shown that \(s_n=-\frac{1}{2}-\cos\left(\frac{(2+n)\pi}{3}\right)\).)
Next, notice that \(\pi/n\) decreases to \(0\) and from calculus, we know that \(\sin x\) is increasing on \([0,\pi/2]\). Combining these observations shows \(b_n,\ n\ge2\), decreases to \(0\).
Now, apply Abel’s Test (Theorem 4.18) to see \(\sum a_nb_n\) converges.
Here is a solution for Problem 24.
When \(\alpha\ge1\), it is easy to see \(\alpha^n n^\alpha\to\infty\), so the series diverges.
It is clear the series converges when \(\alpha=0\). So, suppose \(\alpha\ne0\). \begin{align*} \lim_{n\to\infty}\left|\frac{\alpha^{n+1}(n+1)^\alpha}{\alpha^n n^\alpha}\right| = \lim_{n\to\infty}\left|\alpha\right|\left( \frac{n+1}{n} \right)^\alpha = \left|\alpha\right|. \end{align*} The Ratio Test now implies the series is absolutely convergent when \(\left|\alpha\right|\lt1\).
When \(\alpha=-1\), we get the alternating harmonic series, which converges.
Finally, let \(\alpha\lt-1\), \(\beta=-\alpha\) and define \(f(x)=\beta^x/x^\beta\) for \(x\gt0\). Then \(\beta\gt1\), \(f(1)=\beta\) and \[ f'(x)=\frac{\beta^x \ln\beta}{\beta x^{\beta-1}}\gt0. \]. This shows that for \(n\in\N\) \[ \left|\alpha^n n^\alpha\right| = \frac{\beta^n}{n^\beta}=f(n) \gt \beta = -\alpha \gt1 \] Therefore, the series cannot converge because its terms do not go to \(0\).
In summary, the series converges absolutely when \(-1\lt\alpha\lt1\), converges conditionally when \(\alpha=-1\) and diverges for all other values of \(\alpha\).
Here is a solution for Extra Credit 4..
Assume \(p>1\) and \(n^{p}a_{n}\to L\). Let \(\e>0\). Choose \(M\in\N\) so that \(n^{p}a_{n}\lt L+1\) whenever \(n\ge N\). Note that this is the same as \begin{align} n\ge N \implies a_{n}\lt \frac{1}{n^{p}}(L+1).\tag{1} \end{align} Since \(\series{1/n^{p}}\) converges, the Cauchy criterion guarantees there is an \(N\in\N\) such that \(N>M\) and \begin{align} n>m\ge N\implies \sum_{k=m}^{n}\frac{1}{n^{p}}t \frac{\e}{L+1} \tag{2} \end{align} Let \(n>m\ge N\). Using (1) and then (2), \[ \sum_{k=m}^{n}a_{k} \lt \sum_{k=m}^{n}\frac{1}{n^{p}}(L+1) = (L+1) \sum_{k=m}^{n}\frac{1}{n^{p}} t (L+1)\frac{\e}{L+1} = \e. \] Another call on the Cauchy criterion shows \(\series{a_{n}}\) converges.
When \(p=1\), let \(a_{n}=1/n\), so that \(na_{n}\to1\), but \(\series{a_{n}}=\series{1/n}\) diverges.
Problem 26. (Due Mar 23) Find a sequence of open sets \(G_n\) such that \(\bigcap_{n\in\N}G_n\) is neither open nor closed.
Problem 27. (Due Mar 23) An uncountable subset of \(\R\) must have a limit point.
Here is a solution for Problem 22.
Consider \[ \sum_{n=2}^\infty\frac{2^n}{2^n\left(\ln 2^n\right)^2} = \frac{1}{\ln2}\sum_{n=2}^\infty\frac{1}{n^2} = \frac{\ln2-1}{\ln2}. \] Since \(\frac{1}{n\ln n}\downarrow0\), we may apply the Cauchy Condensation Test to see the original series converges.
Problem 25. (Due Mar 19.) If \(G\) is an open set and \(F\) is a closed set, then \(G\setminus F\) is open and \(F\setminus G\) is closed.
Here is a solution for Problem 21.
Write the repeating decimal as a geometric series. \[ 6.17272727272\cdots = \frac{61}{10}+\frac{1}{10}\sum_{n=1}^\infty\frac{72}{100^n} = \frac{61}{10}+\frac{72}{10}\,\frac{1/100}{1-1/100} = \frac{679}{110}. \] Note that this method can be used to prove that every repeating decimal is a rational number. By noting the number of remainders when you divide two integers is bounded by the size of the divisor, it also follows that every rational number gives rise to a repeating decimal. Therefore, a number is rational if, and only if, it can be represented as a repeating decimal.
Here is a solution for Problem 20.
Since \(x_{n}\in\{0,1,2,3,4,5,6,7,8,9\}\), it follows that \[ \frac{x_{n}}{10^{n}}\le\frac{9}{10^{n}},\ \forall n\in\N. \] Since \(\series{\frac{9}{10^{n}}}\) is a geometric series with \(r=1/10\lt1\), it converges. Therefore, \(\sum_{n=1}^\infty\frac{x_n}{10^n}\) converges by the Comparison Test. Moreover, \[ 0=\sum\frac{0}{10^n}\le\sum\frac{x_n}{10^n}\le\sum\frac{9}{10^n}=1. \] This shows any decimal \(0\,.\,x_1\,x_2\,x_3\cdots\in[0,1]\)
Extra Credit 4. (Due Mar 9.) Prove if \(a_n\ge0\) for all \(n\in\N\) and there is a \(p>1\) such that \(\lim_{n\to\infty}n^pa_n\) exists and is finite, then \(\sum_{n=1}^\infty a_n\) converges. Is this true for \(p=1\)?
Problem 24. (Due Mar 9.) For what values of \(\alpha\) does \(\sum_{n=1}^\infty\alpha^nn^\alpha\) converge?
Problem 22. (Due Mar 7.) Does \(\displaystyle\sum_{n=2}^\infty\frac{1}{n(\ln n)^2}\) converge?
Problem 23. (Due Mar 7.) Does \(\displaystyle \sum\cos\frac{n\pi}{3}\sin\frac{\pi}{n}\) converge?
Here is a solution for Problem 18.
By Theorem 3.22, \(a_n\) converges to some \(A\in\R\). Theorem 3.14 implies \(A=L\).
Here is a solution for Problem 19.
Since \(a_n\) is convergent, Theorem 3.22 implies it is a Cauchy sequence. So, given \(\e>0 \), there is an \(N\in\N \) such that whenever \(m,n>N \), then \(|a_m-a_n|\lt\e \). By assumption, it follows that whenever \(m,n>N \), then \(|b_m-b_n|\le|a_m-a_n|\lt\e \). Therefore, \(b_n \) is a Cauchy sequence. Now, Theorem 3.22 implies \(b_n \) converges.
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.
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 \(a_n\to\ln2\).
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.
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*}
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 \(a_n\to L\).
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.
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 \(a_n\to4/3\).
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.
Problem 14. (Due Feb 16.) Let \(a_1=3\) and \(a_{n+1}=2-1/a_n\) for \(n\in\N\). Analyze the sequence.
Extra Credit 2. (Due Valentine’s Day.) Determine the limit of \(a_{n}=\sqrt[n]{n!}\). (Hint: If \(n\) is even, then show that \(n!>(n/2)^{n/2}\).)
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 \(a=0\).
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.
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 \(\lub(A)=\glb(B)\).
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
\begin{equation}\label{ineqp5.1}
-y\le0\le x\le y. \tag{3}
\end{equation}
If \(x\lt0\), then \(-x=|x|>0\). Multiplying the assumed inequality \(-x=|x|\le y\) by \(-1\) gives
\begin{equation}\label{ineqp5.2}
-y\le x\lt0\le y. \tag{4}
\end{equation}
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, \(|x|\le y\).
Problem 9. (Due Feb 7.) If \(I\) is an interval, then \(\Q\cap I\ne\emptyset\).
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 \(\card{A\cup B}=\aleph_0\).
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 \(\alpha=\lub S\).
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.
Problem 5. (Due Jan 29.) Prove Theorem 2.11(d) from the notes.
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.
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 \(0\lt1\).
To solve Problem 1, let \(S=\{a,\{a\}\}\)
for some \(a\). Then \[ S\cap\powerset{S}=\{a,\{a\}\}\cap\{\emptyset,\{a\},\{\{a\}\}\{a,\{a\}\}\}=\{\{a\}\}\ne\emptyset. \]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 \(h(7)\).
The Schröder-Bernstein theorem was proved.
Problem 1. (Due Jan 17.) Is there a set \(S\) such that \(S\cap\powerset{S}\ne\emptyset\)?
Read §1.1–1.5.
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