# WPR I

**Problem 1.** Prove that $\sqrt{3}$ is irrational.

See textbook, page 11.

**Problem 2.** Show that $(|\vec a|-|\vec b|)(|\vec a|+|\vec b|)=\langle \vec a-\vec b,\vec a+\vec b\rangle$.

Recall that $|\vec a|^2=\langle \vec a,\vec a\rangle$. Expanding the left side gives $|\vec a|^2-|\vec b|^2$. Expanding the right side gives $\langle\vec a,\vec a\rangle + \langle\vec a,\vec b\rangle - \langle\vec b,\vec a\rangle - \langle\vec b,\vec b\rangle = \langle\vec a,\vec a\rangle - \langle\vec b,\vec b\rangle = |\vec a|^2-|\vec b|^2$, since inner products are symmetric. $\blacksquare$

**Problem 3.** For the discontinuous function shown, find an *x* and an $\epsilon>0$ such that for all $\delta>0$ there exists some *x'* such that $|x-x'|<\delta$ and $|f(x)-f(x')|>\epsilon$.

The existence of

xand $\epsilon$ is guaranteed by the definition of continuity (negated). Let $x=1$ and $\epsilon=1$. Then $f(x)=-1$, and if $\delta=1$ then a value $x'\in(1-\delta,1)$ has a function value greater than +1, i.e., $f(x')>1$. Hence, $|f(x)-f(x')| = 2 > \epsilon$. $\blacksquare$

**Problem 4.** What is the difference between the "maximum" and the "Least Upper Bound" of a subset of real numbers?

The maximum may not always exist, while the LUB is guaranteed to exist (by completeness of the reals). For example, if $S=(0,1)$, then

Shas no maximum, whereas $LUB(S)=1$.

**Problem 5.** Given that $\sqrt{5}$ corresponds to the cut $A|B$, write out definitions of *A* and *B*.

The key is to write out the definitions without referencing any nonrationals. Hence, we write $A=\{q\in\mathbb{Q}:q^2<5\}$ and $B=\{q\in\mathbb{Q}:q^2\geq 5\}$.

**Problem 6.** Let $S\subset\mathbb{R}$ be bounded, nonempty, and let $b=LUB(S)$. Prove that for each $\epsilon>0$ there exists $s\in S$ with $b-\epsilon\leq s\leq b$.

We will prove this by assuming that there is no such

sand contradicting the fact thatbis the least upper bound ofS.Assume, by way of contradiction, that no such $s\in[b-\epsilon,b]$ exists. Then $b-\epsilon>t$ for all $t\in S$, so that $b-\epsilon$ is also an upper bound. Since $b-\epsilon<b$, the value

bcannotbe theleastupper bound. This contradiction establishes the proof. $\blacksquare$

**Problem 7.** Given cuts $A|B$ and $C|D$, (i) what does trichotomy imply about *A* and *C*, (ii) prove trichotomy by demonstrating what you stated in (i).

Trichotomy implies that either $A\subsetneq C$, $A\supsetneq C$, or $A=C$.

Suppose that $A\neq C$. Then either there exists $a\in A$ which is not in

Cor there exists $c\in C$ which is not inA. Without loss of generality assume the first. By definition of a cut (see page 11 in the text), that means $a\in D$, and consequently $a>c$ for all $c\in C$. But since if $a>a'$ then $a'\in A$ also, that implies that every $c\in C$ is also contained inA, hence $C\subsetneq A$.

Likewise, if some $c\in C$ is not inA, then $A\subsetneq C$. $\blacksquare$

# WPR II

**Problem 1.** Explain why, at any given time, there is a highest temperature on the surface of the earth. (List all assumptions.)

Assume that the temperature on the earth is a continuous function, whose domain is the surface of the earth. Assume also that the surface of the earth may be taken to be a subset of $\mathbb{R}^3$. As such, by the

Heine-Borel Theoremit is compact, since it is closed and bounded. With these assumptions, theExtreme Value Theoremimplies that the temperature function has a maximum, since it is a continuous function on a compact domain.

**Problem 2.** The middle half Cantor set may be written $C=C_1\cap C_2\cap \cdots$, where $C_1=[0,1]$, $C_2=[0,\frac14]\cup[\frac34,1]$, $C_3=[0,\frac1{16}]\cup[\frac3{16},\frac14]\cup[\frac34,\frac{13}{16}]\cup[\frac{15}{16},1]$, etc. Prove that *C* is compact.

Recall that (i) the finite union of closed sets is closed, and (ii) the infinite intersection of closed sets is closed. By the first, each $C_i$ is closed. By the second, the intersection of the $C_i$ is closed. Hence,

Cis closed, and it is clearly bounded, proving that it is compact (by theHeine-Borel Theorem). $\blacksquare$

**Problem 3.** Prove, directly from the definition of an open set, that the union of two open sets is open.

Let

AandBbe open, and let $p\in A\cup B$. We want to show that there exists an open neighborhood ofpwhich is contained in $A\cup B$. Note that either $p\in A$ or $p\in B$ (or both). Without loss of generality, suppose that $p\in A$. Then, sinceAis open,phas an open neighborhoodNcontained inA, that is $p\in N\subset A$. But then $N\subset A\cup B$ as well, verifying thatphas an open neighborhood contained in the union $A\cup B$. Sincepwas chosen arbitrarily, this completes the proof. $\blacksquare$

**Problem 4.** Let *S* consist of concentric circles of radius $1+\frac1n$ for natural numbers $n\geq 1$, as shown. (i) Describe the limit points of *S* which are not contained in *S*. (ii) Is *S* open, closed, or neither? (iii) Is *S* compact? (iv) What are the interior, closure, and boundary of *S*?

(i) The limit points not contained in

Sare precisely the unit circle.

(ii)Sis not open, since no point contains a neighborhood withinS(consequently, every point ofSis a boundary point). It is not closed, since it does not contain all of its limit points. Hence it isneither.

(iii) SinceSis not closed, it is not compact.

(iv) The interior ofSis empty; the boundary ofSis all ofS; the closure ofSisStogether with the standard unit circle.

**Problem 5.** Let $f(x)=\cos(x)$, and let $(x_n)$ be the sequence defined by setting $x_n=f(n)$ for $n\geq1$. Also, let $(y_n)$ be the sequence $y_n=f(2\pi n)$. (i) Which sequence converges? (ii) Do both sequences have convergent subsequences? Why?

(i) The sequence $(y_n)$ converges, since $\cos(2\pi n)=1$ for all $n\in\mathbb{N}$. The sequence $(x_n)$

does notconverge, since the terms $\cos(1),\cos(2),\ldots$ do not converge.

(ii) Both sequences have convergent subsequences, since they are bothbounded. Indeed, every bounded sequence in $\mathbb{R}$ has a convergent subsequence… this is guaranteed by compactness.

**Problem 6.** Define $x_n=\frac{(-1)^n}{3^n}$. Prove that $(x_n)$ converges, using the $\epsilon$ definition of convergence.

Note first that the limit, provided it exists, would be 0.

Let $\epsilon>0$. We need to find some $N\in\mathbb{N}$ such that for $n>N$, $|x_n-0|<\epsilon$. We can find the appropriate

Nby finding the value at which $|x_p|=3^{-p}=\epsilon$. Taking the log of both sides of this equation, we have $\log(3^{-p})=\log(\epsilon) \Rightarrow -p \log(3) = \log(\epsilon) \Rightarrow p = - \frac{log(\epsilon)}{\log(3)}$. Note that if $\epsilon>0$ is small, this value will be positive. Since we're looking for $N\in\mathbb{N}$, we chooseNto be thesmallestpositive integer greater than $-\frac{\log(\epsilon)}{\log(3)}$. Then if $n> N$, $|x_n|=3^{-n} < 3^{-N} < 3^{-p} = \epsilon$. This establishes convergence to 0. $\blacksquare$

A function *f* defined on a domain *D* is said to be **uniformly continuous** if for every $\epsilon>0$ there exists $\delta>0$ such that for all $x\in D$, if $d(x,y)<\delta$ then $d(f(x),f(y))<\epsilon$.

**Problem 7.**Here is the definition of uniform continuity:

Prove that the function $f(x)=x^2$ is uniformly continuous on the domain $[0,1]$. (**Hint:** $|x^2-y^2|=|x+y|\cdot|x-y|\leq 2|x-y|$ when $x,y\in[0,1]$).

Let $\epsilon>0$, and let $\delta=\frac\epsilon2$. Let $x,y\in D$ be chosen such that $d(x,y)<\delta$. Then, $d(f(x),f(y))=|x^2-y^2|\leq 2|x-y|=2 d(x,y) \leq 2\delta = \epsilon$. $\blacksquare$

# WPR III

**Problem 1.** Using the definition of the derivative, show that the function $f(x)=\begin{cases}x^2\sin\frac{1}{x} & x\neq 0\\ 0 & x=0\end{cases}$ is differentiable at $x=0$, and find the derivative $f'(0)$.

We need to check to see whether $\lim_{h\to0}\frac{f(h)-f(0)}{h}$ exists. This limit is $\lim_{h\to0}\frac{h^2\sin(\frac1h)-0}{h}=\lim_{h\to 0}h\sin(\frac1h)=0$. The last step follows since $\sin$ is a bounded function. Hence, $f'(0)=0$. $\blacksquare$

**Problem 2.** For what values of *p* does the series $\sum_{n=1}^\infty \frac{2008 n}{5 n^p+5}$ converge? (Justify your answer.)

Note that $\frac{2008 n}{5 n^p+5} = \frac{2008}{5} \left(\frac{n}{n^p+5}\right) = \frac{2008}{5}\left(\frac{1}{n^{p-1}+5n^{-1}}\right) < \frac{2008}{5}\left(\frac{1}{n^{p-1}}\right).$ We know by the

power testthat $\sum_{n=1}^\infty\frac{1}{n^{p-1}}$ converges for $p>2$, hence $\frac{2008}{5}\sum_{n=1}^\infty\frac{1}{n^{p-1}}$. Therefore, by thecomparison test, the original series converges for $p>2$ as well. Similarly, it diverges for $p<2$, and we can say nothing about $p=2$.

**Problem 3.** Prove that the function $\sin(x)$ is uniformly continuous. (Hint: use the Mean Value Theorem to obtain the necessary inequality.)

To solve this problem, we need the assumption that $|\cos(x)|\leq 1$. Then the

mean value theorem(with absolute values) states that, given any $x,y\in\mathbb{R}$, there exists some value $c\in(x,y)$ such that $|\sin(x)-\sin(y)|=|f'(c)|\cdot|x-y|=|\cos(c)|\cdot|x-y|\leq|x-y|$.

We now give the proof. Let $\epsilon>0$, and choose $\delta=\epsilon$. Then if $x,y\in\mathbb{R}$ are chosen such that $|x-y|<\delta$, then by the above inequality $|f(x)-f(y)|\leq|x-y|<\delta=\epsilon$, verifying uniform continuity. $\blacksquare$

**Problem 4.** Suppose a function $f(x)$ has a *local maximum* near $x=0$. (i) What does this imply about $f(t)-f(0)$ for *t* near 0? (ii) Show that the sign of $\frac{f(t)-f(0)}{t-0}$ (for *t* near 0) depends entirely on whether *t* is positive or negative. (iii) Why does this prove that $f'(0)=0$?

(i) Since $f(0)$ is the maximum, we must have $f(t)<f(0)$ for

tnear 0, meaning $f(t)-f(0)<0$.

(ii) Since $f(t)-f(0)<0$ for alltnear 0, whereas the sign of the denominator $t-0$ changes depending ont, we see that $\frac{f(t)-f(0)}{t-0}<0$ if $t>0$, and $\frac{f(t)-f(0)}{t-0}>0$ if $t<0$.

(iii) If the derivative exists and $f'(0)=0$, then $\lim_{t\to 0}\frac{f(t)-f(0)}{t-0}$ must exist and be equal to the same value,no matter howtapproaches zero. We have already seen that if $t\to 0$ with positivet, the values in the limit are negative, hence $\lim_{t\to 0}\frac{f(t)-f(0)}{t-0}\leq 0$. Likewise, iftapproaches zero from the negative side, the values are positive, so that $\lim_{t\to 0}\frac{f(t)-f(0)}{t-0}\geq 0$. Hence, the only possible value for the limit is 0. $\blacksquare$

**Problem 5.** The *ratio mean value theorem* states that if *f* and *g* are continuous real functions on $[a,b]$ and differentiable on $(a,b)$, then there is a point $\theta\in(a,b)$ such that $\left(\frac{f(b)-f(a)}{b-a}\right)g'(\theta)=\left(\frac{g(b)-g(a)}{b-a}\right)f'(\theta).$ (i) Provide an interpretation of this theorem in terms of secant and tangent lines. (ii) Verify the theorem for $f(x)=x^2$, $g(x)=x^3$ on the interval $[0,1]$.

(i) Rearranging terms a bit, the theorem says that there is a point at which the ratio of the tangent slopes of

fandgis equal to the ratio of secant slopes offandgbetweenaandb. The statement needs to be altered slightly if a denominator slope is 0, but the flavor is still the same.

(ii) Since $f(0)=g(0)=0$ and $f(1)=g(1)=1$, both secant slopes are equal to1. Hence, the objective is to find a value of $\theta$ for which $\frac{f'(\theta)}{g'(\theta)}=1$ or more simply $f'(\theta)=g'(\theta)$. We have $f'(x)=2x$ and $g'(x)=3x^2$, so that we want to solve the equation $2\theta=3\theta^2 \Rightarrow \frac23=\theta$.