**Proof:** Let *c* be a maximum value. Differentiability implies that $\lim_{t\to\theta}\frac{f(t)-f(\theta)}{t-\theta}$ exists. If $t>c$ then $\frac{f(t)-f(c)}{t-c}\leq 0$ since $f(t)\leq f(c)$. If $t<c$ then $\frac{f(t)-f(c)}{t-c}\geq 0$. Therefore, $f'(c)$ can be written as the limit of (i) a sequence of nonnegative numbers, and (ii) a sequence of nonpositive numbers. The only possible limit is therefore 0. $\blacksquare$

**Proof**: Let $\epsilon>0$ be arbitrary. Since $(x_n)$ converges, there exists a number *N* such that if $n>N$ then $|x_n-x|<\epsilon$, where *x* is the limit of the sequence $(x_n)$. We will show that for the same values of $\epsilon$ and *N*, if $n>N$ then $\big||x_n|-|x|\big|<\epsilon$, which implies the convergence of the sequence of absolute values.

By the triangle inequality, $|x_n-x|+|x| \geq |x_n|$, which implies $|x_n-x|\geq |x_n|-|x|$. Also by the triangle inequality, $|x-x_n|+|x_n| \geq |x|$, which implies $|x_n-x|\geq |x|-|x_n|$.

Since $|x_n-x|\geq |x_n|-|x|$ and $|x_n-x|\geq -\big(|x_n|-|x|\big)$, it follows that $|x_n-x|\geq\big||x_n|-|x|\big|$.

Therefore, with the same values of $\epsilon$ and *N* pertaining to the convergence of the sequence $(x_n)$, if $n>N$ then

\begin{align} \big||x_n|-|x|\big| \leq |x_n-x| < \epsilon. \end{align}

This implies, by the definition of convergence, that the sequence $(|x_n|)$ converges to $|x|$. $\blacksquare$

]]>… sequence $(x_n)$ in

**Problem 2.**

… a sequence $(x_n)$ in *S* for which no subsequence converges to a point in *S*.

**Problem.3.**

A subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded (the Heine-Borel Theorem).

**Problem 4.**

If *S* is a single point *x*, there is only one sequence contained in *S*, the sequence *x*, *x*, … which converges itself.

**Problem 5.**

This fact is implied by the fact that the subsequence converges. In particular, if the subsequence is $(x_{n_k})$, then by definition of convergence for any $\epsilon>0$ every term with $k>N$ for some *N* is in the $\epsilon$-neighborhood of *x*. Since there are an infinite number of these, and they are all elements of the original sequence $(x_n)$, the statement must be true. $\blacksquare$

**Problem 6.**

We must show how to construct a subsequence $x_{n_1},x_{n_2},x_{n_3},\ldots$ which converges to *x*. Let $x_{n_1}$ be any element within the 1-neighborhood of *x* (that is, anything within distance 1 of the limit). Let $x_{n_2}$ be any element with $n_2>n_1$ (that is, some element of the sequence further out than $n_1$) and within the $\frac12$-neighborhood of *x*. Continue in this manner to select $x_{n_i}$ with $n_i>n_{i-1}$ in the $\frac1i$-neighborhood of *x*. Since there are an infinite number of points within each neighborhood of *x* to choose from, we can always find such points.

How can we prove that the subsequence $x_{n_k}$ converges to *x*? It suffices to show that every $\epsilon$-neighborhood of *x* contains some "tail" of the sequence, that is all $x_{n_k}$ with $k>N$ for some *N*. Choose *N* such that $0<\frac1N<\epsilon$. By construction if $k>N$ then $d(x_{n_k},x)<\frac1N<\epsilon$ since everything past $k=N$ is in a $\frac1{N+1}$-neighborhood or smaller. $\blacksquare$.

Thank you.

]]>**Hint 2:** What you are trying to show is that $X$ has both a minimum and maximum. Remember that these are *x* values!

**Hint 3:** How do you get at minimum/maximum? We **can** define the *Least Upper Bound* of $X$ and we know that exists, by the LUB property of the real numbers. Keep in mind, though, that while $LUB(X)$ always exists, *it may not be in the set $X$*.

**Hint 4:** So you can define an *x*-value $x'=LUB(X)$, and if you can show that $x'\in X$, then you've shown that $x'$ is the **maximum** of the set.

I checked the WikiHelp section on the syntax and I can't seem to find the symbol for "Not equal to".

Could you possibly tell me what it is?

Thank you. ]]>

How do you even approach this problem? The best way is to try an example, let's say $x=2^*$. What do you think the inverse should look like in this case? Once your intuition has developed, the trick is finding a way to write it out mathematically.

First, if $x>0^*$, the trick is to notice that all elements in $\frac{1}{B}$ should be in $C$ (except perhaps the smallest element of $B$), as should 0 and all negative numbers. All elements in $\frac{1}{A^+$}$ where $A^+$ is the *positive* elements of $A$, should be in $D$. Let

\begin{align} C=\{q\in\mathbb{Q}:q<\frac{1}{r} \text{ for some } r\in B\}, \end{align}

and let $D=\mathbb{Q}\setminus C$. Why does this work, i.e., why does $(A|B)\cdot(C|D)=1^*$? We expect $C$ to contain all negative numbers and zero, which it certainly does, and if $p\in B$, then $\frac{1}{p}<\frac{1}{r}$ for some $r\in B$, unless $p$ is the smallest element of $B$.

If $x<0^*$, the construction is similar, but this time $C$ should contain only the elements in $\frac{1}{B^-}$, where $B^-$ is the *negative* elements of $B$ (again, exclude the endpoint if it exists). Thus

\begin{align} C=\{q\in\Q:q<\frac{1}{r} \text{ for some } r\in B^-\}, \end{align}

and let $D$ be everything else.

*How do you show something is unique?*- either show that there is only one logical object which satisfies the properties,
*or*show that two objects which satisfy the desired properties must be equal. *How do you show two objects are equal?*- If they are sets, show that every element in one is contained in the other, and vice versa? If the objects satisfy
**trichotomy**, then you might show $a\leq b$ and $b\leq a$.

So there are a number of approaches here. We'll take the *direct* approach:

Suppose that $x=A|B$ is positive. Then $y=C|D$ must be positive, and must satisfy

(3)\begin{align} E=\{r\in\mathbb{Q} : r<0 \text{ or there exist positive $a\in A,c\in C$ such that } r=ac\}=\{r\in\mathbb{Q}: r<1\}, \end{align}

by definition of cut multiplication (bottom of page 14 in the text). Therefore, for any positive $r<1$, there exist positive $a\in A,c\in C$ such that $r=ac$. Therefore, $c\in C$ if and only if there exists a positive $a\in A$ such that $ac<1$, and $C$ is unique. $\blacksquare$

]]>First, $A$ and $A'$ are two subsets of the rational numbers. $A+A'$ represents any rational number which can be expressed as the sum of one in $A$ and another in $A'$. For example, if $A$ consists of all rationals less than 2, and $A'$ of all rationals less than 5, then $A+A'$ consists of all rationals less than 7.

Similarly, $A\cdot A'$ represents the set of all rationals which can be expressed as the product of something in $A$ with something in $A'$. This gets a bit tricky, since two negative numbers multiply to a positive number. Thus, if $A$ is the numbers less than 2 and $A'$ those less than 5, then $A\cdot A'$ consists of *all* rational numbers. Resolving this trickiness with negative numbers is the purpose of one of the homework problems.