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Claim: if $f$ is differentiable on $(a,b)$, continuous on $[a,b]$, and obtains a minimum (or maximum) value at $c\in(a,b)$, then $f'(c)=0$.

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. $\blacksquare$

Example for Homework 10 by elishapetersonelishapeterson, 16 Apr 2008 17:55

Proposition: If a sequence $(x_n)$ converges in $\mathbb{R}$, then the sequence of absolute values $(|x_n|)$ also converges.

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 $|x|$. $\blacksquare$

Just noticed you were trying to figure out how to center the equations.

To center math, type:


This becomes(1)
\begin{equation} x^2=3 \end{equation}

You can refer to the equation by doing the following:

[[math label1]]

Then access the label by typing ([[eref label1]]). For example, to say "In equation (1)…" you would use "In equation ([[eref label1]])..."
How to center math... by elishapetersonelishapeterson, 11 Mar 2008 19:09

Problem 1.
… sequence $(x_n)$ in S there is a subsequence $(x_{n_k})$ which converges to some $x\in S$.

Problem 2.
… a sequence $(x_n)$ in S for which no subsequence converges to a point in S.

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. $\blacksquare$.

No, the radii do not need to be integers.

Remember that if a set S is nonempty and let's say $x\in S$, then the sequence $x,x,x,x,\ldots$ has limit point $x\in S$. So… every point in a set is also a limit point of the set. This means the answer to the second question is no. Rather, all the integers are limit points of the set of integers.

Well I don't want to specifically ask for the answer to this question, but there is no way to get around it. When we are talking about neighborhoods around an integer in the set of integers, do the radii of these neighborhoods necessarily have to be integers themselves. My thoughts would be no since the radius is defined as any $\epsilon$ > 0 but I wanted to make sure. Also, I suppose it is possible for a set to have no limit points, are the integers and example of this. Is a set with no limit points considered closed?

Thank you.

HWK 6, Quest. 6, part b. by bryanjonasbryanjonas, 01 Mar 2008 03:08

Here is the proof that the sum of two functions is continuous.

Let f and g be continuous at some x. Let $\epsilon>0$ be chosen arbitrarily. Since f is continuous, there exists $\delta_f>0$ such that if $|x-u|<\delta_f$ then $|f(x)-f(u)|<\frac{\epsilon}{2}$. Similarly, choose $\delta_g$ such that if $|x-u|<\delta_g$ then $|g(x)-g(u)|<\frac{\epsilon}{2}$.

Now set $\delta=\min\{\delta_f,\delta_g\}$. Then for any u with $|x-u|<\delta$, both $|x-u|<\delta_f$ and $|x-u|<\delta_g$ and so

\begin{align} |f(x)+g(x)-(f(u)+g(u))|=|f(x)-f(u)+g(x)-g(u)| \leq |f(x)-f(u)| + |g(x)-g(u)| < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \quad \blacksquare \end{align}
Re: Problem 19 by elishapetersonelishapeterson, 11 Feb 2008 21:37

Yes, you can assume this. In later chapters, we will prove this fact.

Re: Problem 19 by elishapetersonelishapeterson, 11 Feb 2008 11:48

Sir, can we assume that the sum of two continuous functions is continous? I wrote a proof for part 19.b under this assumption (I let g(x)=f(x)-x).

Problem 19 by Paul FalconePaul Falcone, 11 Feb 2008 02:33

Hint 1: How do you write down the set of x values at which f evaluates to v? The answer is $X=\{x\in [a,b] : \quad?????\quad\}$ for some ??????.

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.

Chapter 1, Problem 38 by elishapetersonelishapeterson, 09 Feb 2008 19:04
elishapetersonelishapeterson 08 Feb 2008 22:36
in discussion Instructor Notes / General Notes » WPR 1

I have posted information about the upcoming WPR at WPR 1. Solutions to the sample exam will be posted at Solutions to WPR 1.

WPR 1 by elishapetersonelishapeterson, 08 Feb 2008 22:36

The symbol is \neq (produces $\neq$).

Also, there's a table of symbols at how-to-edit-pages. Click "+ show the website" to see the table of symbols. You can also go directly to the source of the table at

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.

The problem is to find the multiplicative inverse of a cut $x=A|B$, that is a cut $y=C|D$ such that $x\cdot y=1^*$.

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.

(a) Positive Cuts

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 $B$.

(b) Negative Cuts

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.

(c) Uniqueness

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

\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. $\blacksquare$

Chapter 1, Problem 10 by elishapetersonelishapeterson, 25 Jan 2008 16:12

In the text and in the homework, the notation A+A' and A*A' is used. What does this mean?

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.

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