Math 301 - Introduction To Analysis - Schedule

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(:if name $Name}:) Here are some common symbols we use for standard number systems. We use *$\mathbb{N}=\{1,2,3,\ldots\}$ for the natural numbers, *$\mathbb{W}=\{0,1,2,3,\ldots\}$ for the whole numbers, *$\mathbb{Z}=\{\ldots, -3,-2,-1,0,1,2,3,\ldots\}$ for the integers, *$\mathbb{Q}=\{\frac{p}{q}\mid \text{ $p$ and $q$ are integers with $q\neq 0$\}$ for the rational numbers, and

$\mathbb{R}=(-\infty,\infty)$ for the real numbers.

We'll spend most of the semester looking at collections of real numbers, and these symbols will appear quite often.

Definition (Lower Bound, Upper Bound, Bounded)

Let $S$ be a collection of real numbers (written $S\subseteq \mathbb{R}$, or $S$ is a subset of the real numbers).

A lower bound for $S$ is a real number $m$ such that $m\leq x$ for every $x\in S$. We say that $S$ is bounded below if it has a lower bound.
An upper bound for $S$ is a real number $m$ such that $m\geq x$ for every $x\in S$. We say that $S$ is bounded above if it has an upper bound.
We say that $S$ is bounded if has both a lower and upper bound.

Problem 3: (Practice With Bounded Definitions)

Consider the set $S=[0,4) = \{x\in\mathbb{R}\mid 0\leq x \text{ and } x< 4\}$.

Show that $S$ is bounded below by giving a lower bound. Prove that the number you gave is a lower bound, and then state another lower bound different than the one you gave.
Of all possible lower bounds, which is the greatest lower bound. In other words, produce a lower bound $m$ so that if $m'$ is any lower bound, then we must have $m'\leq m$. Prove your answer.
Show that $S$ is bounded above by giving an upper bound. Justify your answer.
Of all possible upper bounds, which is the least upper bound?

Prediction:

In the previous problem, you found lots of bounds. One of the bounds you found we will call the supremum. Another of these bounds we will call the infimum. Which bound do you think is the supremum? Which do you think is the infimum?

Click to see the formal definition of supremum and infimum.

Definition (Infimum And Supremum)

When a set $S$ is bounded below, there are infinitely many lower bounds. The infimum of $S$ is the greatest lower bound, which we write as $\inf S$. So if $S$ is a set, then we write $m=\inf S$ if and only if both

$m$ is a lower bound for $S$, and
$m$ is the greatest lower bound for $S$ (if $m'$ is another lower bound, then $m\geq m'$).

When a set $S$ is bounded above, there are infinitely many upper bounds. The supremum of $S$ is the least upper bound, which we write as $\sup S$. So if $S$ is a set, then we write $m=\sup S$ if and only if both

$m$ is an upper bound for $S$, and
$m$ is the least upper bound for $S$ (if $m'$ is another upper bound, then $m\leq m'$).

Problem 4: (Practice With Bounded Definitions 2)

Let $S$ be the set $S=\left\{\frac{1}{n}\mid n\in\mathbb{N}\right\}=\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots\} $.

Give two different upper bounds for $S$. Then give the supremum of $S$.
Give two different lower bounds for $S$. Then give $\inf S$.

Remember, any time you make a claim, you must prove that your claim is correct.

In mathematics there are some ground rules that we cannot prove, rather they are just accepted as true without proof. We call these axioms. The following axiom just gives words to a fact that you may think is pretty obvious (something axioms tend to be).

Axiom (The Completeness Axiom)

Every nonempty set of real numbers that is bounded above has a least upper bound.

Problem 5: (Using The Completeness Axiom)

Consider the solutions to the inequality $x^2<2$. We can write this using set builder notation as $S=\{ x\mid x^2<2\}$.

Is $S$ bounded above? Prove your claim.
What does the completeness axiom say about this set?
Give $\sup S$, or explain why there is no supremum of $S$.

One of the skills we need to develop as mathematicians is the ability to encounter a new definition and from that definition start drawing conclusions. As the semester progresses, we'll have lots of opportunities to practice this. The beginnings of analysis and calculus are directly related to the next definition, that of a limit point of a set of real numbers.

Definition (Limit Point Of A Set Of Real Numbers)

Let $S$ be a set of real numbers. We say that a point $p$ is a limit point of $S$ if every open interval $I=(a,b)$ that contains $p$ also contains a point $x$ in $S$ with $x\neq p$.

Problem 6: (A Limit Point Of An Open Interval)

Let $S=(0,1)$ which is the open interval from 0 to 1 that does not include the end points. Prove that $p=1$ is a limit point of $S$. Then state another limit point of $S$.

Problem 7: (A Set With One Limit Point)

Let $\ds S=\left\{\frac{1}{n}\mid n\in \mathbb{N}\right\}$, the collection of fractions of the form $\frac{1}{n}$ where $n$ is a natural number. Prove that $p=0$ is a limit point of $S$.

Prediction:

How does this class differ from your lower level math courses? What's the best way to succeed in this class?

Click to see a possible answer.

In your lower level courses, the focus is often primarily on learning how to mimic an algorithm to obtain a numerically correct answer at the end. In your upper division courses, the focus shifts to the process of how we obtain answers, and the logic behind why the process works. This course prepares you for this shift in focus.

You'll want to spend 2-3 hours between each class period working on the problem set on this course website. The best way to succeed in this course is consistent daily effort. Study groups can be a life saver as well, provided you think of group meetings as working meetings (where everyone works on new material rather than sharing what you already figured out).

Nothing can substitute for diligent consistent effort. As you work, you will hit dead ends. That's a normal part of doing mathematics. Failure is an important part of the learning process. Don't worry if you don't get the solution to every problem before class. Just do your best, and give yourself enough time before class to be able to ask questions of others.

In the work above, we saw several sets of real numbers. We've used sets our whole life to group together objects that have some common property. That is precisely what a set it, a group of things that share something in common.

Definition (Set, Subset, Equality Of Sets)

A set $S$ is a collection of elements that have been grouped together.

We use brackets $\{$ and $\}$ to enclose elements of sets.
We'll write $x\in S$ to say that $x$ is an element of $S$ or $x$ is in $S$. Similarly, we'll write $x\notin S$ to say that $x$ is not in $S$.
We say that a set $B$ is a subset of the set $S$, and we write $B\subseteq S$, if every element in $B$ is also an element of $S$. We also read $A\subseteq B$ as "$A$ is contained in $B$." We'll often write $B\supseteq A$ instead of $A\subseteq B$, and read $B\supseteq A$ as either "$B$ is a super set of $A$" or "$B$ contains $A$."
We say that $B$ is a proper subset of $S$ if $B\subseteq S$ but there is an element of $S$ that is not in $B$.
We say that two sets $A$ and $B$ are equal if $A\subseteq B$ and $B\subseteq A$.

There are two general ways to express elements of a set. We can use the roster method where we list the elements of a set, as in $\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\}$. The roster method requires the reader to guess the remaining elements of the set, and hence can sometimes lead to unclear proofs. To avoid this potential confusion, we use set builder notation. With set builder notation, we express how to obtain the elements, as in $\{\frac{1}{n}\mid \in\mathbb{N}\}$. Any time you see $\{x\mid P(x)\}$ you can read this as the set of $x$ such that $P(x)$ holds.

There are many different ways we can use set builder notation to describe the exact same set, and we need to be able to show when two different ways are equal. For an example, consider the interval $I=(5,9] $. Let $S$ be the collection of upper bounds of $I$, which we can write in set builder notation as $$S=\{x\mid x \text{ is an upper bound of } I\}.$$ I claim that $S=\{x\mid x\geq 9\} $. Halt. This is a claim that two sets, namely $S$ and the set $A=\{x\mid x\geq 9\} $ are equal. From the definition above about equality of sets, to prove this claim is true we must prove that $S\subseteq A$ and that $A\subseteq B$. That's precisely what the next problem has you prove.

Problem 8: (First Proof That Two Sets Are Equal)

Let $I=(5,9] $. Consider the sets $S=\{x\mid x \text{ is an upper bound of } I\}$ and $A=\{x\mid x\geq 9\} $. Prove that $S=A$.

Start by proving that $S\subseteq A$. So show that every element of $S$ is an element of $A$.
Then prove that $A\subseteq S$.

Problem 9: (Second Proof That Two Sets Are Equal)

Let $I=(5,9] $. Consider the sets $T=\{x\mid x \text{ is a lower bound of } I\}$ and $B=\{x\mid x\leq 5\} $. Prove that $T=B$.

An important set that will show up often throughout the semester is the set with nothing in it, which we call the empty set.

Definition (Empty Set)

The empty set is the set $\emptyset = \{\}$ that contains no elements. If we think of a set as a box with elements in it, then the empty set is a box with nothing in it.

Here is another axiom that you have probably used many times in your life without ever realizing it.

Axiom (Well Ordering Principle)

Every nonempty subset $S$ of the natural numbers has a least element. By least element, we mean that there is a natural number $m$ which is an element of $S$ such that $m\leq x$ for every $x$ in $S$.

Problem 10: (Which Dominoes Remain Standing)

Suppose that Jon has set up an infinite number of dominoes, with the dominoes numbered $1,2,3, \ldots$. The dominoes are set up so that if the $k$th domino falls, then the $(k+1)$st domino will also fall. So if the 7th domino falls, then the 8th must fall as well. Jon knocks down the first domino, which starts causing other dominos to fall. Which dominos fall? Which dominoes remain standing? Make sure you prove your result. The well ordering principle will come in handy.

Suggestion: Use set builder notation to help you, so let $F=\{n\in \mathbb{N}\mid \text{domino $n$ fell}\}$ and $S=\{n\in \mathbb{N}\mid \text{domino $n$ remains standing}\}$. Then make some claims and prove they are correct.

For more problems, see AllProblems

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