From Ben: Here is the list of definitions that will appear on the final exam. We've seen all of these now. Our goal is to get really good at working with these definitions.

Use the table of contents to the right to find what you need. I'll periodically make older problems disappear, which you can still view using the link below.

Definition (Invertible Function)

We say that a function $f:D\to C$ is invertible if there exists a function $g:C\to D$ such that $f(g(c))=c$ for every $c\in C$ and $g(f(d))=d$ for every $d\in D$. If such a function $g$ exists, then we use the notation $f^{-1}$ as the name for the function $g$.

Problem 69: (Functions Are Invertible Iff Bijective)

Prove that a function $f:D\to C$ is invertible if and only if $f$ is bijective.

Problem 70: (Proving A Quotient Of Two Quadratic Sequences Converges)

Prove that $\left(\frac{n^2}{n^2+1}\right)$ converges.

Hint: Given $\varepsilon>0$, solve the equality $|a_M-L|=\varepsilon$ to find a value $M$ you can choose to satisfy the definition of converges.

Definition (Image Of A Function)

Let $f:X\to Y$. The image of $f$ is the set $f(X)$.

Some people call the image of $f$ the "range of $f$", while other people use the word "range" to denote the codomain. The word "range" is unfortunately dependent on author. For this reason, we'll stick to the words "codomain" and "image" to separate these two different sets.
Exercise (A Function Is Surjective Iff Codomain And Image Are Equal)

Prove that a function is surjective if and only if the codomain of $f$ and the image of $f$ are equal.

Click to see a solution.

Let $f:X\to Y$. Suppose $f$ is surjective. Clearly the image of $f$ is a subset of the codomain by definition. Let $y$ be an element of the codomain. Since $f$ is surjective, this means we can pick $x$ in the domain such that $f(x)=y$. This shows that $y$ is in the image of $f$, which completes the proof that if $f$ is surjective, then the image of $f$ equals the codomain of $f$.

Now suppose that the image of $f$ equals the codomain of $f$. We need to show that $f$ is surjective. Pick $y$ in the codomain of $f$. Since the image equals the codomain, this means $y$ is in the image of $f$. This means that we can pick $x$ in the domain such that $f(x)=y$. This completes the proof that $f$ is surjective when the image equals the codomain.

Exercise (The Composition Of Surjective Functions Is Surjective Take 2)

Suppose both $f:A\to B$ and $g:B\to C$ are surjective. Prove that $g\circ f:A\to C$ is surjective.

Click to see a solution.

First note that a function is surjective if and only if the image of the domain equals the codomain. We will use both the "if" and "only if" parts of that statment. We now begin the proof. Because the two functions are surjective, we know $f(A)=B$ and $g(B)=C$ (we used the "only if", or implies, part). This means that $(g\circ f)(A)=g(f(A))=g(B)=C$. Since we know $(g\circ f)(A)=C$, this proves that $g\circ f$ is surjective (we used the "if" part).

Definition (The Image Of A Sequence Is A Set)

Given a sequence $(a_n)$ of real numbers, note that the image of the sequence, namely $a(\mathbb{N}) = \{a_n\mid n\in\mathbb{N}\}$, is a subset of the real numbers. Because the image of the sequence is a set of real numbers, we can use any of our previous words that we defined on sets of real numbers, and now apply them to a sequence. Here are some examples:

  • We say a sequence is bounded if the image of the sequence is a bounded set.
  • A lower bound for a sequence is a lower bound for the image of the sequence.
  • The supremum of a sequence is the supremum of the image of the sequence.
  • A limit point of a sequence is a limit point of the image of the sequence.

Problem 71: (Convergent Sequences Are Bounded)

Prove that if a sequence of real numbers converges, then the sequence is bounded.

Problem 72: (Limit Of Product Equals Product Of Limits)

Suppose $(a_n)$ converges to $A$ and $(b_n)$ converges to $B$. Prove that $(a_nb_n)$ converges to $AB$. The triangle inequality should help, along with the fact that convergent sequences are bounded.

Click for a hint.

The key here is to rewrite $|a_nb_n-AB|$ in a new way so that the quantities $a_n-A$ and $b_n-B$ appear. There are lots of ways to do this. One is to just force them to appear by replacing $a_n$ with $a_n-A+A$, giving us $$ |a_nb_n-AB|= |(a_n-A+A)b_n-AB|= |(a_n-A)b_n+Ab_n-AB|= |(a_n-A)b_n+A(b_n-B)|. $$ At this point, the triangle inequality should help to separate things. Then you'll need to pick $M$ large enough so that both $|(a_n-A)b_n|\leq \frac{\varepsilon}{2}$ and $|A(b_n-B)|<\frac{\varepsilon}{2}$. You'll need to guarantee $|a_n-A||b_n|\leq \frac{\varepsilon}{2}$ and $|b_n-B||A|<\frac{\varepsilon}{2}$. You will need an estimate for $|b_n|$ (did you see the bounded part), and use facts about $(a_n)$ and $(b_n)$ converging to get the needed $M$.

Problem 73: (Limits Of Sequences And Limit Points Of Images)

Is a limit point of the image of a sequence equal to the limit of that sequence?

  • Give an example of a sequence $(a_n)$ that converges to $L$, such that $L$ is a limit point of the image of the sequence.
  • Give an example of a sequence $(a_n)$ that converges to $L$, such that $L$ is not a limit point of the image of the sequence.
  • Make a conjecture about when limits of sequences and limit points of images of sequences are equal.
  • Give an example of a sequence $(a_n)$ that does not converge, yet the image of the sequence has one (or more) limit points.
We'll have a class discussion on this one, rather than have someone present it. Please come with examples and a conjecture. If you choose to type this one up, you will need to prove any claims you make (about a sequence converging, or not, to what you say it does (which means you probably want to focus on typing up a different problem, as doing so could be quite time consuming if your examples are non trivial.

Problem 74: (Converges To L Can Be Written As Converges To 0)

Let $(a_n)$ be a sequence of real numbers, and $A$ a real number. Prove that $(a_n)$ converges to $A$ if and only if the sequence $(a_n-A)$ converges to $0$. This will simplify proving that some sequences converge.

Problem 75: (Proving A Rational Sequence Converges)

Consider the sequence $\ds (s_n)=\left(\frac{4n^3-2n^2-7n}{5n^3-3n^2+2n-1}\right)$. In this problem, your job is to prove that $s_n\to 4/5$. You may assume that for all natural numbers, we have $5n^3-3n^2+2n-1>0$.

  1. Show that $\ds(s_n-4/5)= \left(\frac{2 n^2-43 n+4}{5 \left(5 n^3-3 n^2+2 n-1\right)}\right).$
  2. Find a $k_1>0$ and $M_1$ so that $|2 n^2-43 n+4|\leq k_1 n^2$ for all natural numbers $n>M_1$.
  3. Find a $k_2>0$ and $M_2$ so that $|5 \left(5 n^3-3 n^2+2 n-1\right)|\geq k_2 n^3$ for all natural numbers $n>M_2$.
  4. Prove that $(s_n)$ converges to $4/5$.

Problem 76: (Limit Of Quotient Equals Quotient Of Limits)

Suppose $(a_n)$ converges to $A$ and $(b_n)$ converges to $B\neq 0$, and also suppose $b_n\neq 0$ for every natural number $n$. Prove that $(a_n/b_n)$ converges to $A/B$.

Definition (Increasing Decreasing Monotonic Sequences)

Let $(a_n)$ be a sequence of real numbers.

  • We say that $(a_n)$ is (strictly) increasing if $a_n<a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is (strictly) decreasing if $a_n>a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is nonincreasing if $a_n\geq a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is nondecreasing if $a_n\leq a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is monotonic if $(a_n)$ is either nonincreasing or nondecreasing.

You should notice that a strictly decreasing sequence is nonincreasing, and a strictly increasing sequence is nondecreasing.

Problem 77: (Monotonic Sequences Converge If And Only If Bounded)

Let $(a_n)$ be a monotonic sequence. Prove that $(a_n)$ converges if and only if $(a_n)$ is bounded.

Definition (Diverges To Infinity)

Let $(a_n)$ be a sequence of real numbers. We say that $(a_n)$ diverges to infinity if for every $V$, there exists $H$ such that for every $n\in \mathbb{N}$ we know $n>H$ implies $a_n>V$. When $(a_n)$ diverges to infinity, we write $(a_n)\to \infty$.

Think of $V$ as vertical and $H$ as horizontal.

Problem 78: (A Sequence Diverges To Infinity)

Prove that the sequence $(n^2)$ diverges to $+\infty$.

Problem 79: (Diverges To Negative Infinity)

Construct a definition of what it means to diverge to $-\infty$, by appropriately modifying the definition of diverges to $\infty$. Then prove that the sequence $(-n^3+2n)$ diverges to $-\infty$.

From Ben: The problems that remain are all designed to help you become more familiar with working with intersections and unions of arbitrarily many sets.
In the rules for set complements theorem, we prove that the complement of a union is the intersection of the complements, and we proved that the complement of an intersection was the union of the complements. These are often called De Morgan's laws. The next problem has you prove De Morgan's laws are true regardless of the number of sets involved in the union or intersection.

Problem 80: (DeMorgan's Laws For Sets)

Suppose that for each $j$ in some nonempty set $J$ that $A_j$ is a set, and also suppose that $B$ is a set. Pick one of the statements below and prove that it is true. The other is very similar.

  • $\ds B\setminus \left(\bigcup_{j\in J}A_j\right) = \bigcap_{j\in J}\left(B\setminus A_j\right) $
  • $\ds B\setminus \left(\bigcap_{j\in J}A_j\right) = \bigcup_{j\in J}\left(B\setminus A_j\right) $
One of the statements in the next problem is false. I'll let you find it.

Problem 81: (Additional Properties Of Cartesian Products)

Let $A$, $B$, $C$, and $D$ be sets. Prove or disprove each statement.

  1. $(A\times B)\cap(C\times D) = (A\cap C)\times(B\cap D)$
  2. $(A\times B)\cup(C\times D) = (A\cup C)\times(B\cup D)$
Definition (Interior And Closure)

Let $A$ be a subset of the real numbers.

  • We use the notation $A'$ to denote the set of limit points of $A$.
  • The closure of $A$, written $\text{cl}(A)$, is the union of $A$ and its limit points, so $\text{cl}(A) = A\cup A'$.
  • The interior of $A$, written $\text{int}(A)$, is the collection of interior points of $A$.

Problem 82: (The Closure of $A$ Is a Closed Set.)

Let $A$ be a set. Prove that the closure of $A$ is a closed set.

Problem 83: (The Interior Is The Union Of Every Open Set Inside A Set)

Let $S^\circ$ be the union of every open set contained in $S$. Prove that $\text{int} (S)=S^\circ$.

Problem 84: (The Closure Is The Intersection Of Every Closed Set Containing A Set)

Let $\bar A$ be the intersection of every closed set that contains $A$. Prove that $\bar A=\text{cl}(A)$.

Exercise.Standard Induction Argument

We have proven many facts this semester about how to combine two things to obtain something new. There is a very standard induction argument that allows you to take a statement, like any of the ones below, and make the statement true for any finite number of things.

  1. If $f$ and $g$ are surjective, then $f\circ g$ is surjective.
  2. If $f$ and $g$ are injective, then $f\circ g$ is injective.
  3. If $U$ and $V$ are open, then $U\cap V$ is open.
  4. If $U$ and $V$ are closed, then $U\cap V$ is closed.
  5. If $U$ and $V$ are open, then $U\cup V$ is open.
  6. If $U$ and $V$ are closed, then $U\cup V$ is closed.
  7. If $(a_n)$ converges to $A$ and $(b_n)$ converges to $B$, then $(a_n+b_n)$ converges to $A+B$.
  8. If $(a_n)$ converges to $A$ and $(b_n)$ converges to $B$, then $(a_nb_n)$ converges to $AB$.
  9. If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cap A_2)\subseteq f(A_1)\cap f(A_2)$.
  10. If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cup A_2)=f(A_1)\cup f(A_2)$.
  11. If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)$.
  12. If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$.
  13. If $x,y\in\mathbb{R}$ then $|x+y|\leq |x|+|y|$.
  14. If $A,B,C$ are sets, then $(A\cap B)\times C = (A\times C)\cap(B\times C)$.
  15. If $A,B,C$ are sets, then $(A\cup B)\times C = (A\times C)\cup(B\times C)$.
  16. If $A,B,C$ are sets, then $A\setminus(B\cap C) = (A\setminus B)\cup(A\setminus C)$.
  17. If $A,B,C$ are sets, then $A\setminus(B\cup C) = (A\setminus B)\cap(A\setminus C)$.
  18. If $A,B,C$ are sets, then $A\cup (B\cap C)=(A\cup B)\cap(A\cup C)$.
  19. If $A,B,C$ are sets, then $A\cap (B\cup C)=(A\cap B)\cup(A\cap C)$.
  • To continue the problems, please follow this link.