Please Login to access more options.


Today

« July 2018 »

Sun

Mon

Tue

Wed

Thu

Fri

Sat

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

From Ben:

Here is who we'll have present. Please come on time, so we can have you start putting things up when class starts.

  • 71 - Jason
  • 72 -
  • 73 -
  • 75 -
  • 76 -
  • 77 -

The following students were absent

  • Connor
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 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 $-\infty$.

For more problems, see AllProblems