# MATH 409 Lecture 13

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**End Exam 1 content**

## Review for Test

Part 1: Chapter 1 and Appendix A

- Axioms of an ordered field
- is an ordered fields

- Completeness axiom
- Archimedean principle
- Principle of Mathematical Induction
- Binomial Formula (as application of induction)
- Countable and Uncountable sets.

Part 2: Chapters 2.1–2.5 and 3.1–3.4

- Limits of sequences
- Limit theorems for sequences
- Monotone sequences
- Bolzano-Weierstrass theorem
- Cauchy Sequences (i.e. convergent sequences)
- Limits of functions (one-sided, two-sided; at point, at infinity)
- Limit Theorems for functions (analogous to theorems for sequences)
- Continuity of Functions
- Extreme value and Intermediate Value theorems
- Uniform Continuity.

### Axioms of Real Numbers

3 Postulates:

- is a field (+ and · defined with certain properties)
- There is a strict linear order on that makes it into an ordered field
**(Completeness Axiom)**If a nonenmpty subset is bounded above, then has a supremem. Conversely, if is bounded below, then has an infimum.

### Theorems to Know

(will need to prove one)

**Archimedian Principle.** For any real number , there exists a natural number such that .

**Principle of Mathematical Induction.** Let be an assertion depending on a natural variable . Suppose that

- holds, and
- Whenever holds, so does

Then holds for all .

**Theorem.** If , , ... are finite or countable sets, then the union is also finite or countable. As a consequence, , , and are countable.

**Theorem.** is uncountable.

#### Limit Theorems for Sequences

**Squeeze Theorem.**For a sequence , if , and for , then**Comparison Theorem.**if and are convergent sequences, and there exists such that for all , then**Arithmetic Operations Theorems.**Limit of sum/difference is sum/difference of limits; Limit of product/quotient (for nonzero denominator) is product/quotient of limits

**Monotone Convergence Theorem.** Any monotone sequence converges to a limit if bounded, and diverges to infinity otherwise.

**Bolzano-Weierstrass.** Every bounded sequence of real numbers has a convergent subsequence.

**Cauchy Sequences.** Any Cauchy sequence is convergent.

#### Limit Theorems for Functions

**Sequential Characterization of Continuity.** is continuous at if and only if for any sequence of elements of , as implies as .

**Arithmetic Operations Theorems.** (analogous to above)

**Extreme Value Theorem.** If is a closed, bounded interval, then any continuous function is bounded and attains its max and min within .

**Intermediate Value Theorem.**

**Theorem.** Any function continuous on a closed bounded interval [a,b] is also uniformly continuous on [a,b]

## Sample Test

### Problem 1

*15 pts.*

Prove that for any ,

Application of principle of mathematical induction

*Proof.* Let , then . Assume that holds for some . Add to both sides, and

Note, we have just proved that . It is also known that . Therefore for all .

### Problem 2

*30 pts.*

Let be the sequence of Fibonacci numbers and for .

- Show that is increasing while is decreasing.
- Prove that .

*Proof of 1.* Let for . Then

for all . In particular, , , and , and . Notice that

The function is strictly decreasing on the interval and maps it to itself. Therefore its second iteration is strictly increasing on and (a decreasing function reverses the order of , so applying a decreasing function to a decreasing function once again reverses the order to be increasing). We have for . Now it follows by induction on that

(apply to all elements of the inequality for the inductive step).

The proposition holds by transitivity and comparison theorem.

*Proof of 2.* We already know that satisfy inequalities

It follows that is strictly increasing, is strictly decreasing, and both sequences are bounded. Therefore, these sequences are converging to some positive limits: and as . To prove that , it is enough to show that .

For any we obtain

It follows that and as . However, and , which implies that and . Since

Therefore and are roots of the equation . There are two roots because the constant is negative. Moreover, both roots must be of different signs (leading coefficient is positive, and constant is negative). This equation has two roots, . One of the roots is negative, so and are equal to the other root, .

### Problem 3

Prove the extreme value theorem: If is a continuous function on a closed bounded interval , then is bounded and attains its extreme values on .

*Proof.* First let us prove that is bounded. Assume the contrary. Then for every , there exists a point such that . We obtain a sequence .

...

Since is bounded, the image is a bounded subset of . Let , and . For any , the number is not an upper bound of and is not a lower bound of . Hence we can find points such that and . At the same time, ...

...

### Problem 4

*20 pts.*

Consider defined by

and for .

- determine all points at which the function is continuous.
- Is the function uniformly continuous on the interval ? is it uniformly continuous on the interval ?

*Proof of 1.* The polynomial functions and are continuous on the entire real line. Moreover, if and only if . Therefore the quotient is well-defined and continuous on .

Further, the function is continuous on . Since is continuous on , the composite function is continuous on .

Clearly for all . It follows that is continuous on that range as well.

It remains to determine whether the function is continuous at points , , and . Observe that for all . Therefore as , as , and as .

Since the function is continuous at and , we have as and as . Note that since . It follows that as . In particular is discontinuous at .

Further, as . Since , the function has a removable discontinuity at .

All that remains is continuity at . Finally, is not continuous at since it has no limit at . To be precise, let and for all . Then and are two convergent sequences of positive numbers converging to 0. We have and for all . It follows that and as . Hence there is no limit of as .

*Proof of 2.* Any function uniformly continuous on the open interval can be extended to a continuous function on . As a consequence, such a function has a right-hand limit at . However, we already know that has no right-hand limit at . Therefore is not uniformly continuous on (due to oscillation to the right of 0)

The function is continuous at and has a removable singularity at . Changing the value of at to the limit at , we obtain a function continuous on . It is known that every function continuous on the closed interval is also uniformly continuous on . Further, any function uniformly continuous on the set is also uniformly continuous on its subset . Since the redefined function coincides with on , we conclude that is uniformly continuous on

The function is continuous at and has a removable singularity at . Changing the value of at to the limit at , we obtain a function continuous on . It is known that every function continuous on the closed interval is also uniformly continuous on . Further, any function uniformly continuous on the set is also uniformly continuous on its subset . Since the redefined function coincides with on , we conclude that is uniformly continuous on .

### Bonus Problem 5

Given a set , let denote the set of all subsets of . Prove that is not of the same cardinality as .

(paradox: is set of all sets countable? Well, introduce concept of *class*)