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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
quod erat demonstrandum
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
for all
(apply
to all elements of the inequality for the inductive step).
The proposition holds by transitivity and comparison theorem.
quod erat demonstrandum
Proof of 2. We already know that
satisfy inequalities
for all
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,
.
quod erat demonstrandum
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,
...
...
quod erat demonstrandum
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
.
quod erat demonstrandum
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
.
quod erat demonstrandum
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)