Analysis 1 Flashcards
Axioms for the real numbers:
What are the addition axioms?
A1 a + b = b + a [+ is commutative]
A2 a + (b + c) = (a + b) + c [+ is associative]
A3 a + 0 = a [zero and addition]
A4 a + (−a) = 0 [negatives and addition]
Axioms for the real numbers:
What are the multiplication axioms?
M1 a · b = b · a [· is commutative]
M2 a · (b · c) = (a · b) · c [· is associative]
M3 a · 1 = a [the unit element and multiplication]
M4 If a ≠ 0 then a · 1/a = 1 [reciprocals and multiplication]
Axioms for the real numbers: What are the non-addition/multiplication axioms? (2 of them)
D a · (b + c) = a · b + a · c [· distributes over +]
Z 0 ≠ 1 [to avoid total collapse]
Properties of real numbers:
Prove using the axioms
(1) If a + x = a for all a then x = 0 (uniqueness of zero element)
(1) We have
x = x + 0 by A3
= 0 + x by A1
= 0 by the hypothesis, with a = 0.
Properties of real numbers:
Prove using the axioms
(2) If a + x = a + y then x = y (cancellation law for addition, which implies uniqueness of
additive inverse of a).
(2) We have y = y + 0 by A3 = y + (a + (−a)) by A4 = (y + a) + (−a) by A2 = (a + y) + (−a) by A1 = (a + x) + (−a) by hypothesis = (x + a) + (−a) by A1 = x + (a + (−a)) by A2 = x + 0 by A4 = x by A3.
Properties of real numbers:
Prove using the axioms
(3) −0 = 0.
(3) 0 + 0 = 0 by A3 with a = 0. Hence −0 = 0 by A4 and (2).
Properties of real numbers:
Prove using the axioms
(4) − (−a) = a.
(4) We have (−a) + a = a + (−a) by A1 = 0 by A4, and (−a) + − (−a) = 0 by A4. Now appeal to (2) (cancellation law for addition).
Properties of real numbers:
Prove using the axioms
(5) − (a + b) = (−a) + (−b)
proof
Properties of real numbers:
Prove using the axioms
(6) If a · x = a for all a 6= 0 then x = 1 (uniqueness of multiplicative identity)
Use M1-4
Properties of real numbers:
Prove using the axioms
(7) If a 6= 0 and a · x = a · y then x = y (cancellation law for multiplication, which implies
uniqueness of multiplicative inverse of a)
Use M1-4
Properties of real numbers:
Prove using the axioms
(8) If a 6= 0 then 1/ (1/a) = a
Use M1-4
Properties of real numbers:
Prove using the axioms
(a + b) · c = a · c + b · c.
(9) Note that the statement is like Axiom D but with multiplication on the other side. We have (a + b) · c = c · (a + b) by M1 = c · a + c · b by D = a · c + b · c by M1 twice.
Properties of real numbers:
Prove using the axioms
(10) a · 0 = 0
(10) We have a · 0 + 0 = a · 0 by A3 = a · (0 + 0) by A3 = a · 0 + a · 0 by D. Now appeal to (2) (cancellation law for addition)
Properties of real numbers:
Prove using the axioms
(11) a · (−b) = − (a · b). In particular (−1) · a = −a
(11) We have (a · b) + (a · (−b)) = a · (b + (−b)) by D = a · 0 by A4 = 0 by (10). Also (a · b) + (−(a · b)) = 0 by A4. Now appeal to (2).
Properties of real numbers:
Prove using the axioms
· b). In particular (−1) · a = −a.
(12) (−1) · (−1) = 1
(12) We have
(−1) · (−1) = −(−1) by (11)
= 1 by (4)
Properties of real numbers:
Prove using the axioms
(13) If a · b = 0 then either a = 0 or b = 0 (or both). Moreover, if a 6= 0 and b 6= 0 then
1/(a · b) = (1/a) · (1/b)
(13) Assume for a contradiction that a, b ≠ 0 but a · b = 0. Then 0 = (1/a · 1/b) · 0 by (10) = 0 · (1/a · 1/b) by M1 = (a · b) · (1/a · 1/b) by hypothesis = ((b · a) · 1/a) · 1/b by M2 = (b · (a · 1/a)) · 1/b by M2 = (b · 1) · 1/b by M4 = b · (1/b) by M3 = 1 by M4, This contradicts Axiom Z. The second assertion has been proved along the way
There is a subset P (the (strictly) positive numbers) of R such that, for a, b ∈ R,
P1
P2
P3 ???
P1 a, b ∈ P =⇒ a + b ∈ P;
P2 a, b ∈ P =⇒ a · b ∈ P;
P3 exactly one of a ∈ P, a = 0 and −a ∈ P holds
Describe what we mean by > and ≥ using the (strictly) positive numbers set and the non-negative numbers set
There is a subset P (the (strictly) positive numbers) of R such that, for a, b ∈ R
We write a < b (or b > a) iff b − a ∈ P and a ≤ b (or b ≥ a) iff b − a ∈ P ∪ {0}
(the non-negative numbers)
Explain each property of the order on R (reals)
Reflexivity
Reflexivity: a ≤ a.
Proof. a − a = 0 ∈ P ∪ {0}, by A4
Explain each property of the order on R (reals)
Antisymmetry
Antisymmetry: a ≤ b and b ≤ a together imply a = b.
Proof. If a − b = 0 or b − a = 0, then a = b by properties of addition. Otherwise
P 3 a − b and P 3 b − a = −(a − b) (by properties of addition). Now apply P3.
Explain each property of the order on R (reals)
Transitivity:
Transitivity: Assume a ≤ b and b ≤ c. Then a ≤ c, and likewise with < in place of ≤.
Proof. We have c−a = c+ (−a) = c+ 0 + (−a) = c+ (−b) +b+ (−a) = (c−b) + (b−a)
by properties of addition. So the result for < follows from the definition of < and P3.
The proof for ≤ is the same except for the need to allow also for the trivial cases
a = b and/or b = c.
Explain each property of the order on R (reals)
Trichotomy
Trichotomy: Exactly one of a < b, a = b and b < a holds.
Proof. This follows from P3 and the definition of
The following statements hold: (1) 0 < 1 (equivalently, 1 ∈ P), (2) a < b if and only if −b [ ]. In particular a > 0 iff [ ]. (3) a < b and c ∈ R implies a + c [ ]. (4) a < b and 0 < c implies ac < [ ]. (5) a² > 0, with equality iff [ ] (6) a > 0 iff 1/a [ ]. (7) If a, b > 0 and a < b then 1/b [ ] Claims (2)–(4) also hold with ≤ replacing <
(1) 0 < 1 (equivalently, 1 ∈ P),
(2) a < b if and only if −b < −a. In particular a > 0 iff −a < 0.
(3) a < b and c ∈ R implies a + c < b + c.
(4) a < b and 0 < c implies ac < bc.
(5) a² > 0, with equality iff a = 0.
(6) a > 0 iff 1/a > 0.
(7) If a, b > 0 and a < b then 1/b < 1/a.
Claims (2)–(4) also hold with ≤ replacing
Prove the following:
(1) 0 < 1 (equivalently, 1 ∈ P),
(1) By trichotomy, exactly one of (i) 1 < 0, (ii) 1 = 0, (iii) 0 < 1 holds. Axiom Z rules
out (ii). Suppose for a contradiction that 1 < 0. Then −1 = 0 + (−1) ∈ P. We deduce
that (−1) · (−1) ∈ P by P2. But 1 = (−1) · (−1) (by 1.2(12)), so 0 < 1 and we have a contradiction to trichotomy.
Prove the following:
(2) a < b if and only if −b < −a. In particular a > 0 iff −a < 0
(2) By properties of addition,
a < b ⇐⇒ b − a ∈ P
⇐⇒ (−a) − (−b) ∈ P
⇐⇒ (−a) > (−b).
Prove the following:
(3) a < b and c ∈ R implies a + c < b + c
(3) a + c < b + c iff 0 < (b + c) − (a + c) = b − a iff a < b
Prove the following:
(4) a < b and 0 < c implies ac < bc
(4) a < b and c > 0 implies bc − ac = (b − a)c > 0, by P2.
Prove the following:
(5) a² > 0, with equality iff a = 0
(5) Note a² = 0 iff a = 0, by {If a · b = 0 then either a = 0 or b = 0 (or both). Moreover, if a ≠ 0 and b ≠0 then
1/(a · b) = (1/a) · (1/b).}.
Assume a ≠ 0. Then we have a² = a·a = (−a)·(−a).
Since either a > 0 or −a > 0 it follows that a² > 0 by P2.
Prove the following:
(6) a > 0 iff 1/a > 0
Assume for contradiction a > 0 but 1/a < 0. Then −1 = −(a · (1/a)) = a · (−1/a) > 0
by P2 which is impossible by (1). Likewise we obtain a contradiction if a < 0 and
1/a > 0.
Prove the following:
(7) If a, b > 0 and a < b then 1/b < 1/a
Use
a < b and 0 < c implies ac < bc
and
a > 0 iff 1/a > 0
What is Bernoulli’s Inequality?
Let x be a real number with x > −1 and let n be a positive integer. Then
(1 + x)ⁿ ≥ 1 + nx.
Prove Bernoulli’s Inequality
We shall prove the inequality by induction—note that the inequality is trivially true
when n = 1.
Suppose that, for k ∈ N,
(1 + x)ᵏ ≥ 1 + kx
holds for all real x > −1. Then 1 + x > 0 by {a < b and c ∈ R implies a + c < b + c} and kx² ≥ 0 as k > 0 and x² ≥ 0 by
{a² > 0, with equality iff a = 0.}
(1 + x)ᵏ⁺¹ = (1 + x) (1 + x)ᵏ by definition
≥ (1 + x) (1 + kx) by hypothesis and {a < b and 0 < c implies ac < bc} (≤ version)
= 1 + (k + 1) x + kx²
by A1–A4
≥ 1 + (k + 1) x by {Inequality rules}.
Hence the result follows by induction.
Define the modulus
real numbers
The modulus |a| of a ∈ R is defined by
{ a if a > 0
|a| = {0 if a = 0
{ -a if a < 0
Prove the following facts about the modulus:
(1) | − a| = |a|;
(2) |a| ≥ 0
(1) and (2) are immediate from the definition and the fact that a > 0 iff −a < 0
Prove the following facts about the modulus:
(3) |a|² = a²;
We can prove (3) and also (4), by using P3 to enumerate cases; recall too that (−a)(−a) = a² for any a.
Prove the following facts about the modulus:
(4) |ab| = |a||b|;
We can prove (3) and also (4), by using P3 to enumerate cases; recall too that (−a)(−a) = a² for any a.
Prove the following facts about the modulus:
(5) −|a| ≤ a ≤ |a|
For (5), note that
{ −|a| ≤ 0 ≤ a = |a| if a ≥ 0,
{ −|a| = a < 0 ≤ |a| if a < 0
Prove the following facts about the modulus:
(6) if c ≥ 0, then |a| ≤ c iff −c ≤ a ≤ c. if c > 0, then |a| < c iff −c < a < c
Assume first that |a| 6 c, Then, by (5{−|a| ≤ a ≤ |a|} and transitivity of 6, we get
−c ≤ a ≤ c. Conversely, assume −c ≤ a ≤ c. Then −a ≤ c and a ≤ c. Since |a| equals
either a or −a, we obtain |a| ≤ c.
The case with < in place of ≤ is handled similarly.
What is the triangle law/inequality?
Real
(1) Let a, b ∈ R. Then
|a + b| ≤ |a| + |b|.
Prove the triangle inequality
Real
Proof. (1): We have
−|a| ≤ a ≤ |a| and − |b| ≤ b ≤ |b|.
Adding (if a < b, then ac > bc if and only if c < 0) and using properties of addition we get
−(|a| + |b|) ≤ a + b ≤ (|a| + |b|).
Now use 2.5(6){(6) if c ≥ 0, then |a| ≤ c iff −c ≤ a ≤ c. if c > 0, then |a| < c iff −c < a < c}
What is the reverse triangle law/inequality?
Real
(2) Let a, b ∈ R. Then
|a + b| ≥ ||a| − |b||
Prove the reverse triangle inequality
Real
By the Triangle Law,
|a| = |a + b + (−b)| ≤ |a + b| + |(−b)| = |a + b| + |b|,
so |a|−|b| ≤ |a+b|, and likewise, reversing the roles of a and b, we get |b|−|a| ≤ |b+a| = |a+b|.
Now use the fact that |c| is either c or −c always, and apply this with c = |a| − |b|
What are the triangle inequality and reverse triangle inequality for complex numbers?
z + w| ≤ |z| + |w| and |z + w| ≥ ||z| − |w||
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Upper bound
b is an upper bound of S if s ≤ b for all s ∈ S;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Lower bound
b is a lower bound of S if b ≤ s for all s ∈ S;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Bounded above
S is bounded above if it has an upper bound;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Bounded below
S is bounded below if it has a lower bound;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Bounded
S is bounded if it is bounded above and below.
Let S ⊆ R and b ∈ R.
What is the Supremum?
Let S ⊆ R. Then α is the supremum of S, denoted sup S, if
(sup1) s ≤ α for all s ∈ S [α is an upper bound of S]
(sup2) s ≤ b for all s ∈ S implies α ≤ b [α is the least upper bound of S]
Note: sup S is unique if it exists.
What is the Completeness Axiom for the Real Numbers?
Let S be a non-empty subset of R which is bounded above. Then sup S
exists
Note the necessity for the exclusions here. The empty set has no supremum because it has no least upper bound ((sup2) fails). A set which is not bounded above cannot have a supremum because it has no upper bound ((sup1) must fail).
Let S ⊆ R and b ∈ R.
What is the Infimum?
Infimum (= greatest lower bound).
Analogous definitions apply here, by replacing ≤ by > in the definitions above. Let S be a subset of R and α ∈ R. Then α is the infimum of S, written inf S, if
(inf1) α ≤ s for all s ∈ S [α is a lower bound for S]
(inf2) if b ≤ s for all s ∈ S then b 6 α [α is the greatest lower bound of S]
Assume ∅ ≠ S ⊆ R and let s₀ ∈ R
Define the maximum of S
We say s₀ is the maximum of S, and write
s₀ = max S, if
(max1) s₀ ∈ S [s₀ belongs to S]
(max2) s ≤ s₀ for all s ∈ S [s₀ is an upper bound of S]
If a set S is empty or is not bounded above then max S cannot exist
Assume ∅ ≠ S ⊆ R and let s₀ ∈ R
Define the minimum of S
Similarly we say a non-empty set S which is bounded below has a minimum, min S, if there exists s₀ ∈ S such that s₀ ≤ s for all s ∈ S.
Given a, x ∈ R we may interpret |x − a| as ….
Given a, x ∈ R we may interpret |x − a| as the distance from x to a.
For b > 0, and a, x ∈ R,
|x − a| < b ⇐⇒ a − b < x [ ]
For b > 0, and a, x ∈ R,
|x − a| < b ⇐⇒ a − b < x < a + b
For b > 0, and a, x ∈ R,
|x − a| < b ⇐⇒ a − b < x < a + b
prove it
{ if c ≥ 0, then |a| ≤ c iff −c ≤ a ≤ c. if c > 0, then |a| < c iff −c < a < c}
gives |x − a| < b iff −b < (x − a) < b and this holds iff a − b < x < a + b.
So by considering whether this holds for different values of b we can assess how good an
approximation x is to a.
Let S be non-empty and bounded above (so sup S exists). Then, given ε > 0, there exists
sε (in general depending on ε) in S such that
sup S − ε < [ ]
sup S − ε < sε ≤ sup S.
Prove that there exists a unique positive real number α such that α² = 2
End of pg15
Strategy:
Step 1: S ≠ ∅ and S is bounded above, by 2.
Step 2: Assume for contradiction that α² < 2. Note also that α ≥ 1 > 0 since 1 ∈ S. Let
h > 0.
Step 3: Assume for contradiction that α² > 2. Let h > 0.
Explain the incompleteness of Q
The set Q of rationals does not satisfy the Completeness
Axiom with respect to the order it inherits from R. If it did,
T := { q ∈ Q | q > 0 and q² < 2 }
would have a supremum in Q. The proof in 4.10{proof of existence of root 2} works just as well for T as it does for S.
But we know there is no rational square root of 2.
Describe the theorem of the existence of nth roots of real numbers
Any positive real number has a real nth root,
for any n = 2, 3, 4, . . .
Theorem (the Archimedean Property of the Natural Numbers).
(i) N is not bounded above.
(ii) Let ε > 0. Then there exists n ∈ N such that 0 < 1 / n < ε
Prove
Theorem (the Archimedean Property of the Natural Numbers).
(i) Assume for a contradiction that N is bounded above. Then sup N exists by the
Completeness Axiom. By the Approximation Property with ε = 1/2, there exists k ∈ N
with
sup N − 1 / 2 < k ≤ sup N.
But then k + 1 ∈ N and k + 1 > sup N + 1 / 2, a contradiction.
For (ii), exploit the fact that 1/ε cannot be an upper bound for N.
Theorem (compare with the well-ordered property of N)
Minima and maxima of subsets of Z
(i) Every nonempty subset S of Z which is bounded below has a minimum (least element).
(ii) Every nonempty subset S of Z which is bounded above has a maximum (greatest
element) .
Prove:
(i) Every nonempty subset S of Z which is bounded below has a minimum (least element).
(ii) Every nonempty subset S of Z which is bounded above has a maximum (greatest
element) .
Proof. i) We know that inf S exists (by applying the completeness axiom to {−s : s ∈ S}. So by the approximation property with ε = 1 there is some n ∈ S such that
inf S ≤ n < inf S + 1.
It is enough to show that inf S = n, since then inf S ∈ S and so inf S = min S. Assume for
a contradiction that n ≠ inf S, so that n > inf S and hence n = inf S + ε where 0 < ε < 1.
By the approximation property again, there exists some m ∈ S such that
inf S ≤ m < inf S + ε = n.
Since n > m we have n − m > 0 and so n − m ≥ 1 because n − m is an integer, so
n ≥ inf S + 1, which contradicts our first inequality for n.
The proof of (ii) is similar.
What are the density properties?
Density properties.
(i) Given a, b ∈ R with a < b there exists x ∈ Q such that a < x < b.
(ii) Given a, b ∈ R with a < b there exists y ∈ R \ Q such that a < y < b
When are two sets equinumerous?
Let A and B be sets. We say A and B are equinumerous (notation A ≈ B) if there is
a bijection f : A → B. Note ≈ has the properties of an equivalence relation.
What does the symbol A ≤ B? If A and B are two sets
Note:
Not less than or equal to, concave version of the symbol
Given sets A and B we shall write A ≤ B if there is an injection f : A → B. Intuitively
this says that B is at least as big as A.
What is a finite set?
Let A be a set. We call A finite if either A = ∅ or there exists n ∈ N such that
A ≈ {0, . . . , n − 1} (or equivalently if A ≈ {1, . . . , n}).
What is an infinite set?
A set which is not finite is said to be infinite
What is another way to define finiteness of a set, using maps?
Another way to capture the notion of finiteness is to say that A is finite iff every
injective map from A to A is surjective: the Pigeonhole Principle holds for A
Any subset of a finite set is [ ]
finite.
any non-empty finite subset of R must be [ ] (in fact, it contains a [ ])
Hence any subset of R which is not bounded above must be [ ]
bounded above
largest element
infinite
By the Archimedean property, N is not bounded above, and hence is [ ]
infinite
We call a set A
• countably infinite if [ ];
• countable if [ ]
• uncountable if A is [ ]
- countably infinite if A ≈ N;
- countable if A ≤ N; {curvy ≤}
- uncountable if A is not countable
