1. Write down your proof of the "Monotone Convergence Theorem:" that every bounded monotone(-decreasing) sequence converges (to its infimum). Here's the google doc we worked on together.(Links to an external site.) Feel free to add additional explanation of any warrants that you think are missing from the writeup that your group did.

Google Document

Let's work together

Theorem: Every monotoneincreasingsequence that has anupper bound convergesto its supremum.

Proof:

Note: Maybe we could be more careful about set vs sequence by defining A = {a_n}

Let (a_n) be a monotone increasing sequence and suppose that u is an upper bound for the sequence.

For all n in \mathbb{N}, a_n+1 >= a_n, because that's how we defined monotone increasing

Since (a_n) is bounded, it has a supremum because of the axiom of completeness. Let s = sup(a_n).

Let e > 0 and consider s - e.

Since s - e < s, it's not an upper bound of (a_n).

Therefore there exists a term of the sequence that's greater than s-e, call it a_N.

a_N+1 >= a_N > s-e

[Consider the fact that a_n+1 >= a_n, a_n+2 >= a_n+1. By transitivity, a_n+2 >= a_n. Repeating, a_n+b >= a_nfor all natural numbers b.]

[Therefore, if c > d are arbitrary natural numbers, then a_c >= a_d ]

Therefore, if n > N, a_n >= a_N

So for all n > N, a_n > s-e

Since s is the supremum of A (or (a_n) whichever you like better), s >= a_n for any n.

Stringing together inequalities, s-e < a_n <= s < s+e

[s-e n < s+e by transitivity]

d.a.d. lemma: |a_n - s| < e.

Therefore, |a_n - s| < e for all n > N. (All of the terms past N are in the epsilon noodle)

Therefore, (a_n) converges to s.

	Claim (a complete sentence: "this apple is red", but not "this apple" or "is red"	Warrant (general principle: a definition, a theorem ...)	Data (what we know about the objects in our proof)
1	For all n , an+1 >= an	def of monotone increasing	Let (an) be a monotone increasing sequence
2	(an) has a supremum	the axiom of completeness	(an) has an upper bound
3	s-e is not an upper bound of (an)	def of supremum part b: anything smaller isn't an ub.	s-e
4	there exists a term of the sequence that's greater than s-e (s - e < aN)	negation of def of upper bound:	s-e is not an upper bound
5	aN+1 >= aN	def of monotone increasing	(an) monotone increasing
6	aN > s-e	order axioms: x < y means same as y > x	(restating claim 4)
7	an+2 >= an+1	def of monotone increasing	(an) monotone increasing
8	an+2 >= an	transitivity	an+2>= an+1 >= an
9	an+b >= anfor all natural b	"repeated transitivity"	an+b>= an+2
10	Therefore, if c > d are arbitrary natural numbers, then ac >= ad	if c > d are natural, then c= d+b
11	if n > N, an >= aN	substitution	claim 5: (an) monotone increasing
12	for all n > N, an > s-e	transitivity	aN > s-e and an >= aN
13	s >= an for any n	definition of supremum , part a	s is the supremum
14	s-e < an	(claim 12)
15	an <= s	(claim 13)
16	s < s+e	order axioms	e > 0
17	s-e < an < s+e	transitivity	claims 14-16
18	|an - s| < e	d.a.d. lemma	s-e < an < s+e
19	|an - s| < e for all n > N	"summarizing"	18, 12, 13
20	(an) converges to s	def of convergence	19

This what my group did below!

Group

Theorem: Every monotonedecreasingsequence that has alower bound that convergesto its infimum.

Proof:

Let (a_n) be a monotone decreasing sequence and suppose thatl is a lower bound for the sequence.

For all n in {N}, a_n >= a_n+1, because that's how we defined monotone decreasing

Since (a_n) is bounded, it has an infimum because of the axiom of completeness. Leti = inf(a_n).

Let > 0 and consideri + .

Sincei + >i it's not a lower bound of (a_n).

Therefore there exists a term of the sequence that's less thani + , call it a_N.

i + > a_N >= a_N+1

[Consider the fact that a_n >= a_n+1, a_n+1 >= a_n+2. By transitivity, a_n >= a_n+2. Repeating, a_n >= a_n+bfor all natural numbers b.]

[Therefore, if c > d are arbitrary natural numbers, then a_d >= a_c ]

Therefore, if n > N, a_N >= a_n

So for all n > N, a_n <i +

Since i is the infimum of A (or (a_n) whichever you like better), i <= a_n for any n.

Stringing together inequalities,i - < a_n <= i <i +

[i - < a_n <i + by transitivity]

d.a.d. lemma: |a_n -i | < e.

Therefore, |a_n - i| < e for all n > N. (All of the terms past N are in the epsilon noodle)

Therefore, (a_n) converges to i.

	Claim (a complete sentence: "this apple is red", but not "this apple" or "is red"	Warrant (general principle: a definition, a theorem ...)	Data (what we know about the objects in our proof)
1	For all n in \mathbb{N}, an >= an+1	def of monotone decreasing	Let (an) be a monotone decreasing sequence
2	(an) has an infimum	the axiom of completeness	(an) has a lower bound
3	i+e is not a lower bound of (an)	def of infimum part b: anything greater isn't an lb.	i+e
4	there exists a term of the sequence that's greater than i+e (i + e > aN)	negation of def of lower bound:	i+e is not a lower bound
5
6
7
8
9
10