Proof of Convergence Theorem for Decreasing Sequences with Logic Laws

In summary, the conversation discusses the proof of a statement involving a limit and a decreasing sequence. The proof uses the definition of a limit, as well as axioms and definitions for natural numbers and the order relation. The only logical law used is the axiom of trichotomy. A proof by contradiction is also employed.
  • #1
evagelos
315
0
Suppose we are asked to prove :

If [tex]\lim_{n\to\infty}x_{n} = L[/tex] and [tex]x_{n}[/tex] is decreasing,then [tex]\forall n[n\in N\Longrightarrow x_{n}\geq L][/tex].

On the following proof give a list of all the thoerems ,axioms,definitions and the laws of logic that take place in the proof


proof:
Suppose that there exists [tex]n_1\in\mathbb N[/tex] such that [tex]x_{n_1}<L[/tex]. Take [tex]\epsilon=L-x_{n_1}[/tex]. Since [tex]\displaystyle\lim_{n\rightarrow\infty}x_n=L[/tex] there exists [tex]n_2\in\mathbb N[/tex] such that [tex]\|x_n-L\|<\epsilon[/tex] for all [tex]n\geq n_2[/tex]. Take [tex]n_0>\max\{n_1,n_2\}[/tex]. Then we have that [tex]x_{n_1}<x_n<2L-x_{n_1}[/tex] for all [tex]n\geq n_0[/tex] in particular, [tex]x_{n_1}<x_{n_0}[/tex]. But since [tex]n_1<n_0[/tex] and the sequence is decreasing we have that [tex]x_{n_0}<x_{n_1}[/tex] which gives us a contradiction. Hence [tex]x_n\geq L[/tex] for all [tex]n\in\mathbb N[/tex].
 
Physics news on Phys.org
  • #2
Hi evagelos! :smile:

Let's start with

evagelos said:
proof:
Suppose that there exists [tex]n_1\in\mathbb N[/tex] such that [tex]x_{n_1}<L[/tex]. Take [tex]\epsilon=L-x_{n_1}[/tex].

What things do you need for this?
 
  • #3
micromass said:
Hi evagelos! :smile:

Let's start with



What things do you need for this?

No axioms no theorems ,only the definition definition:

[tex]\forall\epsilon[ \epsilon>0\Longrightarrow\exists k(k\in N\wedge\forall n(n\geq k\Longrightarrow |x_{n}-L|<\epsilon))][/tex].

The question now is,what laws of logic
 
  • #4
evagelos said:
No axioms no theorems ,only the definition definition:

[tex]\forall\epsilon[ \epsilon>0\Longrightarrow\exists k(k\in N\wedge\forall n(n\geq k\Longrightarrow |x_{n}-L|<\epsilon))][/tex].

We didn't use convergence yet. So that's not the definition we need.
We do use some definitions and axioms here. For example, you use [itex]\mathbb{N}[/itex], you need to define this in some way and there are quite a lot of axioms that you need to define the natural numbers.

Furthermore, you use the inequality relation < and the difference operation -. In order for these to be well-defined, you had to use some axioms and definitions.

And what about [itex]x_{n_1}[/itex] this is an element of a sequence, that is you have a function [itex]x:\mathbb{N}\rightarrow \mathbb{R}[/itex], but what is a function? You'll need axioms and definitions to precisely define functions. Furthermore, in this function, the codomain is [itex]\mathbb{R}[/itex], and this set needs some axioms and definitions too.

The question now is,what laws of logic

Since we did not do any logical derivations yet, we didn't use much logical laws yet! :smile:
 
  • #5
micromass said:
We didn't use convergence yet. So that's not the definition we need.
We do use some definitions and axioms here. For example, you use [itex]\mathbb{Nitex], you need to define this in some way and there are quite a lot of axioms that you need to define the natural numbers.

Furthermore, you use the inequality relation < and the difference operation -. In order for these to be well-defined, you had to use some axioms and definitions.

And what about [itex]x_{n_1}[/itex] this is an element of a sequence, that is you have a function [itex]x:\mathbb{N}\rightarrow \mathbb{R}[/itex], but what is a function? You'll need axioms and definitions to precisely define functions. Furthermore, in this function, the codomain is [itex]\mathbb{R}[/itex], and this set needs some axioms and definitions too.



Since we did not do any logical derivations yet, we didn't use much logical laws yet! :smile:

O,k then you state the axioms and definitions involved here ,because i am lost.

But surely the application of the laws of logic uppon those axioms and definitions should give us the statement of the proof
 
  • #6
Well, how did we define the natural numbers? And what axioms do you need for that?
An excellent reference for this is "staff.science.uva.nl/~vervoort/AST/ast.ps"[/URL] but you'll need to be able to open .ps files...
 
Last edited by a moderator:
  • #7
micromass said:
Well, how did we define the natural numbers? And what axioms do you need for that?
An excellent reference for this is "staff.science.uva.nl/~vervoort/AST/ast.ps"[/URL] but you'll need to be able to open .ps files...[/QUOTE]

Micromass like that we are getting nowhere.

You mean every time we mention the Natural Nos we have to write down their axioms ??

What for ?
 
Last edited by a moderator:
  • #8
evagelos said:
Micromass like that we are getting nowhere.

You mean every time we mention the Natural Nos we have to write down their axioms ??

What for ?

Judging from your post history, you like it that way. If you don't want it that formal, then you'll have to post from what axioms and definitions you're starting. If you want to take the natural numbers for granted, fine. But then you'll have to say what you're taking for granted and what not...
 
  • #9
micromass said:
Judging from your post history, you like it that way. If you don't want it that formal, then you'll have to post from what axioms and definitions you're starting. If you want to take the natural numbers for granted, fine. But then you'll have to say what you're taking for granted and what not...


O.k

Do you agree that :[tex]\exists n_{1}[n_{1}\in N\wedge x_{n_{1}}< L][/tex] is the negation of the statement:

[tex]\forall n[n\in N\Longrightarrow x_{n}\geq L][/tex] ??
 
  • #11
micromass said:
Yes, of course.

Hence so far we have no axioms or thoerem or definitions involved.

Do you agree??
 
  • #12
Not exactly! The converse of [itex]x_n< L[/itex] is not always [itex]x_n\geq L[/itex]. We have used here that the order relation is total. I.e. that for every a and b, we have

[tex]a\leq b~\text{or}~b\leq a[/tex]

And of course we used axioms and definitions to define [itex]\mathbb{N}[/itex] and to define the order-relation. But I understand that you take these for granted?
 
  • #13
micromass said:
Not exactly! The converse of [itex]x_n< L[/itex] is not always [itex]x_n\geq L[/itex]. We have used here that the order relation is total. I.e. that for every a and b, we have

[tex]a\leq b~\text{or}~b\leq a[/tex]

And of course we used axioms and definitions to define [itex]\mathbb{N}[/itex] and to define the order-relation. But I understand that you take these for granted?


Sorry, i should have mentioned the law of trichotomy (a>b or a=b or a<b),which is the only axiom involved in the proof that:

[tex]\neg\forall n[n\in N\Longrightarrow x_{n}\geq L][/tex] [tex]\Longrightarrow[/tex][tex]\exists n_{1}[n_{1}\in N\wedge x_{n_{1}}< L][/tex]

the rest is logic
 
  • #14
Maybe you can give a list (or reference) to all the axioms you're using? To me, trichotomy isn't really an axiom...
 
  • #15
micromass said:
Maybe you can give a list (or reference) to all the axioms you're using? To me, trichotomy isn't really an axiom...

It is not important whether we consider the trichotomy law is an axiom or a theorem .

It is important that we are using it in proving:



[tex]\neg\forall n[n\in N\Longrightarrow x_{n}\geq L][/tex] [tex]\Longrightarrow[/tex][tex]\exists n_{1}[n_{1}\in N\wedge x_{n_{1}}< L][/tex]

Unless you can produce a proof of the above where you can use different theorems ,axioms e.t.c
 
  • #16
evagelos said:
It is not important whether we consider the trichotomy law is an axiom or a theorem .

It is important in the sense that I need to know where to begin. What axioms can you assume, what theorems can you assume, that's what I need to know first before beginning this question...
 
  • #17
How are we defining "x is decreasing?" (I'm sorry, I don't know how to write LaTex)

Say we define a decreasing function
Xn
such that
n1>n0 iff xn1 < xn0.

So then
~ n > infinity => ~ xn < xinfinity
xinfinity = L,
so it seems it would follow that xn > L.

I guess we need to assume infinity (or maybe that for all natural numbers N, there is a successor of N which is greater), the concept of a limit and a definition of a decreasing function.

And of course predicate logic for the substitution that occurs to move from the definition of the decreasing function to substituting infinity for n0.
 
  • #18
You can't do that. Infinity is not a part of the natural numbers, so you can't talk about infinity and things like xinfinity...
 

1. What is the Proof of Convergence Theorem for Decreasing Sequences?

The Proof of Convergence Theorem for Decreasing Sequences is a mathematical theorem that states that if a sequence is monotonically decreasing and bounded below, then it will converge to its greatest lower bound. This theorem is used in the field of real analysis to prove the convergence of certain sequences.

2. What does it mean for a sequence to be monotonically decreasing?

A sequence is considered monotonically decreasing if each term in the sequence is less than or equal to the previous term. In other words, the sequence is always decreasing as you move from term to term.

3. How is the Proof of Convergence Theorem for Decreasing Sequences used in real analysis?

In real analysis, the Proof of Convergence Theorem for Decreasing Sequences is used to prove the convergence of certain sequences. This is important because it allows us to determine if a sequence will approach a specific value or if it will diverge.

4. What are some examples of monotonically decreasing sequences?

Some examples of monotonically decreasing sequences are the sequence of negative numbers, the sequence of fractions with decreasing denominators, and the sequence of powers of a number less than 1.

5. Can the Proof of Convergence Theorem for Decreasing Sequences be applied to all decreasing sequences?

No, the Proof of Convergence Theorem for Decreasing Sequences can only be applied to sequences that are monotonically decreasing and bounded below. If a sequence is not monotonically decreasing or is not bounded below, this theorem cannot be used to prove its convergence.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
455
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
1K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
3
Views
1K
Replies
14
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
5
Views
2K
  • Calculus
Replies
13
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
2
Views
1K
Back
Top