Proof analysis

  • Thread starter evagelos
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  • #1
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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] wich gives us a contradiction. Hence [tex]x_n\geq L[/tex] for all [tex]n\in\mathbb N[/tex].
 

Answers and Replies

  • #2
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Hi evagelos! :smile:

Let's start with

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
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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
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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
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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
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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...
 
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  • #7
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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 ?
 
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  • #8
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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
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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] ??
 
  • #10
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Yes, of course.
 
  • #11
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Yes, of course.

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

Do you agree??
 
  • #12
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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
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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
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Maybe you can give a list (or reference) to all the axioms you're using? To me, trichotomy isn't really an axiom...
 
  • #15
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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
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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
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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
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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...
 

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