
#1
Jun611, 07:42 AM

P: 308

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=Lx_{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_nL\<\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<2Lx_{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]. 



#2
Jun611, 08:18 AM

Mentor
P: 16,545

Hi evagelos!
Let's start with 



#3
Jun611, 09:25 AM

P: 308

[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
Jun611, 09:47 AM

Mentor
P: 16,545

proof analysisWe 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 welldefined, 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. 



#5
Jun611, 10:23 AM

P: 308

But surely the application of the laws of logic uppon those axioms and definitions should give us the statement of the proof 



#6
Jun611, 10:59 AM

Mentor
P: 16,545

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 but you'll need to be able to open .ps files... 



#7
Jun611, 11:26 AM

P: 308

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



#8
Jun611, 11:38 AM

Mentor
P: 16,545





#9
Jun711, 10:08 AM

P: 308

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
Jun711, 02:59 PM

P: 308

Do you agree?? 



#12
Jun711, 03:02 PM

Mentor
P: 16,545

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 orderrelation. But I understand that you take these for granted? 



#13
Jun711, 05:45 PM

P: 308

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
Jun711, 05:48 PM

Mentor
P: 16,545

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




#15
Jun711, 10:13 PM

P: 308

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
Jun711, 10:22 PM

Mentor
P: 16,545





#17
Jun811, 11:28 PM

P: 2

How are we defining "x is decreasing?" (I'm sorry, I don't know how to write LaTex)
Say we define a decreasing function X_{n} such that n_{1}>n_{0} iff x_{n1} < x_{n0}. So then ~ n > infinity => ~ x_{n} < x_{infinity} x_{infinity} = L, so it seems it would follow that x_{n} > 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 n_{0}. 


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