Proof of Convergence Theorem for Decreasing Sequences with Logic Laws

[evagelos]

Suppose we are asked to prove :

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

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 n_1\in\mathbb N such that x_{n_1}<L. Take \epsilon=L-x_{n_1}. Since \displaystyle\lim_{n\rightarrow\infty}x_n=L there exists n_2\in\mathbb N such that \|x_n-L\|<\epsilon for all n\geq n_2. Take n_0>\max\{n_1,n_2\}. Then we have that x_{n_1}<x_n<2L-x_{n_1} for all n\geq n_0 in particular, x_{n_1}<x_{n_0}. But since n_1<n_0 and the sequence is decreasing we have that x_{n_0}<x_{n_1} which gives us a contradiction. Hence x_n\geq L for all n\in\mathbb N.

 

[micromass]

Hi evagelos!

Let's start with

proof:
Suppose that there exists n_1\in\mathbb N such that x_{n_1}<L. Take \epsilon=L-x_{n_1}.

What things do you need for this?

 

[evagelos]

Hi evagelos!

Let's start with



What things do you need for this?

No axioms no theorems ,only the definition definition:

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

The question now is,what laws of logic

 

[micromass]

No axioms no theorems ,only the definition definition:

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

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 \mathbb{N}, 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 x_{n_1} this is an element of a sequence, that is you have a function x:\mathbb{N}\rightarrow \mathbb{R}, but what is a function? You'll need axioms and definitions to precisely define functions. Furthermore, in this function, the codomain is \mathbb{R}, 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!

 

[evagelos]

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 \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.<br /> <br /> Furthermore, you use the inequality relation &lt; and the difference operation -. In order for these to be well-defined, you had to use some axioms and definitions.<br /> <br /> And what about x_{n_1} this is an element of a sequence, that is you have a function x:\mathbb{N}\rightarrow \mathbb{R}, but what is a function? You&#039;ll need axioms and definitions to precisely define functions. Furthermore, in this function, the codomain is \mathbb{R}, and this set needs some axioms and definitions too.<br /> <br /> <br /> <br /> Since we did not do any logical derivations yet, we didn&#039;t use much logical laws yet! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

<br /> <br /> O,k then you state the axioms and definitions involved here ,because i am lost.<br /> <br /> But surely the application of the laws of logic uppon those axioms and definitions should give us the statement of the proof

 

[micromass]

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...

 

[evagelos]

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 ?

 

[micromass]

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...

 

[evagelos]

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 :\exists n_{1}[n_{1}\in N\wedge x_{n_{1}}&lt; L] is the negation of the statement:

\forall n[n\in N\Longrightarrow x_{n}\geq L] ??

 

[micromass]

Yes, of course.

 

[evagelos]

Yes, of course.

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

Do you agree??

 

[micromass]

Not exactly! The converse of x_n&lt; L is not always x_n\geq L. We have used here that the order relation is total. I.e. that for every a and b, we have

a\leq b~\text{or}~b\leq a

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

 

[evagelos]

Not exactly! The converse of x_n&lt; L is not always x_n\geq L. We have used here that the order relation is total. I.e. that for every a and b, we have

a\leq b~\text{or}~b\leq a

And of course we used axioms and definitions to define \mathbb{N} 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:

\neg\forall n[n\in N\Longrightarrow x_{n}\geq L] \Longrightarrow\exists n_{1}[n_{1}\in N\wedge x_{n_{1}}&lt; L]

the rest is logic

 

[micromass]

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

 

[evagelos]

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:



\neg\forall n[n\in N\Longrightarrow x_{n}\geq L] \Longrightarrow\exists n_{1}[n_{1}\in N\wedge x_{n_{1}}&lt; L]

Unless you can produce a proof of the above where you can use different theorems ,axioms e.t.c

 

[micromass]

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...

 

[asalmon14]

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
n₁>n₀ 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 n₀.

 

[micromass]

You can't do that. Infinity is not a part of the natural numbers, so you can't talk about infinity and things like x_infinity...