Jump to content

Continuity and uncountability


pengkuan

Recommended Posts

26 minutes ago, taeto said:

Not really with anything.

If you say "Assume a set S exists such that 2+2=5", then you reach a contradiction. In which case you just resolve that no such set exists, and you deduce that 2+2 is not 5.  

If you say "Assume a set S exists such that P. Then Q", where P is an undecidable proposition, and you prove your statement, then you have proved an implication: if S exists, then Q is true. This happens very often in algebra, which has lots of undecidable statements, such as P="G is an abelian group" and the likes. Then by showing "P => Q: x and y are in G implies x+y=y+x" you have shown that Q is true given a particular circumstance P.

 

I was really interested in how your diphantine example was a counter example.

1 hour ago, taeto said:

There are some cases, even in ordinary arithmetic, when it can happen that there is no conflict at all with rules or axioms to prevent the existence of the humperdink.

I am distinguishing between rules and axioms and properties.

Don't forget I specified that

2 hours ago, studiot said:

You can define a humperdink, along with its properties, to be wanything you want, so long as there is no conflict.

As you know I can't define a humperdink as a member of the set of Diophantine equations with no solutions.

Link to comment
Share on other sites

20 minutes ago, studiot said:

As you know I can't define a humperdink as a member of the set of Diophantine equations with no solutions.

You are given a Diophantine equation.

A humperdink is defined as a solution to the equation.

You are told that nothing in the rules or axioms prevents that a humperdink exists.

You are told that nothing in the rules or axioms prevents that a humperdink does not exist.

Both are correct.

At this point you cannot assert that a humperdink exists, nor can you assert that none does, based on rules and axioms alone.

Link to comment
Share on other sites

1 hour ago, taeto said:

You are given a Diophantine equation.

A humperdink is defined as a solution to the equation.

You are told that nothing in the rules or axioms prevents that a humperdink exists.

You are told that nothing in the rules or axioms prevents that a humperdink does not exist.

Yes indeed all true.

But the next one is a difficulty.

Suppose I wanted the solution to the equation


[math]\left( {{x^2} - 2 = 0:x \in Q} \right)[/math]

This does not exist, because a solution to this equation is not in Q.

However I could assert that there exists a solution called a humperdink whcih is in the extended rationals, and so long as I didn't contravene any of its properties I could work with my humperdink.

In fact there is a solution in the extended rational field


[math]\left( {r:r = a + b\sqrt {2:} a,b \in Q} \right)[/math]

 

 

 

Link to comment
Share on other sites

1 hour ago, studiot said:

Suppose I wanted the solution to the equation


(x22=0:xQ)

This does not exist, because a solution to this equation is not in Q.

However I could assert that there exists a solution called a humperdink whcih is in the extended rationals, and so long as I didn't contravene any of its properties I could work with my humperdink.

In fact there is a solution in the extended rational field


(r:r=a+b2:a,bQ

 

Your scenario is very different. The question that you ask has to be a valid statement within a particular theory. The variables that you use are constrained to be objects of that theory. If you are working in Peano arithmetic, they have to represent integers. In rational arithmetic they represent rational values, etc. It does not make sense to answer a question about the squareroot of two in rational arithmetic by referring to  the theory of the real ordered field. Exactly in the same way that it makes no sense to answer a question about dividing 1 by 2 within the ring of integers by invoking rationals. 

In fact, you would be given the information that no solution exists, since there is no rational number the square of which is equal to 2. This is correct, and whichever counterassertion you make will be flawed. 

The scenario that I am referring to is one in which you have a clear cut question about an  equation involving integer constants and variables. You are restricted to attempt a solution solely within the realm of the integers. You are given two pieces of information: nothing prevents the existence of a solution, and nothing prevents that there is no solution. In particular, there is no proof of the existence of one, nor a disproof. Whether there is a "solution" outside the given restrictions of having to assign integer values to the variables has no relevance.

Edited by taeto
Link to comment
Share on other sites

4 hours ago, studiot said:

Inconsistent with.

 

Do all the other axioms you are referring to, taken together, preclude the existance of an infinite set?

 

I disagree with your use of the word model.

Models follow they don't precede.

If an assumption would otherwise be inconsistent with the other axioms, making it an axiom leads to an inconsistent system. Your concept of an axiom seems flawed to me.

The other axioms neither imply nor exclude the existence of an infinite set; the axiom of infinity is independent of the other axioms (assuming the usual assumptions of consistency of ZFC).

I am using the word "model" in the precise sense of model theory. The class of (hereditarily) finite sets forms a model for ZFC minus infinity. I don't understand what you mean by "precede" here; I am saying that that class follows the rules of the other axioms.

Link to comment
Share on other sites

1 hour ago, taeto said:

and nothing prevents that there is no solution

I believe I said there was a difficulty with this one, but forgot to say what thet difficulty was.

 

The double negative means that if there is no solution, something prevents there being one.

This means there must be a contradiction with that something.

 

Yes my example was different from yours since there is an integer solution to your example diophantine equation (x=-3, y = -1) as I said.

So displaying a solution prevents there not being one.

 

As a matter of interest my view of number systems is that they arise with the expansion of the search for solutions to ever more complicated equations.

Usually the old number system is a subset of the new one, and I have only appended one additional number to the rationals to form my new one.

 

1 hour ago, uncool said:

If an assumption would otherwise be inconsistent with the other axioms, making it an axiom leads to an inconsistent system. Your concept of an axiom seems flawed to me.

The other axioms neither imply nor exclude the existence of an infinite set; the axiom of infinity is independent of the other axioms (assuming the usual assumptions of consistency of ZFC).

I am using the word "model" in the precise sense of model theory. The class of (hereditarily) finite sets forms a model for ZFC minus infinity. I don't understand what you mean by "precede" here; I am saying that that class follows the rules of the other axioms.

"If an assumption would otherwise be inconsistent with the other axioms, making it an axiom leads to an inconsistent system."
That is what I believe I said, yes.

In what way is my concept of an axiom flawed?

 

A model is a copy of some part of the original.

A blueprint to work from (which I think is your meaning) is better replaced with another word - many have been used - pattern - framework - foundation.

This is, of course, my opinion and you are entitled to yours, but if this is another example of a professional in one corner  taking a word with an already well documented meaning by its originating professionals, and giving it another almost opposite one, then more confusion will surely arise.

 

I am leaving for the coast, early tomorrow morning so I don't think I will have much input for a week or so.

Edited by studiot
Link to comment
Share on other sites

12 minutes ago, studiot said:

In what way is my concept of an axiom flawed?

 

I may be severely misreading you; it seemed to me that you were implying that for something to be an axiom, it would have to contradict/be inconsistent with what came before it. Which is incoherent. If that's not what you meant, then I don't understand why you asked about whether the assumption of infinity would be inconsistent.

12 minutes ago, studiot said:

A model is a copy of some part of the original.

Not in model theory.

In model theory, approximately (I am not a model theorist specifically), there are theories and models. Theories are systems of axioms; models of a theory are things that satisfy those axioms. When talking about a theory, you can talk about statements of that theory; they may be categorically true, categorically false, or neither. In a model, every syntactically valid statement is either true or false. For example, there is the theory of groups, where the axioms are associativity, identity, and inverse(s). A model of that theory is S3, the symmetric group on 3 letters; another is C3, the 3-element cyclic group. Commutativity is a statement; it is true for C3 but not for S3. 

The class of hereditarily finite sets is a model of ZF(C) - infinity; this implies that infinity cannot be derived from the other axioms.

Edited by uncool
Link to comment
Share on other sites

5 minutes ago, uncool said:

Not in model theory.

In model theory, approximately (I am not a model theorist specifically), there are theories and models. Theories are systems of axioms; models of a theory are things that satisfy those axioms. When talking about a theory, you can talk about statements of that theory; they may be categorically true, categorically false, or neither. In a model, every syntactically valid statement is either true or false. For example, there is the theory of groups, where the axioms are associativity, identity, and inverse(s). A model of that theory is S3, the symmetric group on 3 letters; another is C3, the 3-element cyclic group. Commutativity is a statement; it is true for C3 but not for S3. 

The class of hereditarily finite sets is a model of ZF(C) - infinity; this implies that infinity cannot be derived from the other axioms.

As I thought.

What axioms do my great great great grandfather's model "Cutty Sark" satisfy?

When they built the real Cutty Sark they worked from a pattern on the lofting floor which controlled the shape so the real ship could be said to be the model and the pattern the axioms by your definition.

8 minutes ago, uncool said:

I may be severely misreading you; it seemed to me that you were implying that for something to be an axiom, it would have to contradict/be inconsistent with what came before it. Which is incoherent.

I can't see where this might be but I will look back at what I did say and if I misspoke then I apologise and thank you for pointing it out.

Edited by studiot
Link to comment
Share on other sites

8 hours ago, studiot said:

Look again at my explanation of

"There exists."

 

It is not necessary to have a rule which states explicitly there exists and humperdink, if the existance of a humperdink does not conflict with any establish rules or axioms.

You can define a humperdink, along with its properties, to be wanything you want, so long as there is no conflict.

 

You can then correctly assert "there exists a humperdink"

 

Your problem comes when someone comes along and points out that there is, in fact, a conflict as I did with my simple example of a non negative D.

 

My compliments to Strange for some incisive thinking. +1

 

Thanks.

Link to comment
Share on other sites

Ah - studiot, I think I see how I misunderstood what you were asking.

No, the axiom of infinity does not conflict with the other axioms (unless you believe the vanishingly few finitist mathematicians, who usually say they think it can lead to a contradiction but haven't worked it out yet). As I implied, it is a standard part of ZFC set theory - if it led to a known contradictions, set theorists would be up in arms. 

So yes, the existence of infinite sets is an axiom, but it is one pretty universally accepted - and one which you, pengkuan, accepted by allowing the set of natural numbers.

Link to comment
Share on other sites

On 24/10/2018 at 10:59 PM, studiot said:

I believe I said there was a difficulty with this one, but forgot to say what thet difficulty was.

 

The double negative means that if there is no solution, something prevents there being one.

This means there must be a contradiction with that something.

Not really. Uncool gave a nice example in group theory. The statement "G is not abelian" can be formulated in group theory. If G is not abelian, then there are elements \(x,y\) in G for which \(xy \neq yx.\) It is not a theorem in group theory that G is abelian, nor is the negation a theorem. It can happen for a group G that no such \(x,y\) exist, but there is nothing in the rules or axioms which suffices to deduce that this is the case. 

Peano Arithmetic (PA) is much similar. To say that a given Diophantine equation of a particular shape has a solution can be stated in the language of PA, but there is nothing in the rules or axioms to say whether it is true or false. If you argue that if there is nothing in the rules or axioms to prevent it, then it must be true, then you are wrong. If there is something to prevent it, then whatever it is, it is not in the rules or axioms.

True Arithmetic (TA) is different, see https://en.wikipedia.org/wiki/True_arithmetic. Any correctly formulated statement has a proof or a disproof from the rules and axioms. However, TA is mostly avoided for being too cumbersome. In particular it is impossible to formulate TA with finitely many axioms; you need an infinite collection. 

Link to comment
Share on other sites

On 25/10/2018 at 4:22 PM, taeto said:

However, TA is mostly avoided for being too cumbersome. In particular it is impossible to formulate TA with finitely many axioms; you need an infinite collection. 

Is that because of Godel's incompleteness theorem?

Link to comment
Share on other sites

1 hour ago, Strange said:

Is that because of Godel's incompleteness theorem?

Very good point indeed. Though it is more like the other way around. The arithmetical statement that Gödel constructs is really an encoding of the statement "I have no proof from the axioms of arithmetic", which is what creates the sort of dilemma that if it does have a proof then it is false, whereas if it has no proof then it is true, but it is lacking of proof. This does not work for an axiomatic system based on an infinite amount of axioms, because you cannot build a finite statement like that which encodes all the axioms at the same time. 

It turns out that TA is safe from that point of view, from attacks by Gödel type statements. Moreover it is provably consistent and complete. At the cost of a huge amount of axioms.

Link to comment
Share on other sites

  • 3 weeks later...

Graphic of set counting and infinite number


When counting a set, we can plot a graphic that represents the members of the set on the plane (x, y) to observe visually the counting. Also, graphic of counting of infinite set helps us to understand infinite natural number.
PDF Graphic of set counting and infinite number https://pengkuanonmaths.blogspot.com/2018/11/graphic-of-set-counting-and-infinite.html   
 or Word https://www.academia.edu/37766761/Graphic_of_set_counting_and_infinite_number    
 

Link to comment
Share on other sites

"K e is a natural numbers for all n, thus K e has finitely many digits when n increases endlessly."

That's not how any of this works.

 

You are still making the same error, after all these years. 

Edited by uncool
Link to comment
Share on other sites

7 hours ago, pengkuan said:

Can you tell me how any of this works?

I have no doubt that he can do just that.

I just want to remark that one thing which seems confusing about your idea is that a common definition of "finite" says that a set is finite if and only if the number of elements in the set is a natural number. How do you explain to a person who is aware of this definition how one may think that some natural numbers are infinite?

Link to comment
Share on other sites

Among other things:

1) The definition you are attempting to use would make every set finite. 

2) The specific method you are attempting would have to use the definition that a set is finite when there is a natural number corresponding to its size. That does not in any way match the definition you attempt to use to justify your claim there.

Link to comment
Share on other sites

15 hours ago, taeto said:

I just want to remark that one thing which seems confusing about your idea is that a common definition of "finite" says that a set is finite if and only if the number of elements in the set is a natural number. How do you explain to a person who is aware of this definition how one may think that some natural numbers are infinite?

Infinity is not well defined. Is it an object with a value or is it a ever increasing number that is bigger than all?

Does the set of natural number exist? If it exist, then, it contains infinite natural number. If infinite natural number does not exist, then the set of natural number does not exist because it does not contain infinity.

15 hours ago, uncool said:

Among other things:

1) The definition you are attempting to use would make every set finite. 

2) The specific method you are attempting would have to use the definition that a set is finite when there is a natural number corresponding to its size. That does not in any way match the definition you attempt to use to justify your claim there.

I'm not sure if I understand you. What you mean with "That does not in any way match the definition you attempt to use to justify your claim there"?

Link to comment
Share on other sites

8 hours ago, pengkuan said:

Infinity is not well defined.

I don't know of it being defined at all. Do you know what "well defined" means?

8 hours ago, pengkuan said:

Is it an object with a value

Usually it is defined by its properties. You could decide to assign a value to it, but what the value is would not matter.

8 hours ago, pengkuan said:

 is it a ever increasing number that is bigger than all?

That doesn't make any sense.

8 hours ago, pengkuan said:

Does the set of natural number exist? If it exist, then, it contains infinite natural number.

If you mean in set theory, then the set of natural numbers exists. If you do not mean in set theory, then you should not address the notion of a "set".

The set is infinite, and it does not contain as elements anything that is infinite. It contains only natural numbers, and they are all finite, by definition. 

8 hours ago, pengkuan said:

If infinite natural number does not exist, then the set of natural number does not exist because it does not contain infinity.

What do you mean by "infinity"?

Link to comment
Share on other sites

13 hours ago, uncool said:

How do you define a finite set?

A set that has an end.

5 hours ago, taeto said:
14 hours ago, pengkuan said:

Infinity is not well defined.

I don't know of it being defined at all. 

 

5 hours ago, Strange said:

It is defined 

The disagreement between you two shows that infinity is really confusing.

5 hours ago, taeto said:

The set is infinite, and it does not contain as elements anything that is infinite. It contains only natural numbers, and they are all finite, by definition. 

I agree. An infinite set "contains only natural numbers, and they are all finite, by definition. " It is infinite because it does not has end.

Link to comment
Share on other sites

18 minutes ago, pengkuan said:

The disagreement between you two shows that infinity is really confusing.

Maybe we mean different things by "[well] defined" :) 

I'm not sure if "infinity" has a technically, well defined meaning in mathematics (but I am not a mathematician).

Quote

I agree. An infinite set "contains only natural numbers, and they are all finite, by definition. " It is infinite because it does not has end.

And one type of infinity can be defined as the cardinality of that set.

The another infinity is the cardinality of the reals.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.