Sme question about irrational numbers

[matt grime]

Tann, do you even know what a cardinal number is? I don't think you do. And I'm writing as one of the professionals you came here to learn from (your words). You have not shown the slightest inclination to LEARN anything. You have not learned when two sets have the same cardinality, not learned that connectedness and cardinalilty are totally unrelated, and even in your own posts said "[your teacher says] it has the power [cardinality] of the continuum" followed by "but it isn't a continuum" well, look at the first phrase. Does it say it *is* the continuum? Do you think the continuum is uniqely determined merely by its cardinality? If you knew what the *definition* of cardinality is you would see that that is completely and trivially false.

 

[jcsd]

What I say is this.

As much as I know, we have no way to use a 1-1 mapping between any two sets which are uncountable.

We can make an extension in our mind from N to R but (as much as I know) there is no proof that a 1-1 mapping holds in R as it holds in N or Q.

If I am wrong then please show the proof that clearly shows that the 1-1 mapping holds also in uncountable sets.

Well I've already shown you a one-to-one function between two uncountable sets, but here's a simple one:

the set A which is set the set of all real numbers between 1 and o is uncountable and the set B which is the ste of all rela numbers between 1 and 2 is uncountable and f:A-->B f(x) = x+1 is a bijection between them. Or even smple f:R-->R f(x) = x is a bijection.

 

[matt grime]

What on Earth does it even mean for "a mapping to hold"?

Are you sure you're not actually Doron Shadmi? No, you're more coherent than him.


Oh, these "points of view" we're forcing on you are either definitions, or are trivial consequences of the definitions.

 

[Tann]

Tann, do you even know what a cardinal number is? I don't think you do. And I'm writing as one of the professionals you came here to learn from (your words). You have not shown the slightest inclination to LEARN anything. You have not learned when two sets have the same cardinality, not learned that connectedness and cardinalilty are totally unrelated, and even in your own posts said "[your teacher says] it has the power [cardinality] of the continuum" followed by "but it isn't a continuum" well, look at the first phrase. Does it say it *is* the continuum? Do you think the continuum is uniqely determined merely by its cardinality? If you knew what the *definition* of cardinality is you would see that that is completely and trivially false.

Well a cardinal is a size of a set and it can be defined by the 1-1 mapping technique.

Please show that this technique holds also in uncountable sets.

 

[matt grime]

Define "holds". And cardinals are defined by bijections, not , usually, injections. SO, what is a cardinal number then? Approximately.

And, to use your own definition. What has connectedness to do with size?

 

[Tann]

f:R-->R f(x) = x is a bijection.

jcsd, Hurkyl, and matt grime.

I really want to thank you for the time that you spend with me, and sorry again if I was rude in some of my previous posts.

What you wrote jcsd is symbols that express your beliefs (an extension by analogy from N to R) , but I do not see a rigorous proof that clearly show that a 1-1 mapping technique really holds in uncountable sets as it holds in countable sets.

A group of people can agree with this belief but it is not a proof that this is really the case.

If I am wrong then please once for all show a clear proof that a 1-1 mapping technique really holds in uncountable sets.

 

[matt grime]

So, i'll take that as "no I'm not going to tell you what i mean when i say 'holds'"

What is a bijection? can you prove things are and are not bijections? are you going to learn any of these things like you said you would?

Let S be any set, define f a function from S to S by f(s)=s for all s in S.

Does it "hold" that this is a bijection?


Who knows, because yo've nnot deigned to tell us what "holds" means. Until you do you aren'y doing maths, you're doing something weird and expressing your presonality and not mathematics.

and i don't think you want to start on what ir means for things to be true since you don't even appear to want to acknowledge or learn the definitions.


maths really isn't about your opinion. all we have said is self evident from the definitions of those things involved. until such time as you can clearly and mathematically express something we cannot help you.

here's a simple exercise: show that N and Z, the naturals and the integers have the same cardinality. can you do that?

 

[Tann]

'Holds' is 'true'

all we have said is self evident from the definitions of those things involved

And the definitions are based on your beliefs, which you call axioms.

If you say that what is true in N and Q is also true in R and you do not give a proof for this, then all we have is your belief but not your rigorous proof.

And, to use your own definition. What has connectedness to do with size?

Matt, since you are a professional please give us a short explanation that clearly shows why there is no connection between size and connectedness?

 

[Tann]

Meanwhile I wish to ask this question.

Let us say that we define a 1-1 mapping between R and iR where iR is the set of all irrational numbers.

If R has the power of the continuum, then how can we move from a one R member to another, in order to define the next 1-1 mapping and on the same time not to miss any of the members that maybe can be found between any R pairs?

Is there a rigorous proof which clearly gives an answer to this question?

 

[Hurkyl]

I offer it up as an exercise to find an explicit bijection between R and the set of all nonzero reals. As a hint, consider using the bijection between the positive integers and nonnegative integers.

 

[Tann]

I offer it up as an exercise to find an explicit bijection between R and the set of all nonzero reals. As a hint, consider using the bijection between the positive integers and nonnegative integers.

Hurkyl, I do not no how to do that.

Can you please give your answer to post #39?

 

[hello3719]

Sorry to interrupt, but what is a "cardinal number" other than the bogus definition which refers to : "Set A has the same cardinal number as set B if they can be put into 1to1 correpondance with each others"

In every analysis book I've looked up this is the only definition which popups.
You got to admit that it isn't a really rigourous definition since it depends on the existence of "another" set.

 

[Tann]

Sorry to interrupt, but what is a "cardinal number" other than the bogus definition which refers to : "Set A has the same cardinal number as set B if they can be put into 1to1 correpondance with each others"

In every analysis book I've looked up this is the only definition which popups.
You got to admit that it isn't a really rigourous definition since it depends on the existence of "another" set.

Not exactly, the 1-1 mapping can be from a set onto itself (is this the reason that you wrote "another" instead of another. In this case please explain your point of view, thank's).

But please read post #39 and try to give a rigorous answer to it.

 

[Hurkyl]

There are lots of clever ways of dealing with how to define the word "cardinal number".

One useful technique that applies to a great many circumstances is this: you can do most problems within a particular universe U... then we can define an equivalence relation on U saying that two sets are equivalent if they're bijective. (There's a bijection between them) Then, a cardinal number is an equivalence class on U.


However, one doesn't have to resort to such tricks. When using the axiom of choice, cardinal numbers have the "simple" description:

C is a cardinal number if and only if ( C is an ordinal number and there does not exist a bijection between C and any smaller ordinal number ).

So, the cardinality of a set S is simply the smallest ordinal number C that is bijective with S. (There is at least one ordinal number, by the well-ordering principle)


When you don't use the axiom of choice, there is still a way using the "birthday" of a set; the cardinality of a set S is defined to be the one amongst all sets bijective with S that was born first.

 

[hello3719]

yes it can be mapped on itself, but what is the use to do that, prior to the definition of a cardinal number.

il let hurkyl prove you post#39, since I am not a loyal supporter of our "new"(post-cantorian) version of set theory.

when i get the definitions straight i might get into those things.

 

[Tann]

Hurkyl,

Can you please reply to post #39? thank's.

 

[Hurkyl]

If R has the power of the continuum, then how can we move from a one R member to another, in order to define the next 1-1 mapping and on the same time not to miss any of the members that maybe can be found between any R pairs?

This is gibberish. Now aren't you glad you insisted on a response?

 

[Moo Of Doom]

He wants a defined 1 to 1 correspondence between R and F, such that every member of F is paired with a member of R and vice versa.

I myself have no idea how to do this. I only know F defined as all members of R that are not members of Q.

jcsd's post #24 is definitely a 100% solid way to show F has the same cardinality as R, but it isn't a complete bijection. Many members of F are left out.

 

[Hurkyl]

Well, I think the others might be curious, so here's a recipe to constructing a bijection.


First recall that there's a bijection between {0, 1, 2, ...} and {1, 2, 3, ...} simply by adding one to each number. In this way, we've removed an element from {0, 1, 2, 3, ...}. The actual arithmetic doesn't matter -- think of it more in terms of a successor function: we pushed each object over to the next one.


Using your imagination, you should see how to use this to "remove" a single point from any infinite set. (by "remove", I mean to find a bijection between the set and the set with one point removed)

So, to remove countably many points, you just have to find countably many disjoint chains, and you can push along all of them simultaneously.


I'm going to use this:

Let φ map any number of the form: q + n \sqrt{2} to q + (n+1) \sqrt{2}

any leave all other numbers unchanged.

Here, q is any rational number, and n is any nonnegative integer.

φ, if well-defined, is clearly a surjective map from the reals onto the irrationals, so all that's left to do is show that it's injective.

To show it's well-defined, and that it's injective, all boils down to proving that q + n \sqrt{2} = r + m \sqrt{2} implies q = r and n = m.

 

[hello3719]

I guess you won't be able to determine a specific bijection between the irrationnal and real numbers.
Maybe because the irrationnal numbers are uniquely defined as numbers which are "not rationnal numbers", only defined as what it is not.

 

[Hurkyl]

Anything wrong with the φ I gave in my post?

 

[hello3719]

ok ill be more clear, Could you find a way to map EVERY irrationnal number to some real number or the other way around.

Sorry but hadn't took any analysis courses yet, so I don't know the terminology well.

 

[Moo Of Doom]

Could you find a way to map EVERY irrationnal number to some real number or the other way around.

I believe Hurkyl's method was quite successful in doing that.

 

[Hurkyl]

I repeat, what's wrong with φ?

 

[hello3719]

How can we be sure that every irrationnal number is of the form q + (n+1) \sqrt{2}

 

[Hurkyl]

They're not. That's why I added a clause specifying what happens to things not of that form: they're left unchanged.

 

[hello3719]

so those which are not changed are not "associated" to some real number ?

sorry maybe I am skipping something obvious

 

[Hurkyl]

If x is not of the form q + m √2, then φ(x) = x.

 

[Moo Of Doom]

Those that are not changed are associated with themselves. Irrational numbers are real numbers. So something like \pi, which is not in the form q+n\sqrt{2}, is mapped to \pi.

All Hurkyl is doing is eliminating rational numbers by mapping them to irrational numbers, and then mapping those irrationals which they are mapped to to other irrationals, etc.

 

[hello3719]

tx

ok, so they are are mapped to themselves
sorry my logical flaw was that i wanted inconsciously a mapping of every irrationnal number to a general number not described in function of itself