Anoter question on vector spaces an rationnal numbers

[seratend]

I have some questions concerning the rationnal numbers and the vector spaces.
Let's take the set of rational number Q with the usual addition and multiplication.

We can say that (Q,+,.) is a vector space on the Q field. Now, if we add the |x| absolute value, we define have vector space with a norm.

If we add to this space all the points of the cauchy convergent sequence, do we have a complete space (i.e. banach space)?

If yes, we can call this set Q* (may there is already an other name ;). Is (Q*,+,.) a field?

Seratend.

 

[matt grime]

If you complete Q in its ordinary euclidean norm you just get the reals. In fact, that's one of the definitions of the reals.

 

[seratend]

If you complete Q in its ordinary euclidean norm you just get the reals. In fact, that's one of the definitions of the reals.

Thanks a lot.
That is one of my questions and why I am trying to use this model /view to understand some aspects. I am trying to construct the completion of Q with countable series. Thus, how can I get the set of real numbers form this construction if |R is non countable?
I always have interpreted this result (non exsitence of a bijection between |N and |R) as the existence of real numbers that are not the limit of "rationnal" series (i.e. => Q* =/= |R).

May you tell me where I am wrong?

Seratend.

 

[matt grime]

Every real number has a decimal expansion, right? Well, the partial sums are rational, and cauchy and the limit is irrational, and arbitrary.

I've no idea where your interpretation came from, and, as it's not reasonable to me, I can't see how to explain why it's wrong other than to say "it is".

If I can offer a possible anology, why is the power set of N also uncountable, yet the set of all finite subsets countable?

 

[Hurkyl]

The nonexistance of a bijection only says a single sequence can't pass through every real number.

Try counting the number of rational sequences and find the cardinality of that set!

 

[seratend]

The nonexistance of a bijection only says a single sequence can't pass through every real number.

Try counting the number of rational sequences and find the cardinality of that set!

Thanks, for the answer. Now I am trying to think a little bit more.
Ok, I am not an expert of cardinality but I need to understand better the completion of Q, so I prefer to stay with bijection properties bewteen sets.

Therefore, Can we say, we have a bijection between |Nx|Nx ...|Nx ... (the countable infinite product sequence of integer sets) and |R? (I am trying to re-interpret you statement)

Thanks in advance,

Seratend.

 

[matt grime]

The completion of Q is R; it is the "best" definition of what R is.

The completion is an analytic statement, not a set theoretic one.

I don't know if the countable product of N's has cardinality c (card of R). and it isn't important - clearly it is at least as great as c.

 

[seratend]

I don't know if the countable product of N's has cardinality c (card of R). and it isn't important - clearly it is at least as great as c.

Well, I have always missed that point (I am not a specialist – I just want to understand the limit behaviour).
For any countable finite product of |N, call |N^p, we have a bijection between |N and |N^n (we take the odd and even sets of |N and by induction we can generalise to any finite n).

May you give me a demo (or a link, or tell me) why for n-->+oO, this bijection become false (for example by induction the remaining odd/even parts of |N to construct the set product bijection becomes empty)?

Thanks in advance. It helps me in viewing clearer my problem.

Seratend.

 

[matt grime]

Because clearly the infite product of N's has greater cardinality than the infinite product of the two element set {0,1}, which has the cardinality as the power set of N, ie is uncountable.

 

[seratend]

Because clearly the infite product of N's has greater cardinality than the infinite product of the two element set {0,1}, which has the cardinality as the power set of N, ie is uncountable.

Once again thanks, for these answers. However, I do not have spent to much time learning the cardinality of sets. Therefore, I prefer to stay with the application bijection notion between sets (rather than working with the cardinality) to demonstrate/understand better my intial problem.

I understand, that you are using a cardinality results (i think {0,1}^n, n-->+oO is uncountable) however, forgive me, but I would like to get the same result without using this "black box" cardinality and properties.

In fact I would like to understand better the transition between the countable to uncountable sets (based solely on the definition of the non existence of bijection).

For example, how do you proove that {0,1}^n, n--> +oO is uncountable (if I have correctly understood your statement).
However, I prefer a demonstration based on the case |Nx|Nx...x|Nx... as it is closer to my initial problem (understanding the completion limit by the cauchy sequences).

Seratend.

 

[houserichichi]

These may or may not be the methods you had in mind, specifically for the proof. Refer to

This PDF File
(Page 21): Theorem 2.5.1

http://web.umr.edu/~insall/Essays/Set%20Theory/27%20July%202003/Uncountability%20of%20the%20unit%20interval.txt
All about the uncountability of the unit interval

http://www.mtnmath.com/whatth/node29.html
This always seemed the easiest, but it's not the way you're looking for.

Hope at least one of these helps

 

[seratend]

These may or may not be the methods you had in mind, specifically for the proof. Refer to

This PDF File
(Page 21): Theorem 2.5.1

http://web.umr.edu/~insall/Essays/Set%20Theory/27%20July%202003/Uncountability%20of%20the%20unit%20interval.txt
All about the uncountability of the unit interval

http://www.mtnmath.com/whatth/node29.html
This always seemed the easiest, but it's not the way you're looking for.

Hope at least one of these helps

Thanks a lot for these links, especially the first one. However, I've quickly checked them and it is not exactly what I am looking for (anyway I will take more time to check them more carrefully).

In order to be understood better: I am not looking for the uncountability property of the |R set (for example using the decimal representation, cantor diagonal...), I just want to understand better the limit between the countability and uncountability generated from a countable set (i.e. {0,1}^n, n --> +oO or (|Nx|Nx...)_ntimes, n --> +oO).

Seratend

 

[matt grime]

Finding bijections is hard, and completely unnecessary if you ask me. If you insist on such things then I suggest you look up the schroeder bernstien theorem that states two sets X and Y have the same cardinality iff there are two injections one from X and one to Y.

Then {0,1}_{N} which we will use as the notation for the product of the set {0,1} a countably infinite number of times (and can i suggest you stop writing that limit thing?) has the same cardinslity as R is quiet trivial:

there is an obvious injection: send the string (x_i) of 0s and 1s to the real number

0.x_1x_2...
in base 10

the reverse injection is slightly harder to right out but isn't that tricky: every number in base 2 has some representation (disallowing strings ending in an infinite number of 1s)

...y_1y_0.x_1x_2...

which we may assume is infinite in both directions by addding zeroes before and after.

send the x_i to the even places in a string and the y_i to the odd places and yo'uve an injection the other way.


if you are goign to insist on always thinking in only one basic fashion and not broadening to better results you won't get very far.

 

[seratend]

Finding bijections is hard, and completely unnecessary if you ask me. If you insist on such things then I suggest you look up the schroeder bernstien theorem that states two sets X and Y have the same cardinality iff there are two injections one from X and one to Y.

Then {0,1}_{N} which we will use as the notation for the product of the set {0,1} a countably infinite number of times (and can i suggest you stop writing that limit thing?) has the same cardinslity as R is quiet trivial:

there is an obvious injection: send the string (x_i) of 0s and 1s to the real number

0.x_1x_2...
in base 10

the reverse injection is slightly harder to right out but isn't that tricky: every number in base 2 has some representation (disallowing strings ending in an infinite number of 1s)

...y_1y_0.x_1x_2...

which we may assume is infinite in both directions by addding zeroes before and after.

send the x_i to the even places in a string and the y_i to the odd places and yo'uve an injection the other way.


if you are goign to insist on always thinking in only one basic fashion and not broadening to better results you won't get very far.

Thanks for your answer, but, sorry, it is not what I am looking for : (.

Ok, I admit that my questions are not very clear and it is hard for me to formulate them clearly. However, I think if I could define my question clearly, I guess I would get myself the answer : ). Therefore, I make my excuses and thank you, once again, and all the other participants for the answers and I hope to continue to receive some help.

I think that your suggestion on using the existence of injections between the 2 concerned sets rather than defining a bijection sounds pretty good (i.e. we are using the tools of cardinality rather than the cardinality theorems I do not want to use): The schroeder bernstien theorem (I knew it with the name “cantor-bernstein”, I am always surprised with the different theorem names/equations we can encounter all around the world!)

Now, I insist I am trying to see the difference between the countable and uncountable property on a limit point of view (it is a kind of physical point of view if you prefer :) as it can help me to understand/proove better the logical consistence of some physical models. However, I want to keep the mathematic consistency and I understand that it is not the usual mathematical way to do that.
I accept your demonstrations based on the “number labelling”, but they do not show easily the connection at the “limit” (the behaviour I am trying to see or define if you prefer).

I also want to start with the properties of |N and Q that are countable and understand how we get (at the limit, see below) a non countable set.

In order to define a limit, we must have a topology. My hypothesis, that can be wrong, is based on the possibility to define, a priory, a larger set X that contains the limit set (I hope it is not circular and consistent) and to define the topology on it (for example the finest topology consisting of the elements of P(X) or a coarser topology if possible).

With that definition, I may define, if I am not wrong, the limit of product_ntimes {0,1}= {0,1}_{n} (using your notation where I replace the set |N by n) or the limit of product_ntimes {|N}={|N}_{n} as the element of the set X.
In the first case, for example, I may define a topology on the set X= union_n_element_of_|N {{0,1}_{n}} (with, may be, additional {} to be consistent). Thus, I may define logically a sequence of sets and a limit.
The advantage of this complication is only the ability “test”/”see” the evolution of an element of the sequence and the limit itself.
For example, the case of {|N}_{n} where I can define a simple bijection to |N for any n (using inductive odd/even number subsets). Therefore, I may understand better the limit, with the adequate topology, of this sequence. However, if I want to connect this result to my initial question, It is a little bit more complicated because the |Q Cauchy sequences are only a subset of {|Q}_{|N } (problem to define/identify the subset).

For the basic set {0,1} and the sequence {0,1}_{n}, it is more complicated because, for any n, I have no bijection between |N and {0,1}_{n} and it is more difficult to see the evolution from a finite (countable) set to an infinite non countable set.

The ideal should be to find a sequence of infinite countable sets that converge (with the adequate topology) to a non countable set equivalent to the set of all |Q Cauchy sequences or something like that (i.e. a good example to see, how the non countable property comes from the countable property through the limit of a sequence)

Seratend.

 

[matt grime]

consider of sequence of rationals

x_n=\sum _{r=1}^n \frac{\delta_r}{10^r}

where delta_r is either one or zero. This is a cauchy sequence of rationals. Moreoever two different choices of the delta+r produce sequences, the difference of which is cauchy adn does not converge to zero. Hence they represent different equivlance classes of cauchy sequences modulo convergence. RIght? Sp it is a subset of the completion of Q

What is the cardinlity of the set of all possible x_n? It is clearly 2^{aleph-0} or the cardinality of the reals.

Ok? what part of that is troubling?

 

[seratend]

consider of sequence of rationals

x_n=\sum _{r=1}^n \frac{\delta_r}{10^r}

where delta_r is either one or zero. This is a cauchy sequence of rationals. Moreoever two different choices of the delta+r produce sequences, the difference of which is cauchy adn does not converge to zero. Hence they represent different equivlance classes of cauchy sequences modulo convergence. RIght? Sp it is a subset of the completion of Q

What is the cardinlity of the set of all possible x_n? It is clearly 2^{aleph-0} or the cardinality of the reals.

Ok? what part of that is troubling?

I am sorry, for this late answer. I was trying to think on explaining better what I am looking for.

Thanks for this compact example (analogue to the previous one, but in a user friendly form :). It helps me, I think, to remove the un-useful features (for me). I am ok with your demonstration and the previous one (I think I understand them).

We define the symbol “:e:” for “element of” (a set) relation.

Let’s call the Cauchy sequences you define above (xn)_n:e:|N (<=> to the limit of the serie) with xn :e: {0;(10^-r)_r:e:|N*}=S. (I have inserted the n=0 <=> translation by one of your serie label r).

We have a simple bijection between S and |N (thus the same cardinality).
The advantage of (xn)_n:e:|N, (thanks to you), is the cauchy convergence whatever (xn):e:S sequence.
Now let’s call (xq_n)_n:e:|N the q th cauchy sequence (q:e:|N).
We thus have xq_n:e:S for every q => [(xq_n)_nq:e:|N] :e: SxSxS...={S}_{|N}.

The countable set of all Cauchy sequences of the form (xn)_n:e:|N belongs to {S}_{|N}.

You have thus demonstrated that this set {S}_{|N} has the cardinality of |R. In addition, because we have a bijection between S and |N, we also can say that |Nx|Nx|N...= {|N}_{|N} has the same cardinality of |R (bijection between {|N}_{|N} and {S}_{|N}

(at least the {|N}_{|N} form helps me in understanding that the “limit set” {|N}_{|N} is uncountable because |N is completely depleted form the extraction of odd/even subparts at the “limit”– case of {|N}_{2n}.)

Therefore, I can concentrate on a topology model to study the limit of {|N}_{n}, if what I have written above is correct.

Has anyone a simple idea to map, for every n (n=0 is the empty set for example), the value “{|N}_{n}” into a simple topological space? (i.e. a one to one correspondence).

Thanks in advance,

Seratend.

 

[matt grime]

Can I suggest you take a second to learn tex (sticky thread in the physics forum should help) because what you have written takes far too much effort to decipher.

(xn)_n:e:|N (<=> to the limit of the serie) with xn :e: {0;(10^-r)_r:e:|N*}=S

makes no sense at all. I mean, what has teh limit got to do with an if and only iff symbol? you appear to claim something is in \mathbb{N}, what is is that? what is the semicolon in that last bracket? What are the subscript symbols _ for?

how about these symbols

x_n = ?

\mathbb{N}

y \in S

S is the elements of some set of something? what the partial sums x_n?


HOw are you picking the q'th cauchy sequeunce? This presupposes they set of all of such caouchy sequences is countalbe when it clearly isn't.

You appear to be confused by the notion of the elements in the set of partial sums of the cauchy sequence and the set of all such cauchy sequences.


I have not said that there is a bijection between \mathbb{N} producted with itself a countably infinite number of times (though I believe there is one).


What has topology got to do with this? We do not need nor require a topology on the set of all convergent cauchy sequences. This is not a construction of topological spaces. ANd if you introduce topologies on sequences you are going to get some very ugly and at the moment completely useless mathematics.


Here is what I've shown: the space of ALL cauchy sequences is uncountable. If we take the set of ALL cauchy sequences of rationals and define the equivlance relation by x_n ~ y_n iff x_n -y_n converges to zero, then the set of equivalence classes of these is the set of real numbers. Furthermore I have shown that I can pick an uncountable subset of the set of all these equivalence classes - those with only 0s and 1s in the finite partial sums of some numbers base 10.

 

[seratend]

Can I suggest you take a second to learn tex (sticky thread in the physics forum should help) because what you have written takes far too much effort to decipher.

You are completely right for tex (I will make the effort ;).

You appear to be confused by the notion of the elements in the set of partial sums of the cauchy sequence and the set of all such cauchy sequences.

“(<=> to the limit of the serie)” is only a comment. I should have replaced the mathematical symbol by the word equivalent.

I am working with the limit of your sequence (the series), not with the partial sum.
I am just defining a map between the limit of your sequence sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}} and the definition of the sequence (x_n)_n_\in_\mathbb{N} (i.e. the limit of your sequence is a series that is equal to the sum of all the (x_n)_n_\in_\mathbb{N} (i.e. I have defined a one to one correspondence between the limit of your sequence and one peculiar (x_n)_n_\in_\mathbb{N}.
(I should have labelled (x_n)_n_\in_\mathbb{N} with another letter in order not to confuse with your partial sum xn.

sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}}\equiv (x_n)_n_\in_\mathbb{N} \ {(with \ the \ sum)}

To obtain the full equivalence, we define the set S form where the x_n elements belong to.
S=\{0;\{10^-^r\}_r_\in_\mathbb{N*}\} (previously S={0;(10^-r)_r:e:|N*}).

We have a one to one correspondence between the elements of S and \mathbb{N} (i.e. the label of the elements of S).

I have not said that there is a bijection between \mathbb{N} producted with itself a countably infinite number of times (though I believe there is one).

It is what I am trying to show in my previous post using your demonstration example and using the equivalence between the set S and |N.

We have a simple bijection between S and |N, thus we have the same cardinality.

The advantage of the (x_n)_n_\in_\mathbb{N} is that the elements x_n belong to S forcing in an explicit way the Cauchy convergence of sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}}.

Therefore, to every (x_n)_n_\in_\mathbb{N} \ {;} \ x_n \in S corresponds a cauchy convergent series (sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}} i.e. your cauchy sequence on the partial sums).

Now, we can label by q, each sequence (x_n)_n_\in_\mathbb{N} by (x_q_n)_n_\in_\mathbb{N} we have the following result (and I correct what I've said: we do not know, a priori if the label q is countable or not, q is just a label to distinguish 2 different sequences):

Each countable cauchy convergent series, (x_q_n)_n_\in_\mathbb{N} belongs to the set {S}_{|N}=\prod_{n \in \mathbb{N}^*}S (definition) by construction. We do not know if the set of different labels is countable or not (in fact it is what we want to know).

You have thus demonstrated that the set {S}_{|N} (i.e. the set of all cauchy convergent sums) has the cardinality of |R. Therefore, because we have a bijection between S and |N, we also can say that the set “|Nx|Nx|N...”= “{|N}_{|N}” (definition) has the same cardinality as {S}_{|N} and |R (i.e. we have a bijection between {|N}_{|N} and {S}_{|N} through the bijection between |N and S).

Here is what I've shown: the space of ALL cauchy sequences is uncountable. If we take the set of ALL cauchy sequences of rationals and define the equivlance relation by x_n ~ y_n iff x_n -y_n converges to zero, then the set of equivalence classes of these is the set of real numbers. Furthermore I have shown that I can pick an uncountable subset of the set of all these equivalence classes - those with only 0s and 1s in the finite partial sums of some numbers base 10.

You are perfectly right and I agree. Now, I think that I have also demonstrated that we have the same cardinality for the set I have labelled {|N}_{|N}.

What has topology got to do with this? We do not need nor require a topology on the set of all convergent cauchy sequences. This is not a construction of topological spaces. ANd if you introduce topologies on sequences you are going to get some very ugly and at the moment completely useless mathematics.

I know that a topology structure will add some complications, but currently, it is one of the tools that help me to “see” the limit from the finite product of |N sets and the infinite product of |N sets (i.e. that has the same cardinality of |R). Recall that I want to try to “see” the transition between the countable and uncountable property and the limit is a powerful tool and if I have such a tool, I can use it to understand better the coherence of some other physics problems.

Therefore, I have tried to prove that the cardinality of {|N}_{|N} is the same as |R using your demonstrations (and the last one you gave was perfect for that). It allows me to restrict the search of a topology to the case of the {|N}_{|N} set limit which is simpler than using a more compact subset (the other examples you give):

I can concentrate on a topology model to study the limit of {|N}_{n} (a finite product of |N sets) to study the limit n --> +oO {|N}_{n}={|N}_{|N } (the infinite countable product of |N sets).

Has anyone a simple idea to map, for every n (n=0 is the empty set for example), the value “{|N}_{n}” (the finite product of |N sets) into a simple topological space?.

Thanks in advance,

Seratend.

 

[ktpr2]

seratend - you might want to check out http://mathworld.wolfram.com/ and drill down into every topic of interest.

 

[matt grime]

Let \mathbb{N}_r = N_r be the product of r copies of the natural numbers for any ordinal r.

Then N_r has many topolgies itself. Discrete, finite, cofinite, the product topology from any topology defined on N such as the topoogy of arithemetic progressions.

The topology does not tell you why the set of all equivalence classes of cauchy sequences of rationals is uncountable.

The fact that I've shown you how to define an uncountable number of inequivalent classes mean nothing?


If you just want an explanation as to why \mathbb{N}_{\omega} is uncountable (the w, omega, just means indexed by a countable infinite set) whilst the finite ones are countable?

Ok, but you aren't going to like it.

Firstly, you have to understand what you mean when you say limit. The are two kinds of limit, the one you have in mind gives you the 'finite power set' if we think in terms of {0,1}s instead of N,then this the set of all strings that are eventually 0. This is a countable set. This roughly correpsonds to taking the direct limit, or the coproduct. The one you need to consider behaves as the inverse limit, or product, and it is the set of all strings.

THis is simply the difference between direct product and coproduct (union)

 

[seratend]

seratend - you might want to check out http://mathworld.wolfram.com/ and drill down into every topic of interest.

Thanks a lot. The problem with wolfram is that most of topics are described shortly (most of the time with the name solely knowed by the specialist of the subject), therefore it is difficult to find what you are searching, especially when you have some problems to define exactly what!
However, It is a good idea to look at it, if I can explain what I am looking for with this limit approach.

The topology does not tell you why the set of all equivalence classes of cauchy sequences of rationals is uncountable.

The fact that I've shown you how to define an uncountable number of inequivalent classes mean nothing?

I will answer later to the other points of your interesting post (I have to think a little on it, before answering).
I agree with you and your demonstrations and that the topology won't tell me what is uncountable or countable as the countable property is only the existence of a bijection between |N and another set. It is not the intention of the tool I am trying to construct/define (even if I am not clear in what I really want :).

I need to construct a "simple" topology in order to attempt describing the *transition* between a sequence of countable sets towards an uncountable set. In a topological space, I must have defined, before any deduction, the open sets and thus the countable and uncountable sets. It is not important in a first step if this topology only applies to a single limit if it allows to describe the transition.

I repeat that my first intention is not to demonstrate that a set is uncountable on a given topological space, just how we can describe the *transition*, through a limit (in this attempt), from a countable set sequence towards an uncountable set.

Now, I will work on the other points of your post (and by the way, you already have helped me a lot to clarify what I may need with your examples and demonstrations : ).

Do not hesitate to add some other comments or help in the meanwhile! (and also the other members of this forum !)

Seratend.

 

[matt grime]

Well, you will need to learn category theory and look up the definiton of the inverse limit of a diagram, just to warn you, seeing as you don't as yet have a good notion of limit, it appears. NB. in no sense do you NEED to construct simple topologies, since they have little to do with "transition" indeed I can give you different topologies with wildly differing "limit" behaviour.


To see why I'm warning you off this you might wish to learn about profinite groups and derived lim^1 and Mittag-Leffler conditions.

I also think you should learn about different topologies too.

 

[seratend]

Ok, but you aren't going to like it.

Firstly, you have to understand what you mean when you say limit. The are two kinds of limit, the one you have in mind gives you the 'finite power set' if we think in terms of {0,1}s instead of N,then this the set of all strings that are eventually 0. This is a countable set. This roughly correpsonds to taking the direct limit, or the coproduct. The one you need to consider behaves as the inverse limit, or product, and it is the set of all strings.

THis is simply the difference between direct product and coproduct (union)

Ok, I am trying to re state your post in order to allow you to correct me if I am wrong.
Let's call, for each p \in \mathbb{N}^* the finite power of {0,1} as {0,1}_{p}=\prod _{n=1}^p \{0,1\}.
we have, for each p \in \mathbb{N}^, {0,1}_{p} is the set of all strings of length p (with either 0 or 1 values) that is a countable set.
e.g. (1,0,0,1,1,1) \in {0,1}_{6}
If we take a “limit” in p of {0,1}_{p} we obtain a countable set we can call {0,1}_{p=oO} and I agree with that.

Now, we can consider the finite set of all finite strings with a length lower than p (call it {{0,1}_{p}}). This is a countable set (even if in the previous post I have wrongly said that we have the countable property if we have a bijection between sets, forgive me : (.
Let’s define this set:

for each p \in \mathbb{N}^ the finite set of all finite strings with a length lower than p {{0,1}_{p}}=\prod _{n=1}^p {0,1}_{n}.

If we take the “limit” of {{0,1}_{p}}, we obtain a non countable set. Therefore, the topological space, I am interested in is the one on the sets {{0,1}_{p}} not {0,1}_{p}.

Do you agree with that? (I hope I have made no mistakes this time ; )

Seratend

 

[seratend]

Well, you will need to learn category theory and look up the definiton of the inverse limit of a diagram, just to warn you, seeing as you don't as yet have a good notion of limit, it appears. NB. in no sense do you NEED to construct simple topologies, since they have little to do with "transition" indeed I can give you different topologies with wildly differing "limit" behaviour.


To see why I'm warning you off this you might wish to learn about profinite groups and derived lim^1 and Mittag-Leffler conditions.

I also think you should learn about different topologies too.

You are perfectly right. But, my brain as a very limited capacity to learn all these topics and I think (I hope or I pray if you prefer; ) I just need a single example with the "smoothest" possible transition.
This is why I am trying to define a single example: just to use as little as possible the properties of other theories (even if in this example we re-demonstrate/redecover properties of other theories). It is a very pragmatic approach for a single need (one description of a "as smooth as possible" "transition").

Seratend.

P.S. I have looked quickly at the category theory and I prefer to restrict the context to ZF of ZFC sets or even more restrictive as well as the limits (for example using continuous bijective functions between topologies if their exist).

 

[matt grime]

The first step you MUST take is to say what you think you mean by taking the limit of sets, second you ought to define what you think transition means. And then you should clearly state what is you're trying to ask, because it isn't clear at all (to me, at least).

We have a sequence

\mathbb{N} \leftarrow \mathbb{N}_2 \leftarrow \mathbb{N}_3 \ldots

the object \mathbb{N}_{\omega} is the inverse limit of this diagram, the infinite product of N with itself a countable number of times.

This is an uncountable set.

The topologies we could place on the objects have nothing to do with this set theoretic statement.

But how does this relate to your original question about being troubled that the completion of Q in the euclidean norm was R.

 

[seratend]

The first step you MUST take is to say what you think you mean by taking the limit of sets, second you ought to define what you think transition means. And then you should clearly state what is you're trying to ask, because it isn't clear at all (to me, at least).

You are right and I know that I am not very clear, but I am trying to make it clearer thanks to you. I am now thinking on a concrete example using the information you and others have already provided (e.g. a topology example for the "limit" \mathbb{N}_{\omega}). I will submit it to you and the other members of this forum for approval and comments (I hope that this example, if I succeed in building it, leads to a common basis of comprehension in order to proceed further).
By the way, I am not forgetting my initial question concerning the completion of Q and vector spaces, it is directly related to the evolution of this thread (I will give later precisions on this, once I am clearer with my statements, otherwise I risk to add other confusions to this thread).

Thanks in advance,

Seratend.

 

[seratend]

Here is the (basic) example proposal to define the limit of the countable product of |N sets (in order to develop the subject and to add precisions):

Let's call, for each p \in \mathbb{N}^* the finite power p of |N sets as {|N}_{p}= \prod _{n=1}^p \mathbb{N}.
And the infinite countable product of |N sets: {|N}_{ω}= \prod _{n=1}^\infty \mathbb{N}

(I use this specific notation, in order not to confuse with other existing definitions)

Let's define the topology space T_|N= (X_|N, OpenSets), where the open sets are the ones generated by the following countable basis of open sets: B(O_|N):
for each p \in \mathbb{N}^*, we define the open set O_|N(p)={{|N}_{p}}= \{ \prod _{n=1}^p \mathbb{N}\} and also the set O_|N(+oO)={{|N}_{ω}}= {\prod _{n=1}^\infty \mathbb{N}}.
Any union of these sets are open sets for this topology.
e.g. {{|N}_{1}, {|N}_{60}} is an open set of T_|N.

We have X_|N= \bigcup_{p\in\mathbb{N^*}}{{|N}_{p}} \bigcup {|N}_{ω} and thus OpenSets=P(X_|N) (the finest topology where the "points" of this set are the finite and countable infinite product of |N sets).

If we create a one to one mapping between the points of X_|N set and the points of the set |N U{+oO}, we have an other possibility to look at this basic topology (homeomorphism).

However, with the topology T_|N, we can now define the sequence u_n and the limit:

For each n \in \mathbb{N}^*, u_n= {|N}_{n} \in X_|N

{|N}_{+oO}=\lim_n_\rightarrow_\infty u_n = {|N}_{ω}

\Longleftrightarrow \forall G\in V(\{\mathbb{N}\}\_\{\omega\}) \ \exists N_G \in \mathbb{N}^*; \ \forall n \geq N_G, \ u_n \in G


Do you agree with that? i.e. We can see {|N}_{ω} in this context (the topology) as the limit point of the sequence u_n (i.e. the function u(n) is continuous on the point {|N}_{ω}).

Does someone have a simple solution to introduce a metric on this topology or on an equivalent one (using an isomorphism or an inclusion)? (for example, it could be interesting, using a bijection, to “move” the “infinite point” to a finite location).

Seratend.

 

[matt grime]

Firstly, what set is this a topolgoy on? N+p? N_w? Nope, none of those. It is at best a topology on the set whose points are the countable products of copies of N, and it is the discrete topology. It isn;t a topology on any of the N_p.

Moreover it is the discrete topology on that set and thus only eventually constant sequences are Converegent. Also, this topology tells you nothing about any of the sets N_p, or N_w

All you#'ve done is define the discrete topology on an infinite countable set. Thsi doesn't tell you anything about the elements in the set.


I thought you might atleast be defining a topology on N_p or something, because this deosn't tell you the first thing about teh infinite product of copies of N.

 

[seratend]

Reflexivity of Banach spaces.

(I am revisiting quickly some properties of the banach spaces, trying to connect them later to this thread).

Let’s assume that (E, || ||) is a banach space,
Let’s call J_E the canonical application between E and its bidual E**:
J_E: E --> E**
x --> x**=J_E(x) \in B(E*, \mathbb{K}) (the continuous linear applications of E, i.e. the bounded applications)
where for all x* \inE*, x**(x)=[J_E(x)](x*)=x*(x).

A banach space is said reflexive iff J_E is a bijection.
(I hope it is the correct English term. ; ).

I have read that l1(|N) and loO(|N) are non-reflexive banach spaces while Hilbert spaces are reflexives. I want to understand this statement .

For each banach space (E, || ||), we can define an associated hermitian product <,>:
4<y,x>=||x+y||² - ||x-y||²+i||x+iy||²-i||x-iy||² for all x,y \inE.

Thus, each Banach space is a Hilbert space for the associated hermitian product of the norm. Therefore, I think I can say that l1(|N) and loO(|N) are Hilbert spaces for their associated hermitian product, and therefore are reflexive banach spaces which seems in contradiction with what I have read (i.e. we have a one to one correspondance between H and H**).

The question is : where am I wrong?

 

[matt grime]

Can I suggest you try to find better notation, ie learn latex, because it's hard to read your technical definitions of objects.

Working on a better reply about the maths.

Right,

the polarization identity only gives an inner product under certain conditions and does not always give an inner prooduct. the banach norm must sttisfy the parallelogram idnetity

|u+v|^2 + |u-v|^2 = 2( |u|^2 + |v|^2)

 

[seratend]

the polarization identity only gives an inner product under certain conditions and does not always give an inner prooduct. the banach norm must sttisfy the parallelogram idnetity

|u+v|^2 + |u-v|^2 = 2( |u|^2 + |v|^2)

Thanks a lot. I have missed the linearity and antilinearity pb of <x,y> based on the polarisation identity.

Seratend.

 

[seratend]

Can we say that in a non separable hilbert space there isn't any injective compact bounded linear applications on this space (and thus, no compact bounded bijections)?

Seratend.

 

[matt grime]

I doubt that that is true. Can you save me the effort and give me some examples of non-separable hilbert spaces?

having said that, you mean functional by application don't you?

 

[seratend]

I doubt that that is true. Can you save me the effort and give me some examples of non-separable hilbert spaces?

having said that, you mean functional by application don't you?

Yes, I mean a continuous linear function f: H --> H that is compact (i.e. closure[f(B)] is a compact set in H where B is the unit radius ball in H).

Example of a non separable hilbert space: l^2(I) where I is an uncountable set (for example a |R interval). It is the hilbert space of uncountable real number sequences (x_i)_i_\in _I such that \sum _{i_\in _I}|x_i|^2 &lt; \infty

Seratend

 

[seratend]

Hint for my question: It is an extrapolation from the spectral theory of compact continuous linear functions (i.e. only Ker(f) seems to be the only subspace that can be non separable).

Seratend.

 

[matt grime]

a linear functional is not the same as a linear function from H to itself.

 

[seratend]

a linear functional is not the same as a linear function from H to itself.

Ok, I am sorry for my bad mathematical terms (sometimes they are direct translations).
I mean a linear operator (call it f), if you prefer, "acting" on the hilbert space H (an endomorphism if I am right):
f: H --> H
x\in H --> f(x) \in H

such that f(x1+k.x2)=f(x1)+k.f(x2)

we choose f as a bounded compact operator on H. My question stays: can we say that there isn't any injective bounded compact linear operator on a non separable hilbert space?

 

[seratend]

We can either restrict this question to the normal subset of these linear operators (i.e. an operator such that ff*=f*f).

Seratend.

 

[seratend]

I doubt that that is true. Can you save me the effort and give me some examples of non-separable hilbert spaces?

having said that, you mean functional by application don't you?

In order to help the members of this forum to answer to this question (and also to believe that the answer is probably yes), we can already say that there isn't any bijective compact bounded operator on infinite dimensional Hilbert spaces (separable or not).
Here is the demonstration:
We know that the closed unit ball cannot be a compact set in an infinite dimensional Hilbert space. However, if we have a compact isomorphism between Hilbert spaces that are continuous, the image of closed set is a closed set and thus because we have a compact operator, the image of the closed unit ball is a compact set (the closure of the set is equal to the set, because the set is closed). Thus, the image of this compact set under the inverse continuous operator is a also compact set. Thus, the unit ball is a compact set for a compact continuous isomorphism. QED. So it is easy to demonstrate the impossibility of compact isomorphisms on infinite dimensional Hilbert spaces (separable or not).

However, if we reduce the case to the injective compact continuous operators case, we just have the property that the closure of the image of a closed unit ball is a compact set. Therefore, the unit ball is not a compact set if the image is not compact and the question is still valid.

Seratend.

 

[seratend]

The answer:

There is no compact (continuous) injective operator on a non separable hilbert space.

Quick demo: H the hilbert space. T the compact injective operator => Closure[T(H)] is a separable space (compacity property) => there is no injection between a non separable hilbert space and the separable one.

Moreover, we have a weaker formulation of this impossibility: there is no injective compact operator between a non separable Hilbert space and any banach space. (i.e. we can have an injective compact operator between 2 banach spaces, even if the banach spaces are non separable).

Seratend.