Some quetions about set theory.

[MathematicalPhysicist]

1)let P={p1,p2...} be the set of all prime numbers for every n natural, present n as a representation of its prime factors, n=product(p_k^a_k)
1<=k< \infty (where a_k=0 besdies to a finite number of ks).
and now define F:N->Q+ by: F:n=product(p_k^a_k)->r=procudt(p_k^f(a_k)), and show that's a bijection from N onto Q+.

2)let X be a set of sequences with the next property: given a sequence (c_n:n>=1), then there exist a sequence (b_n:n>=1) which belongs to X and lim n-> \infty c_n/b_n=0.
prove that the set X isn't countable.

3)find the cardinality of the set of all the straight lines in a plane.

about the last question i got that, that this set is the plane itself, i.e R^2, so |RxR|=|R||R|= \aleph*\aleph is this the right answer?

 

[HallsofIvy]

What do you mean by \aleph? Presumably not \aleph_0 since |R| is not \aleph_0.

 

[MathematicalPhysicist]

i mean this:
http://mathworld.wolfram.com/Aleph.html

 

[matt grime]

Well, you shouldn't mean that. It is like saying the cardinality of the set {1,2} is n, when it is 2. aleph is the generic symbol for the cardinality of any infinite set.

In Q1 in the definition of F you introduce f without explaining what it is. Is it suppose to be F too and this is some recursive definition of F?

I've never seen Q2 before. I think it seems reasonable. If it is a countable set then enumerate it, any of the b_n's appears at a finite position in the enumeration, produce a c_n using this where c_n/b_n does not converge.

 

[AKG]

For question 3, your approach is wrong. How is the set of lines in the plane the same as the plane itself? The set of lines has lines as its ELEMENTS. The plane has points as its ELEMENTS, and lines as (a special type of) SUBSET. Yes, the union of all lines in the plane equals the plane, but the union of all lines is not the same as the set of all lines. The set of lines is a set of sets, since each line is a set. The plane itself is a set of points.

For question 2, the thing you're asked to prove seems false. Define:

X = {(n^m)_n>1 | m > 1}

X is countable, as it is enumerated by m > 1, i.e. the positive integers. Given any sequence (n^m)_n>1, the sequence (n^m+1)_n>1 dominates it in that:

lim_{n\to\infty}\frac{n^m}{n^{m+1}} = \lim_{n\to\infty}\frac{1}{n} = 0

 

[matt grime]

I don't think c_n is supposed to be in X, though it isn't clear.

 

[MathematicalPhysicist]

Well, you shouldn't mean that. It is like saying the cardinality of the set {1,2} is n, when it is 2. aleph is the generic symbol for the cardinality of any infinite set.

In Q1 in the definition of F you introduce f without explaining what it is. Is it suppose to be F too and this is some recursive definition of F?

I've never seen Q2 before. I think it seems reasonable. If it is a countable set then enumerate it, any of the b_n's appears at a finite position in the enumeration, produce a c_n using this where c_n/b_n does not converge.

about Q2 how do i produce c_n, i mean what b_n's from the enumeration should i take as c_n?
about Q1 you are right, but this is how it is written in my text, perhaps it's connected to the first question in y book which define f as such:
f(n)=(-1)^(n+1)[0.5(n+1)] where [] is i think you call it the floor function, and n is natural.

and akg,then what is the answer to my last question?

 

[matt grime]

Q1 use some common sense. f(n) is not some universally known function, and we don't have your book (or at least we don't know what your book is).

Q2 I didn't say you take any of the b_n from the enumeration. You don't. As AKG pointed out, if c_n is restricted to being one of the b_n then the question is false. I can't think of a good explanation of the answer that isn't just giving you the answer, but that's because I've only just figured out the answer and therefore haven't worked out what it's 'really' asking.

Let me try it this way, firstly there is no harm in assuming all entries in the b_n are strictly positive for all sequences in X:

pick some enumeration of the b_n, let b(m)_n mean the m'th one in the list.

I need to find a c_n such that c_n/b(m)_n does not converge for any m, right?

I can easily do this for b(1)_m, by just letting c_n = 1/b(m)_n. The quotient is always 1. Call this sequence c(1)_n

Now, if I look at c(1)_n/b(2)_n this may or may not converge.

What I could do is increase c(1)_n so that c(1)_n/b(2)_n =1 (and this c(1)_n/b(1)_n>1) at the n where c(1)_n/b(2)_n is less than 1.

This might involve changing all of the c(1)_n. Call the new sequence c(2)_n

I can repeat for b(3) and so on to make c(3)_n etc, but there is no guarantee that the c(m)_n converges to a sequence of real numbers s m tends to infinity.

Can you try to correct this so that it does converge?

 

[MathematicalPhysicist]

"What I could do is increase c(1)_n so that c(1)_n/b(2)_n =1 (and this c(1)_n/b(1)_n>1) at the n where c(1)_n/b(2)_n is less than 1."

by increasing, do you mean the limit of c(1)_n/b(2)_n converges to 1?

 

[matt grime]

No, I meant exactly what I said apart from in the brackets the word 'this' should be 'then'.

 

[MathematicalPhysicist]

the i don't understand how the term c(1)_n/b(2)_n can equal 1 and be less of 1. (without calling out the weakest equality <=).
care to explain?

 

[matt grime]

You make c(1)_n, right, using b(1)_n, now you look at c(1)_n/b(2)_n and changing the c(1)_n as required you make c(2)_n so that c(2)_n/b(1)_n and c(2)_n/b(2)_n are all at least 1 by replacing any of the c(1)_n with *larger numbers* (ie increasing them) if required.

This is supposed to give you an idea of how to do a 'diagonal' argument for yourself, it is not a proof as I explain later in the post. So what have you done to solve this problem?

There are better ideas you can use, since all you're trying to do is find a c_n where c_n/b_n does not tend to zero if the b_n were countable (you can even invoke the fact that there are infinitely many primes to prove this) though they all (the ones I've thought of) involve the idea that a double indexed sequence a(n,m) can be made to converge to many different things depedning on how you let m and n tend to infinity.

 

[MathematicalPhysicist]

after overlooking your post eight, i ask this question: by changing the terms of c_n up to max(1/b_mn) you actually getting that: c_n/b_mn=1, so you suppose that X is countable and thus also b_n, but the limit doesn't converge to 0 and thus we have a contradiction, and X isn't countable.
if I am right here, why in you first post you said that it needs to approach infinity (or as common to say, diverge)?

 

[matt grime]

What? This makes no sense:

"if I am right here, why in you first post you said that it needs to approach infinity (or as common to say, diverge)?"

as a sentence. I have not said anything 'needs to diverge', merely that something 'might' diverge, which is very different.

I never use the notation b_mn, either.

I did mistype something in my post, which I will now correct.

Just enumerate them, ie put them in a list, now try to make a c_n that doesn't satisfy the hypothesis, that is all i''m trying to make you do, and something you should be trying to do yourself: there are different ways to do this, what have *you* tried?

I gave you the naive first approximation to how to do this, now try to make it work.

 

[AKG]

I see, for question 2, (c_n)_n>1 need not be in X. Okay, there's some set S for which there's a famous argument showing that it is uncountable, by assuming that it is countable, then constructing a new element, contradicting the assumption of countability. What is this set S, and what is this famous argument? If you know this argument, then its not hard to apply this idea to question 2. Note that if you assume X to be countable, and you construct a sequence (c_n)_n>1 such that for EVERY (b_n)_n>1 in X, the limit of c_n/b_n as n goes to infinity is nonzero, then you're done. In particular, you don't need c_n/b_n to converge as n goes to infinity, you just need it to not converge to 0.

and akg,then what is the answer to my last question?

I'm not going to just tell you the answer. You haven't really made an effort at it yet. The solution you posted suggest you weren't even thinking of the set of lines in the correct way. I told you how to think of them (sets of lines, not a union of lines), but you'll have to do your own work.

 

[MathematicalPhysicist]

but akg, the set of all the straight lines in a plane isn't it the union of the sets?
if a straight line is set in itself then the set of all the stragiht lines is a set of union of the stragiht lines iteself.

but let's say I'm wrong here, then by your reply the answer should be 2^\aleph

cause we have a set of sets, which is equal to the power set.

 

[MathematicalPhysicist]

"I can easily do this for b(1)_m, by just letting c_n = 1/b(m)_n. The quotient is always 1. "

which quotient do you say equals 1, does it c_n/b(m)_n or 1/b(m)_n in both cases you are setting that b(m)_n=1 in the first case 1/b^2(m)_n and second 1/b(m)_n.

i must say that i don't understand, if i try first to to show that c_n/b_n equals 1, and afterwards by showing that the subsequences also equal 1 by increasing c_n then i need to prove that also in the end it converges to 1, by changing c_n,right?

if this is correct then i can set c_n=1+b_n and when n appraoches infinity also b_n appraoches infinity and therefore the limit equals 1 in the end.

 

[matt grime]

It shuold have been: let c(1)_n=b(1)_n not 1/b_n(1).

But you don't have to prove things converge to 1, just that they do not converge to zero, and one way to do this it to produce a c_n such that c_n/(b(m)_n =>1 for all n and for any m, where b(m) is the m'th sequence in an enumeration. I gave you a starting idea, now try to make it actually work.

 

[matt grime]

but akg, the set of all the straight lines in a plane isn't it the union of the sets?

it is, but I think he was objecting to you asserting the union of the set of lines was the plane.

but let's say I'm wrong here, then by your reply the answer should be 2^\aleph

cause we have a set of sets, which is equal to the power set.

No, that is not true. He says it is a set of sets (all sets are sets of sets in some sense except the empty set), this does not make it the power set of anything.

 

[MathematicalPhysicist]

then how should i appraoch set of sets, if the sets are infinite (not null aleph, because you can do a bijection from the lines onto the set [0,1)) then is the number of all straight lines in plane equals \aleph, because there's a bijection as i mentioned in the brackets.

 

[matt grime]

then how should i appraoch set of sets, if the sets are infinite (not null aleph, because you can do a bijection from the lines onto the set [0,1)) then is the number of all straight lines in plane equals \aleph, because there's a bijection as i mentioned in the brackets.

do you really mean to say that? Because if you have indeed got a bijection onto [0,1) then you know the cardinality, so why ask the question?

And stop using aleph, you don't mean aleph, that is the generic symbol for any infinite cardinal (we've already brought this up at least twice). The cardinality of the real numbers is denoted c (for continuum), and if we accept the continuum hypothesis this is the same as \aleph_1

 

[MathematicalPhysicist]

i just need an approval that my bijection is correct.
let's take the set of all straight lines as L={m and n belong to R|y=mx+n for y and x which belong to [0,1)} then there is a function f:L->[0,1) by y=f(x)=mx+n, is this right?

 

[MathematicalPhysicist]

by the way, my book states that the cardinality of [0,1) is aleph (and uses this notation), and also R has the same cardinality of [0,1) and this is why i use the same notation.

 

[matt grime]

by the way, my book states that the cardinality of [0,1) is aleph (and uses this notation), and also R has the same cardinality of [0,1) and this is why i use the same notation.

Well, it (the use of alpeh) is not standard, just like f was not standard notation.

i just need an approval that my bijection is correct.
let's take the set of all straight lines as L={m and n belong to R|y=mx+n for y and x which belong to [0,1)} then there is a function f:L->[0,1) by y=f(x)=mx+n, is this right?

No, this isn't right. How does it even define a map between [0,1) and the set of all lines in the plane, never mind a bijective one? What the heck is f? How does that take an element in the set of all linesand produce an element of [0,1)? Take the line y=2x+1, what is the element of [0,1) this corresponds to? What about the line y=3x-2?

 

[MathematicalPhysicist]

perhaps i shouldve had done a bijection between the lines and R, because then there's one to one between any pair of (x,y) and RxR, right?

 

[matt grime]

What are x and y? Suggestively they are the coordinates of RxR, but you don't appear to be using them for that at all.

What is the easiest parametrization of the set of lines in the plane? Slope and intercept, right? y=mx+c, call that line L(m,c), this gets all lines except the vertical ones.

 

[MathematicalPhysicist]

i thought it was clear that (x,y)=(x,mx+n), this way we define the set of lines as a set of a set of points.

 

[matt grime]

What does that have to do with what you wrote earlier, where you restricted x and y to lie in the interval [0,1)? And where your map f from the lines to the interval [0,1) acted as y=f(x)? That just makes no sense. x,y and coordinates, nothing more. Why don't you start again. If L is the set of lines what is this map f to [0,1) that you claim to have? Where does it send the line y=mx+c? Where does 2x+1 go? What about 3x+8? Where does it send the line x=2? Why is it a bijection?

 

[MathematicalPhysicist]

my first answer is incorrect as you suggested, this is why i suggested that perhaps the bijection between the set of all lines in a plane can be cosntructed as L(x,y)->RxR by the equality y=mx+n from (x,y)->(x,mx+n)

to prove that it's one to one let (x0,y0) != (x1,y1) then (x0,mx0+n)=(x0,y0) != (x1,y1)=(x1,mx1+n).

 

[matt grime]

This does not define a map from the set of points in a plane to the set of lines in a plane. What are m and n for heaven's sake? Are they fixed? are they allowed to vary? what are they? (I know what they ought to be, but you don't appear to be using them at all.)

So, I take the the line with equation y=2x+1 and I map this to the 'point' in RxR (x,2x+1), well that isn't a point in the plane, is it, because it has the variable x in it? Why are you doing anything with x and y going anywhere? They are just variables, coordinates of points on the lines, they are not parametrizing the set of lines. m and n are the things that parametrize lines in the plane, slope and intercept, (of course we haven't gotten round to vertical lines).

Here, let's show that the cardinality of the set of vertical lines in the plane is the same as the cardinality of R.Let L_r be the vertical line x=r for r in R, note every vertical line is of this form, then the map that sends L_r to r is a bijection from the set of vertical lines to R. I didn't touch the x, because that is just the coordinate of the line.

 

[MathematicalPhysicist]

so if i define as you suggeted:
(m,n)->RxR for y=mx+n is that enough?
as you said the vertical lines which make the coordinate system aren't considered here, this is why i opted to choose the point (x,y) and therefore considering the set of all lines as a set of sets of points which consist the lines (somehow i feel this will not suffice here too).

p.s
i think that in order to include the vertical lines, if you cannot use your option of (m,n) then why not the pair (x,y) which will include all the points?

 

[matt grime]

Nope, you're still not making sense.

"i opted to choose the point (x,y) and therefore considering the set of all lines as a set of sets of points which consist the lines"

choose the point (x,y) where? You can't pick a point lying on the line to parametrize the line since anyone point in the plane (where we are considering the lines) lies on uncountably many different lines. It is easy to write down a bijection between the set of non-vertical lines and RxR, the line L(m,n) which is the line with equation y=mx+n goes to..., and it can be modified to take account of the vertical lines too; a vertical line can be thought of as a line with slope \infty, so there is a bijection between the lines and (\mathbb{R}u\{\infty\})\times \mathbb{R} and \mathbb{R}u\{\infty\} has the same cardinality as R.

 

[AKG]

but akg, the set of all the straight lines in a plane isn't it the union of the sets?

No. The set of lines is the set of lines. The union of lines is the plane.

if a straight line is set in itself then the set of all the stragiht lines is a set of union of the stragiht lines iteself.

HUH!?

but let's say I'm wrong here, then by your reply the answer should be 2^\aleph

WHAT?!

cause we have a set of sets, which is equal to the power set.

None of what you're saying even begins to make sense.

 

[HallsofIvy]

You are not being precise enough. "A set of sets" is not necessarily a power set. The power set of A is specifically the collection of all subsets of A. Given a set A, it would be possible to select a "set of subsets of A" that is not the power set of any set- just don't include the empty set!

In particular, if A= {1, 2, 3}, with cardinality 3, it's power set is {{}, {1}, {2}, {3},{1,2},{1,3},{2,3},{1,2,3}} which has cardinality 2³= 8. But A= {{1},{2},{3}} is also a "set of sets" and has cardinality 3.