# Some quetions about set theory.

1. Apr 25, 2006

### 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?

Last edited: Apr 25, 2006
2. Apr 25, 2006

### HallsofIvy

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

3. Apr 25, 2006

### MathematicalPhysicist

4. Apr 25, 2006

### 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.

Last edited: Apr 25, 2006
5. Apr 25, 2006

### 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 = {(nm)n>1 | m > 1}

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

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

6. Apr 26, 2006

### matt grime

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

7. Apr 26, 2006

### MathematicalPhysicist

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?

8. Apr 26, 2006

### 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?

Last edited: Apr 26, 2006
9. Apr 26, 2006

### 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?

10. Apr 26, 2006

### matt grime

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

11. Apr 26, 2006

### 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?

12. Apr 26, 2006

### 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.

Last edited: Apr 26, 2006
13. Apr 26, 2006

### 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 isnt countable.
if im right here, why in you first post you said that it needs to approach infinity (or as common to say, diverge)?

14. Apr 26, 2006

### matt grime

What? This makes no sense:

"if im 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.

Last edited: Apr 26, 2006
15. Apr 26, 2006

### AKG

I see, for question 2, (cn)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 (cn)n>1 such that for EVERY (bn)n>1 in X, the limit of cn/bn as n goes to infinity is nonzero, then you're done. In particular, you don't need cn/bn to converge as n goes to infinity, you just need it to not converge to 0.
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.

16. Apr 27, 2006

### 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.

17. Apr 27, 2006

### MathematicalPhysicist

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.

18. Apr 27, 2006

### 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.

Last edited: Apr 27, 2006
19. Apr 27, 2006

### matt grime

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

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.

20. Apr 27, 2006

### 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.