- Joined
- 10 Dec 06
- Moves
- 8528
The post that was quoted here has been removed
If you have trouble understanding my criticism of DeepThought's analogy,
it may be because you have a (hidden) default assumption that everything
must belong to the real numbers.
Actually, it was because I was confusing "I" with "C"
So "R" is incomplete with respect to "C"
But "I" must have the same cardinality as "R" for "C" to be everywhere dense.
To me it seems that your argument ( and I may be misinterpreting this ) is that "I" is a subset of "R". So how can a subset of "R" have the same cardinality as "R"? It seems like an impossibility to me. Maybe I'm misunderstanding the definition of incomplete/complete sets?
There are many notions of "completeness" in mathematics. The two most common ones, in my experience, are:
1) Completeness of a totally ordered set S.
2) Completeness of a metric space (X,d).
The set of complex numbers, while complete in the second sense (with respect to the usual Euclidean metric), cannot be complete in the first sense because it is not even a totally ordered set. By definition a set S is totally ordered by a relation, often denoted by ≤ (less than or equal to), if the following properties hold for all x,y,z ∈ S:
P1) x ≤ y or y ≤ x.
P2) If x ≤ y and y ≤ x, then x=y.
P3) If x ≤ y and y ≤ z, then x ≤ z.
The set of real numbers R, with the conventional notion of "less than or equal to," is a totally ordered set. That R is complete is simply taken as axiomatic (it cannot be proven using the other axioms that define R).
As for the set of integers Z, I believe it can be proven to be complete with the help of the Completeness Axiom for R. Let S be a set of integers with an upper bound. Since S is a set of real numbers (all integers are real), by the Completeness Axiom for R there exists a (real-valued) least upper bound for S. Call it u. So for all n ∈ S we have:
n ≤ u.
That's because u is an upper bound for S. Also, for every ε>0, there exists some n ∈ S such that:
u-ε < n ≤ u.
That's because u is the least upper bound for S. Letting ε=½ in particular, it follows that there must exist some m ∈ S such that
u-½ < m ≤ u.
Suppose there exists some n ∈ S such that n>m. Since m and n are integers, we must have n ≥ m+1. But then we get
n ≥ m+1 > ( u-½ )+1 = u+½,
which shows n to be greater than the upper bound u for S -- a contradiction. So there is no element of S greater than m, which means m is an upper bound for S.
Now, since m is an upper bound for S, m ≤ u, and u is the least upper bound for S, it must be that m=u.
Therefore u is an integer. We conclude that every set of integers with an upper bound must possess a least upper bound that is an integer. By definition this means the set of integers Z is complete.
EDIT: I found a way to put the symbols ≥ and ≤ into this post, and also epsilon ε, and "is an element of" ∈. I hope they work for everyone. In case they don't, below is the same post without the fancy symbols.
There are many notions of "completeness" in mathematics. The two most common ones, in my experience, are:
1) Completeness of a totally ordered set S.
2) Completeness of a metric space (X,d).
The set of complex numbers, while complete in the second sense (with respect to the usual Euclidean metric), cannot be complete in the first sense because it is not even a totally ordered set. By definition a set S is totally ordered by a relation, often denoted by <= (less than or equal to), if the following properties hold for all x, y, and z in S:
P1) x <= y or y <= x.
P2) If x <= y and y <= x, then x=y.
P3) If x <= y and y <= z, then x <= z.
The set of real numbers R, with the conventional notion of "less than or equal to," is a totally ordered set. That R is complete is simply taken as axiomatic (it cannot be proven using the other axioms that define R).
As for the set of integers Z, I believe it can be proven to be complete with the help of the Completeness Axiom for R. Let S be a set of integers with an upper bound. Since S is a set of real numbers (all integers are real), by the Completeness Axiom for R there exists a (real-valued) least upper bound for S. Call it u. So for all n in S we have:
n <= u.
That's because u is an upper bound for S. Also, for every e>0, there exists some n in S such that:
u-e < n <= u.
That's because u is the least upper bound for S. Letting e=1/2 in particular, it follows that there must exist some m in S such that
u-½ < m <= u.
Suppose there exists some n in S such that n>m. Since m and n are integers, we must have n >= m+1. But then we get
n >= m+1 > (u - 1/2) + 1 = u + 1/2,
which shows n to be greater than the upper bound u for S -- a contradiction. So there is no element of S greater than m, which means m is an upper bound for S.
Now, since m is an upper bound for S, m <= u, and u is the least upper bound for S, it must be that m=u.
Therefore u is an integer. We conclude that every set of integers with an upper bound must possess a least upper bound that is an integer. By definition this means the set of integers Z is complete.
Originally posted by joe shmoWhen we are dealing with sets and subsets that have an infinite number of elements the simple notions that all of us have of cardinality just break down.
So how can a subset of "R" have the same cardinality as "R"? It seems like an impossibility to me.[/b]
Deep down cardinality is about making one-to-one correspondences between two sets:
{a,b,c}, {d,e,f} have the same cardinality because you can make a one-to-one correspondence like a->d; b->e c->f. The set {a,b,c} has a bigger cardinality than the set {g,h} because when you make a one-to-one correspondence between the elements of these two sets one element of the first set has no correspondence in the second set.
Hence if you can show that you can make a one-to-one correspondence between a set and a subset of its then they have the same cardinality.
Take the natural numbers and the integer numbers. Naively it seems that the integers are a "bigger" set than the naturals, but
0->0; 1-> -1; 2->1; 3-> -2; 4->2; etc.
and you can make a very simple one-to-one correspondence between these two sets. Again infinity plays a trick on our naive minds just because we were consistent with a simple definition. Since a one-to-correspondence exists between these two sets they have he same cardinality.
It is is easy to show that Q=Z=N in terms of cardinality by a simple one-to-one correspondence from Q to Z.
Now what about R. Is it R=Q in terms of cardinality? After all of this fact wouldn't be a surprise to us if it was so. But it isn't! Cantor showed us by his beautiful diagonal argument that the real numbers have elements that can't be put in a one-to-one correspondence with the real numbers. Hence the cardinality of R is bigger than the cardinality of Q.
One can also show that the cardinality of the irrational numbers (R\Q) is the same as R. Which is to say that when you look at the real number line almost all of its elements are irrational! In fact if you pick a random number in the real number line the probability of it being rational is 0 while the probability of it being irrational is 1.
Which is to say the naive image that we have of a set that is dense being a set that has no holes in it is utterly and totally wrong! The rational numbers are dense in the real number line but there are holes everywhere!!!
Originally posted by adam warlockI have always found that somewhat non-intuitive. There are infinitely more irrationals than rationals, yet between every two irrationals there is a rational. Intuitively one would think that either rationals and irrationals alternate, or there are some irrationals side by side without rationals in between them, but neither of these is true.
Which is to say the naive image that we have of a set that is dense being a set that has no holes in it is utterly and totally wrong! The rational numbers are dense in the real number line but there are holes everywhere!!!
Originally posted by twhiteheadOne has to truly see it to believe it, since this result is highly non-intuitive at all levels
I have always found that somewhat non-intuitive. There are infinitely more irrationals than rationals, yet between every two irrationals there is a rational. Intuitively one would think that either rationals and irrationals alternate, or there are some irrationals side by side without rationals in between them, but neither of these is true.
This reminds me of an old tale about a man that planned his suicide (don't ask me why). He was a very organized man so he planned his last days so that after his death there would be no loose end and he'd take care of everything he needed to take care of. He even had his hour of death perfectly scheduled.
In his last days he was even more efficient and he was able to take care of his business hours before schedule. With some time to spare he decided to go to a library and to immerse himself in reading. Since he wasn't very acquainted with mathematics he decided to pick up Euclid's Elements. After a reading for a bit he was surprised by a proposition whose validity e doubted. Such an absurd result had to be false. After reading the proof of the proposition and understanding it he had no option but to accept it. After a few more pages there was in front of him another result that had to be false. In no way could such a contrived result be true. But it was. After a few more experiences were is naive intuition was disproved by the cold sobriety of mathematical rigour he just had to keep on reading.
When the good man decided to take a break he noticed that his suicide time was past due. Merrily the man decided to keep on going and not to commit suicide. Mathematics saved his life and if I recall correctly after that the man decided to study mathematics more seriously (I'll have to get the book where I read that).
That long story of mine has point: the point is that my love affair with mathematics is feed by two main emotions:
1 - The feeling of finding about new things by myself (it doesn't matter if
other already did that before me, what matters is that I did it by myself)
2 - Being constantly humbled by results that go way beyond my naive and simple imagination. Time and time again Physics and Mathematics force me to look deeper, broader and farther than what I'd have looked by self and that experience is just exhilarating.
Originally posted by twhiteheadYeah, that one does my head in too.
I have always found that somewhat non-intuitive. There are infinitely more irrationals than rationals, yet between every two irrationals there is a rational. Intuitively one would think that either rationals and irrationals alternate, or there are some irrationals side by side without rationals in between them, but neither of these is true.
Originally posted by FabianFnasI disagree, if someone knows, or thinks of, another proof then it might be fun to see it. There are over 370 proofs of Pythagoras' theorem from the other Euclidean axioms. Although it is neater to use it as an axiom and drop the parallel lines axiom.
And by this proof we have concluded the fact. Nothing more about 0.999... = 1 need to be said.