This is making me go 1N54N3!!

**ThePerfectHacker** · February 8th 2006, 06:47 PM

I already asked the following 2 question but they were not answer thus I shall ask them again.

1)Are the cardinal numbers countable?

2)Are the cadinal numbers dense; meaning, for $\aleph_n<\aleph_m$ there exists a $\aleph_p$ such as $\aleph_n<\aleph_p<\aleph_m$ . For example, the rationals and the reals are dense? (Perhaps this is connected to the continuum hypothesis?)

3)Finally we get to a question to is making me go insane! I was thinking about this when I was falling asleep. Consider the set of all FINITE sets. What is the cardinality of this set?!?! I was able to prove (although not formally but you can consider it to be a proof) with the property of the power set, that I can make the cardinality of this set as large as I like!!!!! Thus, there is no cardinal number for this set!!!! When I was falling asleep a solution entered my mind. Who says that any infinite set must have a cardinal number? Perhaps, that reasoning is not true. And this is such a case. Thus, I decided to this the "super-cardinal number". Just as a cardinal number always excedes any natural number (cardinality of finite sets) so too the supercardinal number excedes any cardinal number (cardinality for SOME infinite sets). What happened?

**CaptainBlack** · February 8th 2006, 10:53 PM

Originally Posted by ThePerfectHacker

3)Finally we get to a question to is making me go insane! I was thinking about this when I was falling asleep. Consider the set of all FINITE sets. What is the cardinality of this set?!?!

Try something simpler. What is the Cardinality of all sets with ONE element?

RonL

**ThePerfectHacker** · February 9th 2006, 12:41 PM

I did, and I approach to the same conclusion. This set can be made larger than any cardinal number!

Did I find something new?

**CaptainBlack** · February 9th 2006, 12:50 PM

Originally Posted by ThePerfectHacker

I did, and I approach to the same conclusion. This set can be made larger than any cardinal number!

Did I find something new?

I'm being elliptic. I promise to be more direct in future.

When I ask about the cardinality of all sets with one element I am trying
to get you to think about "sets of what?". I want you to do this because
I suspect you have in mind something like a set of all sets.

The reason this is something to think about is because such an entity is
not well defined or unparadoxical.

RonL

**albi** · February 9th 2006, 12:57 PM

Originally Posted by ThePerfectHacker

I already asked the following 2 question but they were not answer thus I shall ask them again.

1)Are the cardinal numbers countable?

2)Are the cadinal numbers dense (...)

3)Finally we get to a question to is making me go insane! I was thinking about this when I was falling asleep. Consider the set of all FINITE sets. What is the cardinality of this set?!?! I was able to prove (...) that I can make the cardinality of this set as large as I like!!!!!(...)

Ad 2. It is the continuum hypothesis (generalised). It states that there is no cardinal number between $\aleph_n$ and $2^{\aleph_n}$ thus the set of cardinal numbers is not dense.

Ad 3. Yes if we assume that generalised continuum hypothesis is true. Let us imagine sequence that it's first element is 0, the second element is $\aleph_0$ , the third 1, the fourth $2^{\aleph_0}$ and so on... it is
$ x_1 = 0, \quad x_2 = \aleph_0 $

$ x_n = x_{n-2}+1 \quad for \quad n = 2k+1 > 2 $

$ x_n = 2^{x_{n-2}} \quad for \quad n = 2k > 2, $

$ where \quad k \in \mathbb{N} $
Due to continuum hypothesis we can't miss any infinite cardinal, and because natural numbers are countible we can't miss them either.

Ad 3.
And it is very interesting. What is that proof?
Just as the set of all sets does not exist, maybe the set of all finite sets also. But I cannot imagine the proof of this fact is like this one Russel did. Maybe the fact that it does not really have cardinallity could be such proof. Interesting.

**albi** · February 9th 2006, 01:01 PM

Hmm... Looks like you are already at this forum. Greetings to you

**ThePerfectHacker** · February 9th 2006, 05:44 PM

My proof is not going to be very formal (and I do not need the countinuum hypothesis

) I am going to be working with a set of sets having single elements. This is what CaptainBlack proposed because I think it would be easier.

Let $\mathbb{Z^+}=\{1,2,3,...\}$ the natural numbers. Thus, $P(\mathbb{Z^+})$ is larger (the property of a power set). Thus, $\bigcup\{x\} ,x\in P(\mathbb{Z^+})$ having the cardinality greater than $\aleph_0$ . But the previous demonstration is the the uninon of all sets having single elements. Further, by the properties of the power set we have that $P(P(\mathbb{Z^+}))$ is even larger. Thus, $\bigcup\{x\},x\in P(P(\mathbb{Z^+}))$ is the union of sets having a single element. This is even larger then the previous one. Thus, using this construction we can construct the cardinality of "the sets of all finite sets" to be as large as we like. Thus there is no such cardinal number. Now the conclusion I draw from this that some sets can be so large that you cannot use cardinals numbers anymore, and thus there is no problem with this set.

I asked some people about this problem, and they have no idea. I hope someone finds a solution to my problem. Perhaps, the problem is that no one ever considered this set.

CaptainBlack, you mention a "well-defined set" I heard this term many times before and wanted to ask what it mean? Further I happen to know that the "Super-set" the set of all sets is not well defined and thus we cannot use it. Further, my set is the set of all FINITE sets and it itself is infinite, thus it is not an element of itself. I see no problem with my set.

**CaptainBlack** · February 9th 2006, 10:36 PM

Originally Posted by ThePerfectHacker

My proof is not going to be very formal (and I do not need the countinuum hypothesis

) I am going to be working with a set of sets having single elements. This is what CaptainBlack proposed because I think it would be easier.

Let $\mathbb{Z^+}=\{1,2,3,...\}$ the natural numbers. Thus, $P(\mathbb{Z^+})$ is larger (the property of a power set). Thus, $\bigcup\{x\} ,x\in P(\mathbb{Z^+})$ having the cardinality greater than $\aleph_0$ . But the previous demonstration is the the uninon of all sets having single elements. Further, by the properties of the power set we have that $P(P(\mathbb{Z^+}))$ is even larger. Thus, $\bigcup\{x\},x\in P(P(\mathbb{Z^+}))$ is the union of sets having a single element. This is even larger then the previous one. Thus, using this construction we can construct the cardinality of "the sets of all finite sets" to be as large as we like. Thus there is no such cardinal number. Now the conclusion I draw from this that some sets can be so large that you cannot use cardinals numbers anymore, and thus there is no problem with this set.

I asked some people about this problem, and they have no idea. I hope someone finds a solution to my problem. Perhaps, the problem is that no one ever considered this set.

CaptainBlack, you mention a "well-defined set" I heard this term many times before and wanted to ask what it mean? Further I happen to know that the "Super-set" the set of all sets is not well defined and thus we cannot use it. Further, my set is the set of all FINITE sets and it itself is infinite, thus it is not an element of itself. I see no problem with my set.

Suppose such a set of all sets exists, then each of its subsets is an element
of it, hence it contains its Power Set. Hence the Cardinality of its Power
Set is less than or equal to its Cardinality.

But you will have seen a proof that the Cardinality of the Power Set of a Set
is strictly greater than the cardinality of the Set itself - Contradiction

RonL

**albi** · February 10th 2006, 01:39 PM

So, we have all to prove that the set of all one-element sets does not exist. Let $\mathfrak{U}_1$ denote the set of all one-element sets.
Using Cantor theorem we have that the cardinality of $\mathcal{P}(\mathfrak{U}_1)$ is greater than the cardinality of $\mathfrak{U}_1$ .

Suppose that $X \in \mathcal{P}(\mathfrak{U}_1)$ The set {X} satisfies ${X} \in \mathfrak{U}_1$ by the definition thus there exist an injection from $\mathcal{P}(\mathfrak{U}_1)$ to $\mathfrak{U}_1$ . This way: $card (\mathcal{P}(\mathfrak{U}_1)) \leq card( \mathfrak{U}_1)$ . It is in contradiction with Cantor theorem thus the set $\mathfrak{U}_1$ does not exist.

I've heared that there is the sollution of your problem. The set of all sets does not exist, however there exists the class of all sets. The class theory was founded by von Neumann and Bernays. In thirties Kurt Godel have created axiomatic system of class theory. More details I don't know.

**ThePerfectHacker** · February 13th 2006, 02:30 PM

Originally Posted by CaptainBlack

Suppose such a set of all sets exists, then each of its subsets is an element
of it, hence it contains its Power Set. Hence the Cardinality of its Power
Set is less than or equal to its Cardinality.

But you will have seen a proof that the Cardinality of the Power Set of a Set
is strictly greater than the cardinality of the Set itself - Contradiction

RonL

AH!!!!!!
But you are assuming that it has a CARDINALITY!!!!
Not necessarily it has a cardinal number, did Canotr prove that each infinite set has a cardinal number???
I demonstrated that this is false by construction the set of all finite sets.

**ThePerfectHacker** · February 13th 2006, 02:35 PM

Originally Posted by albi

So, we have all to prove that the set of all one-element sets does not exist. Let $\mathfrak{U}_1$ denote the set of all one-element sets.
Using Cantor theorem we have that the cardinality of $\mathcal{P}(\mathfrak{U}_1)$ is greater than the cardinality of $\mathfrak{U}_1$ .

Suppose that $X \in \mathcal{P}(\mathfrak{U}_1)$ The set {X} satisfies ${X} \in \mathfrak{U}_1$ by the definition thus there exist an injection from $\mathcal{P}(\mathfrak{U}_1)$ to $\mathfrak{U}_1$ . This way: $card (\mathcal{P}(\mathfrak{U}_1)) \leq card( \mathfrak{U}_1)$ . It is in contradiction with Cantor theorem thus the set $\mathfrak{U}_1$ does not exist.

I've heared that there is the sollution of your problem. The set of all sets does not exist, however there exists the class of all sets. The class theory was founded by von Neumann and Bernays. In thirties Kurt Godel have created axiomatic system of class theory. More details I don't know.

How can a set not exists

I was reading on Wikipedia that first we have Naive set theory which rested on intuition then Axiomatic set theory came along and removed all the paradoxes in set theory. Thus, I understand that a set (even though undefined) must satisfy some condition to be a set, right? Thus, my set does not satisfiy that.

Important, you set my problem was the set of all sets, and I know that no such set exists (called superset), but my problem is more reasonable that it is a set of all finite sets. Again, I am not speaking about the Superset overhere, taking about a completely different set.

**CaptainBlack** · February 14th 2006, 04:50 AM

Originally Posted by ThePerfectHacker

AH!!!!!!
But you are assuming that it has a CARDINALITY!!!!
Not necessarily it has a cardinal number, did Canotr prove that each infinite set has a cardinal number???
I demonstrated that this is false by construction the set of all finite sets.

No I'm not, Cardinality denotes a relationship between sets.

We say set $A$ has a greater than or equal cardinality to set
$B$ if there exists a one-one mapping from $B$ into $A$ .

We say set $A$ has a cardinality equal to that of a set $B$ if there
exists a one-one mapping from $B$ into $A$ , and there exists a
one-one mapping from $B$ into $A$

The proof that the Cardinality of the Power Set of a set is strictly greater
than the cardinality of the set itself is a proof the there is a one-one
mapping form the set into the Power Set, but that there is no such one-one
mapping from the Power Set into the Set itself.

Thus our contradiction stands.

RonL

(note into in this context includes onto)

**albi** · February 14th 2006, 08:51 AM

Originally Posted by ThePerfectHacker

How can a set not exists

Important, you set my problem was the set of all sets, and I know that no such set exists (called superset), but my problem is more reasonable that it is a set of all finite sets. Again, I am not speaking about the Superset overhere, taking about a completely different set.

Read again. I'm not talking about universum (or superset as you are saying). I am talking about $\mathfrak{U}_1$ , it is the set of all ONE-ELEMENT sets.

Originally Posted by ThePerfectHacker

AH!!!!!!
(...), did Canotr prove that each infinite set has a cardinal number???

I don't know what Cantor was thinking about cardinal numbers. However I have seen two (similar!) attitudes to this problem in my Set Theory book.

We can simply introduce an axiom that every set has cardinal number. (For every set there exists cardinal number such as $card(A) = card(B)$ if and only if $A$ has the same number of elements as $B$ )

The second one was to introduce something called "relation type" (it is the direct translation from Polish, I was unable to find anyting about it in Wikipedia or MathWorld).

Axiom. For any relation system $<A, \mathcal{R}>$ , $\mathcal{R} \subset A \times A$ there exists "relation type".
Two relation systems $<A, \mathcal{R}>$ and $<B, \mathcal{S}>$ have the same "relation type" iff they are isomorphic.

The cardinal number of the set $A$ is "relation type" of the system $<A, A \times A>$ .

**ThePerfectHacker** · February 14th 2006, 02:23 PM

Originally Posted by CaptainBlack

No I'm not, Cardinality denotes a relationship between sets.

We say set $A$ has a greater than or equal cardinality to set
$B$ if there exists a one-one mapping from $B$ into $A$ .

We say set $A$ has a cardinality equal to that of a set $B$ if there
exists a one-one mapping from $B$ into $A$ , and there exists a
one-one mapping from $B$ into $A$

The proof that the Cardinality of the Power Set of a set is strictly greater
than the cardinality of the set itself is a proof the there is a one-one
mapping form the set into the Power Set, but that there is no such one-one
mapping from the Power Set into the Set itself.

Thus our contradiction stands.

RonL

(note into in this context includes onto)

Ah, but then how did Cantor proof the the fundamental property of the power set if it fails in this case with the Superset!!!???

**CaptainBlack** · February 14th 2006, 09:18 PM

Originally Posted by ThePerfectHacker

Ah, but then how did Cantor proof the the fundamental property of the power set if it fails in this case with the Superset!!!???

Look at the ZF axioms. The axiom of regularity effectivly rules that
your "set of all sets" is not a set in ZF set theory.

RonL

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