Deeper than primes

[doronshadmi]

No amount of points along a line completely covers it !

Any attempt to completely cover an infinitely long straight line by points, is doomed to fail, because:

1) The all pairs of points (marked by red points in the following diagram) along an infinity long straight line are the result of circles with unique curvature degrees, as shown in the following diagram:

2) Being a circle is based on a measurable constant, known as pi=circumference/diameter.

3) By fact (2) the common center point of the circles along the line, is inaccessible to the set of all circles along the line, because pi does not exist at the center point.

4) By fact (2) the common infinitely long straight line along the set of all circles is inaccessible to this set, because pi does not exist at the straight line state.

 

[laca]

Any attempt to completely cover an infinitely long straight line by points, is doomed to fail, because: <snipped garbage>

Oh my, you're back to that again? Please, do show a part of a line where there is no point.

 

[zooterkin]

3) By fact (2) the common center point of the circles along the line, is inaccessible to the set of all circles along the line, because pi does not exist at the center point.

So you're saying that there is no point at the common centre point of the circles?

 

[doronshadmi]

So you're saying that there is no point at the common centre point of the circles?

No, I say that this point is inaccessible to the set of circles (and theirs associated points along the line) which are based on the fact that in order to be considered as a circle, pi must be a constant property of the considered object.

As for the question about a part of a line that is not covered by points, the answer is very simple:

Any arbitrary pair of points is associated with the collection of all circles, which can't completely cover an infinitely long straight line, as clearly shown in http://www.internationalskeptics.com/forums/showpost.php?p=6985580&postcount=14581 .

 

[doronshadmi]

http://www.mathacademy.com/pr/prime/articles/cantor_theorem/index.asp is a clear example of Cantor's theorem as a proof by contradiction, which leads to contradiction if one tries to define mapping between an explicit P(S) member and S member, because of the construction rules of the explicit P(S) member (the member of S must be AND can't be a member of the explicit P(S) member, according to the construction rules of the explicit P(S) member, under Cantor's theorem).

Also since P(S) is not less then S (because it is trivial to show that all S members (for example {a,b,c,d,...}) are at least mapped with {{a},{b},{c},{d},...} P(S) members), then by using this fact and the contradiction shown above, one must conclude that P(S) is a larger set than S.

----------------------------------------------

But this is not the only way to look at this case, for example, we are using Cantor's construction method to systematically and explicitly define P(S) members, for example:

By using the trivial mapping between {a,b,c,d,...} S members and P(S) {{a},{b},{c},{d},...} P(S) members, we explicitly define P(S) member {}.

Also by using the mapping between {a,b,c,d,...} S members and {{},{a},{b},{c},...} P(S) members, we explicitly define P(S) member {a,b,c,d,...}.

Actually by using Cantor's construction method independently of Cantor's theorem, we are able to explicitly define the all P(S) members between {} and {a,b,c,d,...}.

As a result, there is a bijection between S and P(S) members, as follows:

a ↔ {}
b ↔ {a,b,c,d,...}
c ↔ some explicit P(S) member, which is different than the previous mapped P(S) members

...

etc. ... ad infinitum.

Please be aware of the fact that this construction method has nothing to do with Cantor's theorem exactly because the construction is used independently of Cantor's theorem and therefore it is not restricted to the logical terms of Cantor's theorem.

jsfisher does not understand the mapping between the infinite sets S and P(S) as a bijection between a set and its proper subset, because he gets this subject only in terms of Cantor's theorem and its logical conditions, which ,again, have nothing to do with Cantor's construction method of explicit P(S) members and their bijection with S (which is a proper subset of P(S)) members.

 

[The Man]

They are all circles, yet the exact amount of them is unsatisfied (it is permanently changing), or in other words, the set of all circles has no exact size, which is a fact that you ignore and as a result you do not understand the inherent property of "+1" (permanently next ...) of the concept of cardinality.

Here is another example of the unsatisfied exact amount of an infinite set.

{...} is the set of all differences.

{...} is a member of {...}, such that {{...},...} where {...} is different than the rest ,... of {{...},...}, but then

{{...},...} is a member of {{...},...} such that {{{...},...},{...},...} where {{...},...} is different than the rest ,{...},... of {{{...},...},{...},...} but then

... etc. ... ad infinitum ... such that the size of the set of all differences is inherently unsatisfied.

So the set of all circles includes all of your circles? What do you mean "the exact amount"? All denotes, well, all, regardless of the amount, "exact" or otherwise.

 

[The Man]

Any attempt to completely cover an infinitely long straight line by points, is doomed to fail, because:

1) The all pairs of points (marked by red points in the following diagram) along an infinity long straight line are the result of circles with unique curvature degrees, as shown in the following diagram:

[qimg]http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg[/qimg]

2) Being a circle is based on a measurable constant, known as pi=circumference/diameter.

3) By fact (2) the common center point of the circles along the line, is inaccessible to the set of all circles along the line, because pi does not exist at the center point.

4) By fact (2) the common infinitely long straight line along the set of all circles is inaccessible to this set, because pi does not exist at the straight line state.

Points aren't circles. So your claim is now that an infinitely long straight line can't be completely covered by concentric circles? How about circles that aren't all, well, concentric?

 

[jsfisher]

Wrong, B is defined by the opposite terms of f(), such that "B exists AND f() does not exits" is a true proposition.

Doronetics may allow nonsense such as this, but Mathematics does not.

{ x ∈ A : x ∉ f(x) } lacks meaning if f() does not exist, just as { x ∈ A : x ∈ f(x) } lacks meaning when f() does not exist. Any expression that includes an undefined term is, well, undefined.

 

[epix]

Wrong, B is defined by the opposite terms of f(), such that "B exists AND f() does not exits" is a true proposition.

You do not understand this part ( http://en.wikipedia.org/wiki/Cantor's_theorem ):

and the fact that B is a placeholder for |P(A)| P(A) explicit members, that are mapped with explicit A members, which are members of A that is a proper subset of P(A), where both A and P(A) are infinite sets, and it is well-known that there is a bijection between infinite sets and the members of some of their proper subset.

This fact is shown between, for example, the set of natural numbers and its proper subset of, for example, even numbers,

This fact is shown between, for example, the set of rational numbers and its proper subset of natural numbers (the n/1 form).

This fact is shown between, for example, the set P(N) and its proper subset of N members, such that, by using Cantor's construction method, one explicitly defines (without exceptional) the P(N) members, which enables him\her to define the bijection between the explicit P(N) members, and the explicit N members, where N is a proper subset of P(N), and such a bijection is a fact among infinite sets and any arbitrary given proper subset of them.

Sure, there is a condition where it is impossible to point a finger at an instance where you have a member of a power set of an infinite set on the right side and a blank space on the left where the natural bijector should be without creating certain characteristic within the whole construct. Suppose that you ask a mathematician to show you a particular member of a power set for which there is no natural correspondent to complete the bijection. That's impossible to do. But there is a condition where it is possible and the condition depends on the organization of data in the power set. Let's start bijecting the members of N and P(N) which are organized in a comprehensive manner to rule out random distribution . . .

N............P(N)

1 <-----> {1}
2 <-----> {2}
3 <-----> {3}
4 <-----> {4}
5 <-----> {5}
6 <-----> {6}
.
.
.

After some 258 trillion years of successful bijecting, it will dawn on the thinkers, who subscribe to the school of parallel reasoning, that the chance that there will ever be a bijection between a natural number and a member of the power set that contains more than one element, such as {a,b}, is running rather slim due to the infinity of

{1}, {2}, {3}, {4}, {5}, {6}, . . .

In order to register the first instance of {#,#} = {1,2}, the last {#} must be known, which is not possible, coz there is no last {#} -- the sequence of the power set members containing just one element is infinite. So given the particular data organization, the power set is not denumerable, or is not countable. Can we conclude with

This is a contradiction[, and the existence of this contradiction shows that no element of X can be matched with this subset. Our match-up cannot be complete. And since we cannot make a one-to-one match-up between X and P(X), and since we already saw that P(X) cannot be smaller than X, the only possible conclusion is that P(X) must be larger than X. This completes the proof of Cantor’s Theorem.

???

The set of natural numbers features two well-known subsets: subset ODD and subset EVEN. In order to render N uncountable (say what?), the set is defined as

N = {1, 3, 5, 7, 9, . . .}

where the first member of the EVEN subset appears right after the sequence of the odd numbers ends. That will happen when Ruckenbaur III spells "Cantor" backward without making a mistake.

 

[doronshadmi]

Doronetics may allow nonsense such as this, but Mathematics does not.

{ x ∈ A : x ∉ f(x) } lacks meaning if f() does not exist, just as { x ∈ A : x ∈ f(x) } lacks meaning when f() does not exist. Any expression that includes an undefined term is, well, undefined.

EDIT:

jsfisher, if B (the member of P(S)) does not exist, Cantor's theorem can't provide a P(S) member, which is not mapped with some S member, and therefore it can't conclude that |P(S)| > |S|, so also by your reasoning (B exists) AND (f() as "only a mapping between S and P(S)", does not exist), such that there is f() that is not a mapping between S and P(S) (where in this case B is always an explicit member of P(S), which is a fact that Cantor's theorem ignores).

Please tell to the posters of this thread why do you ignore http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 ?

 

[doronshadmi]

So the set of all circles includes all of your circles? What do you mean "the exact amount"? All denotes, well, all, regardless of the amount, "exact" or otherwise.

The set has only circles and yet it is inherently and permanently changing, such that the smallest and the biggest circles do not exist NOW at present continuous state, and as a result the amount of such a set is unsatisfied NOW at present continuous state.

 

[doronshadmi]

Points aren't circles.

Yet, as I show, any possible pair of points along an infinitely long straight line is associated with a circle with unique curvature, where the collection of all circles can't completely cover an infinitely long straight line, and so is the set of all points that are associated to the set of all circles.

So your claim is now that an infinitely long straight line can't be completely covered by concentric circles? How about circles that aren't all, well, concentric?

Since the smallest circle does not exist, then also the set of all non-concentric circles can't completely cover an infinitely long straight line.

Furthermore, since a set is a collection of different members (any given member is different than the rest of the members), then the set of all concentric circles and the set of all non-concentric circles, is actually the same set.

 

[doronshadmi]

Sure, there is a condition where it is impossible to point a finger at an instance where you have a member of a power set of an infinite set on the right side and a blank space on the left where the natural bijector should be without creating certain characteristic within the whole construct. Suppose that you ask a mathematician to show you a particular member of a power set for which there is no natural correspondent to complete the bijection. That's impossible to do. But there is a condition where it is possible and the condition depends on the organization of data in the power set. Let's start bijecting the members of N and P(N) which are organized in a comprehensive manner to rule out random distribution . . .

N............P(N)

1 <-----> {1}
2 <-----> {2}
3 <-----> {3}
4 <-----> {4}
5 <-----> {5}
6 <-----> {6}
.
.
.

After some 258 trillion years of successful bijecting, it will dawn on the thinkers, who subscribe to the school of parallel reasoning, that the chance that there will ever be a bijection between a natural number and a member of the power set that contains more than one element, such as {a,b}, is running rather slim due to the infinity of

{1}, {2}, {3}, {4}, {5}, {6}, . . .

In order to register the first instance of {#,#} = {1,2}, the last {#} must be known, which is not possible, coz there is no last {#} -- the sequence of the power set members containing just one element is infinite. So given the particular data organization, the power set is not denumerable, or is not countable. Can we conclude with

???

The set of natural numbers features two well-known subsets: subset ODD and subset EVEN. In order to render N uncountable (say what?), the set is defined as

N = {1, 3, 5, 7, 9, . . .}

where the first member of the EVEN subset appears right after the sequence of the odd numbers ends. That will happen when Ruckenbaur III spells "Cantor" backward without making a mistake.

epix, it is well known that there is a bijection between a set and its proper subset, if the set is infinite, exactly as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6985470&postcount=14579 (which is a fact that is ignored by you).

 

[doronshadmi]

In http://www.tcs.tifr.res.in/~raja/publications/online/yapct08.pdf there are several ways to prove that |P(S)| > |S|, yet all of them ignore the fact that Cantor's construction method of P(S) members is not restricted to the logic that is used by each one of the given proofs.

 

[The Man]

The set has only circles and yet it is inherently and permanently changing, such that the smallest and the biggest circles do not exist NOW at present continuous state, and as a result the amount of such a set is unsatisfied NOW at present continuous state.

Changing? How? What specifically is changing about your set of all circles? What circle was not a circle before you changed your set of all circles to include that circle? "the smallest and the biggest circles do not exist NOW at present continuous state"? If that "present" "state" is "continuous" then it can't, well, change. You still simply have nothing to add to your set of all circles which will change that set "NOW" or at anytime.


Again the question was...

So the set of all circles includes all of your circles?

not 'does your set of all circles have only circles'?


Again as well...

So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?

 

[The Man]

Yet, as I show, any possible pair of points along an infinitely long straight line is associated with a circle with unique curvature, where the collection of all circles can't completely cover an infinitely long straight line, and so is the set of all points that are associated to the set of all circles.

Not all circles are concentric Doron, you have simply limited the circles you are considering to the set of all concentric circles and then seem stupefied (and expect us to be as well) as to why you can't pair the center point with another point on some circle. Please learn some basic geometry.

Since the smallest circle does not exist, then also the set of all non-concentric circles can't completely cover an infinitely long straight line.

Wrong, it has absolutely nothing to do with "the smallest circle". Try non-concentric circles of any fixed diameter (the distance between each of your pairs of points).

Furthermore, since a set is a collection of different members (any given member is different than the rest of the members), then the set of all concentric circles and the set of all non-concentric circles, is actually the same set.

Wow, that is really a bizarre claim even for you. So you don't understand the difference between concentric and non-concentric circles? Let me try to make it simple for you. The circles of the set of all concentric circles all have the same center point (that's what makes them, well, concentric). So the center point is specifically not an aspect by which any of the members of that set will differ from any other member of that set. However, the set of all non-concentric circles (even just all non-concentric circles of some fixed diameter) specifically only includes circles that do not have the same center point as any other circle in that set. So the center point is specifically an aspect by which all of the members of that set will differ from any other member of that set. You assert "since a set is a collection of different members (any given member is different than the rest of the members)" yet obviously and capriciously fail to take note of exactly where the members of each set differ from other members of that set and where they do not.

 

[jsfisher]

EDIT:

jsfisher, if B (the member of P(S)) does not exist, Cantor's theorem can't provide a P(S) member

Cantor's Theorem doesn't "provide a P(S) member". Cantor's Theorem establishes that |S| < |P(S)| for any set, S. The proof of the theorem relies on showing the very set you are so fascinated by does not exist.

 

[doronshadmi]

Not all circles are concentric Doron, you have simply limited the circles you are considering to the set of all concentric circles and then seem stupefied (and expect us to be as well) as to why you can't pair the center point with another point on some circle. Please learn some basic geometry.



Wrong, it has absolutely nothing to do with "the smallest circle". Try non-concentric circles of any fixed diameter (the distance between each of your pairs of points).




Wow, that is really a bizarre claim even for you. So you don't understand the difference between concentric and non-concentric circles? Let me try to make it simple for you. The circles of the set of all concentric circles all have the same center point (that's what makes them, well, concentric). So the center point is specifically not an aspect by which any of the members of that set will differ from any other member of that set. However, the set of all non-concentric circles (even just all non-concentric circles of some fixed diameter) specifically only includes circles that do not have the same center point as any other circle in that set. So the center point is specifically an aspect by which all of the members of that set will differ from any other member of that set. You assert "since a set is a collection of different members (any given member is different than the rest of the members)" yet obviously and capriciously fail to take note of exactly where the members of each set differ from other members of that set and where they do not.

The Man, once again your poor reasoning is demonstrated.

In this case you are unable to understand that, whether different circles are organized around a single point or not, these are the same circles, where their organization method is not significant.

Furthermore, the set of all different circles (which is a one and only one set, no matter how its members are organized) is incomplete exactly because it does not have the smallest or the biggest circle.

Indeed your poor reasoning has to be upgraded in order to deal with real Geometry or real Set theory.

 

[doronshadmi]

Cantor's Theorem doesn't "provide a P(S) member". Cantor's Theorem establishes that |S| < |P(S)| for any set, S. The proof of the theorem relies on showing the very set you are so fascinated by does not exist.

And in oder to establish that |S| < |P(S)| for any set, S, it must show an explicit P(S) member, which is not in the range of S.

 

[jsfisher]

And in oder to establish that |S| < |P(S)| for any set, S, it must show an explicit P(S) member, which is not in the range of S.

No, it does not. Even if I translate your gibberish into something meaningful like "an explicit P(S) member that is not in any mapping from S to P(S) must be shown", it is still wrong.

No explicit member is shown. All that is shown, all that needs to be shown is that for any mapping from S to P(S), any mapping at all, there must exist at least one element of P(S) not mapped by an element of S.