> > > The clusters of the original I_n intervals will continue to exist as
> > > clusters regardless. How you move around your A-B-intervals, will not
> > > change that.
> > There are no clusters.
> In order for there not to be any clusters, your I_n intervals would have
> to be pairwise disjoint, which can only happen if no interval of
> positive length contains more than one rational number.
Joint intervals form an interval. Under cluster we understood
sequences of disjoint intervals. If you like to call joint intervals
clusters, we can do so. But that does not change anything about my
proof.
> > There are points on the real axis. This proof
> > concerns geometry - nothing else.
> Your argument is far short of being a proof
The question is only whether a coincidence of endpoints is possible or
not. I do not see any obstacle. That results in a configuration having
all endpoints and only countably many intervals with and without
rationals.
> > It creates all endpoints of the I_n construction.
> But in the process, destroys all properties of the A_k-B_j ordering in
> order to create the I_n construction. There need not be any two of the
> I_n intervals having a common endpoint but seclusive intervals in your
> A_k-B_j sequence all do, there is no way to make one into the other
> without destroying the first's structure.
You are in error. I showed it. Look it up. The endpoints can move in a
completely arbitrary way. Nothing with the A-B-intervals changes. Here
you can see it again. Observe point 5:
> Which is hardly enough. Even all points being there, by having been
> moved discontinuously is not enough.
Look at the drawing above.
> > If there is anything else in the original
> > arrangement of the I_n, then it is not explicitly stated and does not
> > belong to mathematics.
> If it is deducible from the original arrangements by mathematicians,
> even if not by WM, then it does belong to mathematics, and is binding,
> even on non-mathematicians like WM.
There is nothing deductible from the original arrangement. I know it
because I composed it.
> The conversion from WM's A_K-B_j arrangement to his I_n arrangement
> has not been show capable of being done by only continuous deformations,
> so that none of what WM claims for it has been shown to hold.
If you have not yet understood, then look again at the drawing above.
> 1)
> Define a sequence of points p_n in the unit interval
> p_n = 1/n.
> These points define intervals
> A_k = [1/n, 1/(n+1)] for odd n
> and B_j = [1/n, 1/(n+1)] for even n.
> The intervals of sort A ==== and B ---- are alternating. If the points
> are denoted by n, we have something like the following configuration.
> ...7--6==5--4====3---------2===========1
> Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> number of intervals remains the same.
> ...7--6==5--3====4---------2===========1
> The intervals remain alternating. And in particular the number of
> intervals cannot increase.
From this Theorem we immediately get:
If any finite set of the p_n are moved then the
number of intervals does not increase.
Look! Over There! A Pink Elephant
If any set of the p_n are moved then the number
of intervals does not increase.
> > > > The clusters of the original I_n intervals will continue to exist as
> > > > clusters regardless. How you move around your A-B-intervals, will not
> > > > change that.
> > > There are no clusters.
> > In order for there not to be any clusters, your I_n intervals would have
> > to be pairwise disjoint, which can only happen if no interval of
> > positive length contains more than one rational number.
> Joint intervals form an interval. Under cluster we understood
> sequences of disjoint intervals. If you like to call joint intervals
> clusters, we can do so. But that does not change anything about my
> proof.
> > > There are points on the real axis. This proof
> > > concerns geometry - nothing else.
> > Your argument is far short of being a proof
> The question is only whether a coincidence of endpoints is possible or
> not.
In which case you can say nothing about anything but those points themselves.
While it is trivially possible to ennumerate the endpoints of the I_n intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the properties of those intervals, so that WM's enumeration also does prove change anything.
> I do not see any obstacle.
That WM does not see an obstacle is hardly proof that there is none.
> That results in a configuration having
> all endpoints and only countably many intervals with and without
> rationals.
That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves nothing.
> > > It creates all endpoints of the I_n construction.
> > But in the process, destroys all properties of the A_k-B_j ordering in
> > order to create the I_n construction. There need not be any two of the
> > I_n intervals having a common endpoint but cosecutive intervals in your
> > A_k-B_j sequence all do, there is no way to make one into the other
> > without destroying the first's structure.
> You are in error.
WRONG!
> I showed it.
Also WRONG!
> Look it up. The endpoints can move in a
> completely arbitrary way. Nothing with the A-B-intervals changes.
Everything of the A_B_ intervals changes or everything of the I_n intervals changes, or both.
In any case you have not, and cannot prove that "nothing changes", which proof is necessary before you can claim that nothing changes
> > Which is hardly enough. Even all points being there, by having been
> > moved discontinuously is not enough.
> Look at the drawing above.
> > > If there is anything else in the original
> > > arrangement of the I_n, then it is not explicitly stated and does not
> > > belong to mathematics.
> > If it is deducible from the original arrangements by mathematicians,
> > even if not by WM, then it does belong to mathematics, and is binding,
> > even on non-mathematicians like WM.
> There is nothing deductible from the original arrangement. I know it
> because I composed it.
Your claim of knowledge about things you claim to have constructed is not evidence and is not proof.
Consideration of WM's misrepresentation of, say, complete infinite binary trees, justifies rejection of WM's claim to be able to claim things without actually proving them.
> > The conversion from WM's A_K-B_j arrangement to his I_n arrangement
> > has not been show capable of being done by only continuous deformations,
> > so that none of what WM claims for it has been shown to hold.
> If you have not yet understood, then look again at the drawing above.
I see it, but it proves nothing except that WM is a fairly sloppy typist.
All I see is that WM is able to swap the positions of two adjacent labels.
Since this, however often done. does not affect the order type of the A-B set of points (all of which orderings will be well-ordered, and not any other ordering) and the standard indexing of endpoints of the I_n are certainly not a well-ordering of their positions in [0,1].
Thus WM again claims what he would be incapable of proving even if it were true.
--
> On Jul 22, 6:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > Matheology § 092
> > My second best proof contradicting set theory
> > 1)
> > Define a sequence of points p_n in the unit interval
> > p_n = 1/n.
> > These points define intervals
> > A_k = [1/n, 1/(n+1)] for odd n
> > and B_j = [1/n, 1/(n+1)] for even n.
> > The intervals of sort A ==== and B ---- are alternating. If the points
> > are denoted by n, we have something like the following configuration.
> > ...7--6==5--4====3---------2===========1
> > Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> > number of intervals remains the same.
> > ...7--6==5--3====4---------2===========1
> > The intervals remain alternating. And in particular the number of
> > intervals cannot increase.
> From this Theorem we immediately get:
> If any finite set of the p_n are moved then the
> number of intervals does not increase.
That means the increase of intervals can be put into a sequence:
0, 0, 0, ....
that is 0 for every finite n.
> Look! Over There! A Pink Elephant
The limit of above sequence is uncountable.
No, I do not look over there but I use mathematics, in particular
analysis. The limit of a sequenece of zeros is zero.
By the way, why do you believe in the equinumerousity of N and Q?
For every finite initial segment we have equinumerousity. Look! Over
there!
Or why do you believe in uncountability?
For every finite initial segment of the list we can exclude the
presence of the diagonal number. Look! Over There!
Do you recognize that your argument disqualifies all of Cantor's?
On 10 Aug., 00:55, Virgil <vir...@ligriv.com> wrote:
> > The question is only whether a coincidence of endpoints is possible or
> > not.
> In which case you can say nothing about anything but those points
> themselves.
The A and B intervals show that the space between the points is filled
with countably many intervals.
> While it is trivially possible to ennumerate the endpoints of the I_n
> intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
> properties of those intervals, so that WM's enumeration also does prove
> change anything.
But there are the A-B-Intervals fitting just the same situation.
> That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
> nothing.
You forget that for every move of a point the number of intervals
remain the same.
> > > A_k-B_j sequence all do, there is no way to make one into the other
> > > without destroying the first's structure.
> > You are in error.
> WRONG!
Proof for one point:
...
7-6=5-----4=====3--------------------2================================1
...
7-6===5-4=====3--------------------2================================1
...
7-6=====4---5==3--------------------2================================1
...
7-6=====4---------3==5--------------2================================1
...
7-6=====4---------3===========2---------5==========================1
Induction proves this for every finite number of points. The increase
of intervals is always 0.
Analysis shows that the sequence of interval-increases
0, 0, 0, ... has the limit 0.
> > > The question is only whether a coincidence of endpoints is possible or
> > > not.
> > In which case you can say nothing about anything but those points
> > themselves.
> The A and B intervals show that the space between the points is filled
> with countably many intervals.
Your A and B intervals would require that every space between points be filled with only finitely many intervals, but the space between any two points is filled with infinitely many I_n intervals, and your moving one point at a time will never change that.
> > While it is trivially possible to ennumerate the endpoints of the I_n
> > intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
> > properties of those intervals, so that WM's enumeration also does prove
> > change anything.
> But there are the A-B-Intervals fitting just the same situation.
Note that every I_n interval overlaps infinitely many other I_n intervals in an interval whereas no A-B-interval overlaps any other A-B-interval.
Further, almost all I_n intervals have infinitely many predecessors (I_n intervals starting closer to 0, so no such interval can ever become one of your A-B-intervals by your one at a time endpoint swaps.
> > That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
> > nothing.
> You forget that for every move of a point the number of intervals
> remain the same.
And the number of I_n's unaccounted for remain the same, infinitely many of them. You one at a time processing never exhausts an infinite task.
After each of your "steps", as much remains to be done as before that step, so the job can never be diminished that way.
WM <mueck...@rz.fh-augsburg.de> wrote:
> On 9 Aug., 23:19, William Hughes <wpihug...@gmail.com> wrote:
> > On Jul 22, 6:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Matheology § 092
> > > My second best proof contradicting set theory
> > > 1)
> > > Define a sequence of points p_n in the unit interval
> > > p_n = 1/n.
> > > These points define intervals
> > > A_k = [1/n, 1/(n+1)] for odd n
> > > and B_j = [1/n, 1/(n+1)] for even n.
> > > The intervals of sort A ==== and B ---- are alternating. If the points
> > > are denoted by n, we have something like the following configuration.
> > > ...7--6==5--4====3---------2===========1
> > > Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> > > number of intervals remains the same.
> > > ...7--6==5--3====4---------2===========1
> > > The intervals remain alternating. And in particular the number of
> > > intervals cannot increase.
> > From this Theorem we immediately get:
> > If any finite set of the p_n are moved then the
> > number of intervals does not increase.
> That means the increase of intervals can be put into a sequence:
> 0, 0, 0, ....
> that is 0 for every finite n.
> > Look! Over There! A Pink Elephant
> The limit of above sequence is uncountable.
If uncountable , then not zero after all!
> No, I do not look over there but I use mathematics, in particular
> analysis. The limit of a sequenece of zeros is zero.
What WM claims is mathematics is not recognized as such by actual mathematicians anywhere outside of his own little world of WMytheology.
> By the way, why do you believe in the equinumerousity of N and Q?
> For every finite initial segment we have equinumerousity. Look! Over
> there!
If one can construct a bijection between two sets, as has often been done for N and Q, then they satisfy the standard mathematical definition of equinumerousity.
Equivalently if one can construct injections from each to the other, as has often been done for N and Q, then they also satisfy the standard mathematical definition of equinumerousity.
> Or why do you believe in uncountability?
Because we can find a set which allows injection of N into it but neither a surjection of N onto it nor an injection from it to N.
Which is the standard test for a set being uncountable.
At least outside of WMytheology.
> For every finite initial segment of the list we can exclude the
> presence of the diagonal number. Look! Over There!
Since your "over there" looks remarkably like WMytheology, I prefer to continue looking only at standard mathematics.
> Do you recognize that your argument disqualifies all of Cantor's?
I do not recognize as true in standard mathematics those things that are true, if at all, only in WM' WMytheology.
And WM has yet to show any part of Cantor's results to be false in standard mathematics, or, indeed, anywhere outside of his own WMytheology.
--
> On 9 Aug., 23:19, William Hughes <wpihug...@gmail.com> wrote:
> > On Jul 22, 6:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Matheology § 092
> > > My second best proof contradicting set theory
> > > 1)
> > > Define a sequence of points p_n in the unit interval
> > > p_n = 1/n.
> > > These points define intervals
> > > A_k = [1/n, 1/(n+1)] for odd n
> > > and B_j = [1/n, 1/(n+1)] for even n.
> > > The intervals of sort A ==== and B ---- are alternating. If the points
> > > are denoted by n, we have something like the following configuration.
> > > ...7--6==5--4====3---------2===========1
> > > Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> > > number of intervals remains the same.
> > > ...7--6==5--3====4---------2===========1
> > > The intervals remain alternating. And in particular the number of
> > > intervals cannot increase.
> > From this Theorem we immediately get:
> > If any finite set of the p_n are moved then the
> > number of intervals does not increase.
> That means the increase of intervals can be put into a sequence:
On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > That means the increase of intervals can be put into a sequence:
> > 0, 0, 0, ....
> > that is 0 for every finite n.
> Thus the limit of the number of changes is 0
> Look! Over There! A Pink Elephant
> Thus the number of changes of the limit is 0
There is no "change of limit".
There is the number of changes of the interval number for 1, 2,
3, ..., n, ... moving points.
And there is the limit of changes of the interval number, namely for
infinitely many moving points.
This sequence can be understood mathematically by so-called analysis:
The terms of the sequence consist of numbers and the sequence has a
number as its limit
0, 0, 0, ... --> 0
The limit of these numbers 0 is 0 and that is the number of the limit.
No escape possible.
On 10 Aug., 22:17, Virgil <vir...@ligriv.com> wrote:
> > The A and B intervals show that the space between the points is filled
> > with countably many intervals.
> Your A and B intervals would require that every space between points be
> filled with only finitely many intervals,
No. Only in the initial configuration the space between eps and 1 is
filled with finitely many A-B-intervals. But every space between two
points will, in the final configuration, contain infinitely many
endpoints and thus infinitely many intervals.
Compare Cantor's enumeration of the rationals: Up to every natural
number n, the number of rationals q_1 to q_n is finite.
> but the space between any two
> points is filled with infinitely many I_n intervals, and your moving one
> point at a time will never change that.
Why should I move one point at a time? I may employ infinitely many
ants.
Why should moving one point at a time never change that? Compare
Tristram Shandy. Compare the increasing skill of my ants, such that
the n-th move requires 1/2^n years only.
You should recognize that these arguments of yours are invalid. Try
again.
> Note that every I_n interval overlaps infinitely many other I_n
> intervals in an interval whereas no A-B-interval overlaps any other
> A-B-interval.
Why do you mention that? If infinitely many I_n overlap or include
each other, then in this union of intervals, there are infinitely many
endpoints of the A-B-intervals - but not a single irrational number of
the complement.
> Further, almost all I_n intervals have infinitely many predecessors (I_n
> intervals starting closer to 0, so no such interval can ever become one
> of your A-B-intervals by your one at a time endpoint swaps.
Again, we have an infinity of moves during the first year. And if that
turns out insufficient, then we have infinitely many years. Note: Only
the *possibility* that my ants will accomplish the required
configuration of endpoints invalidates the basis of set theory.
> After each of your "steps", as much remains to be done as before that
> step, so the job can never be diminished that way.
That is a very good argument. Why not apply it to Cantors counting of
rationals? Or to his "proof" of uncountability? But no simultaneity is
required. Therefore I have hired aleph_0 little ants. Each one is
instructed to put, on command, its point on an endpoint of the
intervals I_n. (You must know: According to Cantor set theory also
covers biology: Der dritte Theil bringt die Anwendungen der
Mengenlehre auf die Naturwissenschaften: Physik, Chemie, Mineralogie,
Botanik, Zoologie, Anthropologie, Biologie, Physiologie, Medizin etc.
Ist also das, was die Engländer „Natural philosophy" nennen. 1912,
Sept. 20, Cantor to Hilbert)
> But after each step there are still infinitely many steps to go, so the
> job remaining cannot even be diminished, much less completed.
On 10 Aug., 22:36, Virgil <vir...@ligriv.com> wrote:
> > By the way, why do you believe in the equinumerousity of N and Q?
> > For every finite initial segment we have equinumerousity. Look! Over
> > there!
> If one can construct a bijection between two sets, as has often been
> done for N and Q, then they satisfy the standard mathematical definition
> of equinumerousity
If equinumerousity exists by definition, then that is not enough to
imply bijectability?
If I define that there are aleph_0 endpoints of the I_n and aleph_0
endpoints of the A-B-intervals then that is not sufficient to prove
bijectability, i.e., that they can be paired?
So you insist that a bijection implies equinumerousity, but that
equinumerousity does not imply bijectability?
> I prefer to
> continue looking only at standard mathematics.
In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
> > > By the way, why do you believe in the equinumerousity of N and Q?
> > > For every finite initial segment we have equinumerousity. Look! Over
> > > there!
> > If one can construct a bijection between two sets, as has often been
> > done for N and Q, then they satisfy the standard mathematical definition
> > of equinumerousity
> If equinumerousity exists by definition, then that is not enough to
> imply bijectability?
Equinumerousity between two sets is only known to exist when a bijection between those sets has been proved to exist.
> If I define that there are aleph_0 endpoints of the I_n and aleph_0
> endpoints of the A-B-intervals then that is not sufficient to prove
> bijectability, i.e., that they can be paired?
Proof of equinumerousity does not prove anything else.
It does not prove anything about the order properties of the sets in question. And it does not prove that your alleged method of reordering of the endpoints of the I_n is possible.
> So you insist that a bijection implies equinumerousity, but that
> equinumerousity does not imply bijectability?
No I do not!
> > I prefer to
> > continue looking only at standard mathematics.
> In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
Which is totally irrelevant to WM's problem, of proving his false claims.
There is no way to rearrange the endpoints of the I_n to you A-B pattern without destroying what the I-n were designed to do.
Among other things: (1) the cumulative length of the I_n intervals is less than 1/9 where the cumulative length of the A-B intervals is 1.
(2) the midpoints, and very likely the endpoints, of the I_n intervals are dense in [0,1] but the A_B interval endpoints have only one accumulation point.
Yet WM claims that, by a sequence of operations each moving one point at a time or swapping the position of two points, he can convert one to the other. The problem being that such an infinite process always has infinitely farther to go before it gets any closer to being finished.
For example, in removing the rationals, one at a time, from [0,1], what is left after any finite number of steps remains dense in [0,1].
Thus the proper "limit", if any, would have to be a dense in [0,1].
> > > The A and B intervals show that the space between the points is filled
> > > with countably many intervals.
> > Your A and B intervals would require that every space between points be
> > filled with only finitely many intervals,
> No. Only in the initial configuration the space between eps and 1 is
> filled with finitely many A-B-intervals. But every space between two
> points will, in the final configuration, contain infinitely many
> endpoints and thus infinitely many intervals.
Claimed but not proven possible.
> Compare Cantor's enumeration of the rationals: Up to every natural
> number n, the number of rationals q_1 to q_n is finite.
> > but the space between any two
> > points is filled with infinitely many I_n intervals, and your moving one
> > point at a time will never change that.
> Why should I move one point at a time? I may employ infinitely many
> ants.
That is what you described doing!
> Why should moving one point at a time never change that? Compare
> Tristram Shandy. Compare the increasing skill of my ants, such that
> the n-th move requires 1/2^n years only.
> You should recognize that these arguments of yours are invalid. Try
> again.
No more invalid than your arguments.
> > Note that every I_n interval overlaps infinitely many other I_n
> > intervals in an interval whereas no A-B-interval overlaps any other
> > A-B-interval.
> Why do you mention that? If infinitely many I_n overlap or include
> each other, then in this union of intervals, there are infinitely many
> endpoints of the A-B-intervals - but not a single irrational number of
> the complement.
> > Further, almost all I_n intervals have infinitely many predecessors (I_n
> > intervals starting closer to 0, so no such interval can ever become one
> > of your A-B-intervals by your one at a time endpoint swaps.
> Again, we have an infinity of moves during the first year. And if that
> turns out insufficient, then we have infinitely many years. Note: Only
> the *possibility* that my ants will accomplish the required
> configuration of endpoints invalidates the basis of set theory.
> > After each of your "steps", as much remains to be done as before that
> > step, so the job can never be diminished that way.
> That is a very good argument. Why not apply it to Cantors counting of
> rationals? Or to his "proof" of uncountability? But no simultaneity is
> required. Therefore I have hired aleph_0 little ants. Each one is
> instructed to put, on command, its point on an endpoint of the
> intervals I_n. (You must know: According to Cantor set theory also
> covers biology: Der dritte Theil bringt die Anwendungen der
> Mengenlehre auf die Naturwissenschaften: Physik, Chemie, Mineralogie,
> Botanik, Zoologie, Anthropologie, Biologie, Physiologie, Medizin etc.
> Ist also das, was die Engländer “Natural philosophy" nennen. 1912,
> Sept. 20, Cantor to Hilbert)
> > But after each step there are still infinitely many steps to go, so the
> > job remaining cannot even be diminished, much less completed.
WM <mueck...@rz.fh-augsburg.de> wrote:
> On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > That means the increase of intervals can be put into a sequence:
> > > 0, 0, 0, ....
> > > that is 0 for every finite n.
> > Thus the limit of the number of changes is 0
> > Look! Over There! A Pink Elephant
> > Thus the number of changes of the limit is 0
> There is no "change of limit".
> There is the number of changes of the interval number for 1, 2,
> 3, ..., n, ... moving points.
> And there is the limit of changes of the interval number, namely for
> infinitely many moving points.
As soon as any one endpoint of the I_n has been moved, it is no longer known whether all rationals are still covered, at which point your argument fails.
> This sequence can be understood mathematically by so-called analysis:
> The terms of the sequence consist of numbers and the sequence has a
> number as its limit
> 0, 0, 0, ... --> 0
Actually the real terms of your A-B sequence are a sequence of intervals with consecutive endpoints and remain so forever, merely being renamed does not change that structure.
> The limit of these numbers 0 is 0 and that is the number of the limit.
> No escape possible.
There is no need to escape from a place one has never been and can never be.
--
> On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > That means the increase of intervals can be put into a sequence:
> > > 0, 0, 0, ....
> > > that is 0 for every finite n.
> > Thus the limit of the number of changes is 0
> > Look! Over There! A Pink Elephant
> > Thus the number of changes of the limit is 0
> There is no "change of limit".
> There is the number of changes of the interval number for 1, 2,
> 3, ..., n, ... moving points.
The phrase
"number of changes of the interval number" is nonsense.
There are a number of possible things you might mean by
this ambiguous phrase. Since a popular MO with you is to
use two different interpretations and claim a contradiction
I will not guess.
> On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> > > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > That means the increase of intervals can be put into a sequence:
> > > > 0, 0, 0, ....
> > > > that is 0 for every finite n.
> > > Thus the limit of the number of changes is 0
> > > Look! Over There! A Pink Elephant
> > > Thus the number of changes of the limit is 0
> > There is no "change of limit".
> > There is the number of changes of the interval number for 1, 2,
> > 3, ..., n, ... moving points.
> The phrase
> "number of changes of the interval number" is nonsense.
My construction has aleph_0 endpoints and aleph_0 intervals (of type A
and B) between them.
The number of changes of the number of intervals is zero for every
move of every endpoint in every direction.
> There are a number of possible things you might mean by
> this ambiguous phrase.
There is no number of things because every possible change is zero.
But your "Look over there is" is nonsense. If a sequences of sets
could increase in another manner than its cardinality, then we would
possible have for the set of initial segments (1, 2, ..., n) of the
natural numbers:
The limit oo of the sequence of cardinalities need not be identical to
the cardinality of the limit N of the sequence. There was no
justification at all that aleph_0 was the first transfinite
cardinality. It could as well be 2^2^2^aleph_0 or -10 or something
else..
> I will not guess.
You will stop with your silly "Look over there" until you think that
the serious objection against countability implied by your arguing
will have been forgotten, and then you will start again with that
nonsense.
> > > So you insist that a bijection implies equinumerousity, but that
> > > equinumerousity does not imply bijectability?
> > No I do not!
> > > > I prefer to
> > > > continue looking only at standard mathematics.
> > > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
> > Which is totally irrelevant to WM's problem, of proving his false claims.
> > There is no way to rearrange the endpoints of the I_n to you A-B pattern
> I need only equinumerousity to prove that a bijection is possible,
> because (few lines above) you answered the question:
> > So you insist that a bijection implies equinumerousity, but that
> > equinumerousity does not imply bijectability?
> by
> No I do not!
> > without destroying what the I-n were designed to do.
> The I_n remain what they are. There is nothing changed and, therefore,
> nothing destroyed.
Then also nothing newe is proved by WM's arguments.. The "equinummreoisity" or "bijectability" of the sets of points in question has effectively nothing to so with the order properties of those sets, since we already know, e.g., that the naturals and the rationals are equinumerous but are quite different as ordered sets
Note that each of the set of rationals in [0,1] and the set of irrationals in [0, 1] is dense in the other even though there is no bijection between them and thy are not equinumerous.
--
WM <mueck...@rz.fh-augsburg.de> wrote:
> On 11 Aug., 19:46, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> > > > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > That means the increase of intervals can be put into a sequence:
> > > > > 0, 0, 0, ....
> > > > > that is 0 for every finite n.
> > > > Thus the limit of the number of changes is 0
> > > > Look! Over There! A Pink Elephant
> > > > Thus the number of changes of the limit is 0
> > > There is no "change of limit".
> > > There is the number of changes of the interval number for 1, 2,
> > > 3, ..., n, ... moving points.
> > The phrase
> > "number of changes of the interval number" is nonsense.
> My construction has aleph_0 endpoints and aleph_0 intervals (of type A
> and B) between them.
> The number of changes of the number of intervals is zero for every
> move of every endpoint in every direction.
You are effectively arguing that all but one endpoint of one of the I_n intervals must be an upper endpoint of exactly one interval and the lower endpoint of another such interval, which is nonsense. or that their order properties are irrelevant, which mis even ore nonsensical.
> > There are a number of possible things you might mean by
> > this ambiguous phrase.
> There is no number of things because every possible change is zero.
The no changes have occurred, and your argument is nothing.
> But your "Look over there is" is nonsense. If a sequences of sets
> could increase in another manner than its cardinality,
It is not that the NUMBER of I_n intervals differs from the NUMBER of intervals A-B intervals, but that order properties of those two sets of intervals differ and that difference is not destroyed by your 'bijection'.
> The limit oo of the sequence of cardinalities
There are only two equal cardinalities involved, but their equality is irrelevant to other properties of the set of I_n intervals.
> You will stop with your silly "Look over there"
Wy should he stop when you are the one claiming that your argument proves something else quite different and irrelevant to it.
> until you think that
> the serious objection against countability implied by your arguing
> will have been forgotten, and then you will start again with that
> nonsense.
Countability is NOT the issue. The isue is whether countability proves anything else, and in this cans it does NOT prove that the set of I_n intervals behaves as WM claims it behaves.
WM wishes to claim that after removing the countably many I_n intervals, with I_n of length 1/10^n, from [0,1], one will have only countably many points remaining.
> > > > So you insist that a bijection implies equinumerousity, but that
> > > > equinumerousity does not imply bijectability?
> > > No I do not!
> > > > > I prefer to
> > > > > continue looking only at standard mathematics.
> > > > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
> > > Which is totally irrelevant to WM's problem, of proving his false claims.
> > > There is no way to rearrange the endpoints of the I_n to you A-B pattern
> > I need only equinumerousity to prove that a bijection is possible,
> > because (few lines above) you answered the question:
> > > So you insist that a bijection implies equinumerousity, but that
> > > equinumerousity does not imply bijectability?
> > by
> > No I do not!
> > > without destroying what the I-n were designed to do.
> > The I_n remain what they are. There is nothing changed and, therefore,
> > nothing destroyed.
> Then also nothing newe is proved by WM's arguments.
Nothing is new, but unknown to many.
> The
> "equinummreoisity" or "bijectability" of the sets of points in question
> has effectively nothing to so with the order properties of those sets,
Therefore it is clear that the number of intervals does not increase
when the endpoints are moved.
> since we already know, e.g., that the naturals and the rationals are
> equinumerous but are quite different as ordered sets
We know that there are as many intervals of sort A and B as there are
endpoints. The moving of endpoints does not change this
equinumerousity, does it?
> Note that each of the set of rationals in [0,1] and the set of
> irrationals in [0, 1] is dense in the other even though there is no
> bijection between them and thy are not equinumerous.
Since my proof shows the contrary, your claim is invalid.
> > > > > So you insist that a bijection implies equinumerousity, but that
> > > > > equinumerousity does not imply bijectability?
> > > > No I do not!
> > > > > > I prefer to
> > > > > > continue looking only at standard mathematics.
> > > > > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
> > > > Which is totally irrelevant to WM's problem, of proving his false > > > > claims.
> > > > There is no way to rearrange the endpoints of the I_n to you A-B > > > > pattern
> > > I need only equinumerousity to prove that a bijection is possible,
> > > because (few lines above) you answered the question:
> > > > So you insist that a bijection implies equinumerousity, but that
> > > > equinumerousity does not imply bijectability?
> > > by
> > > No I do not!
> > > > without destroying what the I-n were designed to do.
> > > The I_n remain what they are. There is nothing changed and, therefore,
> > > nothing destroyed.
> > Then also nothing newe is proved by WM's arguments.
> Nothing is new, but unknown to many.
> > The
> > "equinummreoisity" or "bijectability" of the sets of points in question
> > has effectively nothing to so with the order properties of those sets,
> Therefore it is clear that the number of intervals does not increase
> when the endpoints are moved.
But how do you move endpoints in the A-B intervals so as to leave uncovered all the points not covered by the I_n intervals? Unless you can show how to do that, your arguments fail.
> > since we already know, e.g., that the naturals and the rationals are
> > equinumerous but are quite different as ordered sets
> We know that there are as many intervals of sort A and B as there are
> endpoints. The moving of endpoints does not change this
> equinumerousity, does it?
Irrelevant! Unless you can show how to move your A_B endpoints so as to leave uncovered precisely those points left uncovered by the I_n intervals, your arguments fail.
And you certainly have not done so yet!
> > Note that each of the set of rationals in [0,1] and the set of
> > irrationals in [0, 1] is dense in the other even though there is no
> > bijection between them and thy are not equinumerous.
> Since my proof shows the contrary, your claim is invalid.
You "proof" is not a proof until you can at least show how to move your A_B endpoints so as to leave uncovered every point left uncovered by the I_n intervals
--
In article <ff5249cb-54cf-4274-8957-e7ef7331e...@n13g2000vby.googlegroups.com>,
WM <mueck...@rz.fh-augsburg.de> wrote:
> On 12 Aug., 20:47, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > The number of changes of the number of intervals is zero for every
> > move of every finite set of endpoints in every direction
> > Look! Over There! A Pink Elephant!
> If infinitely many points move (or if one point moves infinitely
> often?), the number of intervals-changes is not only infinite but
> uncountable.
> There are some rules of matheology that have to be learned peu a peu.
WW is the only one who can ever 'learn' any of the rules of his matheology, as he makes them up as he goes along.
Whereas re the rules of standard mathematics, WM does not even learn them peu a peu.
Note that WM's A-B intervals cover all but one point of [0,1], whereas his I_n intervals leave at least 8/9 of [0,1] uncovered, which even in WMs world works out to be at least countably many points uncovered..
And WM has not yet explained how his method of moving the endpoints of the A-B intervals can ever leave more than just one point of [0,1] uncovered, much less countably many, which is itself much less than 8/9 of the points in [0,1].
--
On Aug 13, 11:29 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 12 Aug., 20:47, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > The number of changes of the number of intervals is zero for every
> > move of every finite set of endpoints in every direction
> > Look! Over There! A Pink Elephant!
> > > The number of changes of the number of intervals is zero for every
> > > move of every endpoint in every direction.
> If infinitely many points move (or if one point moves infinitely
> often?), the number of intervals-changes is not only infinite but
> uncountable.
Your problem is that a change from your initial state
to your final state requires the movement
of an infinite number of points. A theorem that only deals with
the movement of finite numbers of points
(even if it deals with every finite number of points)
is not applicable.
