Re: 1-1/2+1/3-1/4+1/5-1/6+1/7

On Jan 16, 12:56 pm, tommy1729 <tommy1...@xxxxxxxxx> wrote:
now the set containing N but not "Square" is L
rearrange the set N : containing both L and "Square" elements.
this set contains all non-negative integers and is called T.
T = [0,1,4,9,16,25,...,2,3,5,6,7,8,10,...]
now since the "one to one correspondance" we can count oo long for the squares and thus the limit of our series can skip the non-square elements.

OK, I think I see what the confusion is now.

Here tommy1729 is making an assumption which, while admittedly
intuitive, is false in ZFC. The assumption is that the composition of
infinitely many permutations must itself be a permutation. While it
is definitely true that the composition of finitely many permutations
must be a permutation, it's not true if there are infinitely many.

Since tommy1729 keeps bringing up Hilbert's Hotel I'll do likewise.

Here is a restatement of tommy1729's question in terms of this
hypothetical hotel. Suppose all the rooms are occupied, and the
guest in room n has 1/n dollars. (For this question, just ignore the
fact that a real dollar isn't divisible when n > 100.) Is there a way
to
rearrange the guests in the rooms so that the total amount of
money in the hotel is finite?

The standard answer is that since the harmonic series diverges,
there must always be an infinite amount of money in the hotel.

But tommy1729 argues that it's possible. For he argues that we
can rearrange the guests as follows: At 1/2^n seconds before
noon (shades of the ball/vase problem again?!?) the guest with
1/n^2 dollars moves down to room n, and all the guests whose
room numbers are between n and the one previously occupied
by the 1/n^2-dollar guest move up a room. Then the guest in
room n at noon has exactly 1/n^2 dollars. Add it all up, and the
total is exactly zeta(2) (= pi^2/6) dollars, so it's now finite.

So who's right? It's vague, because it depends on an exact
definition of "rearrangement."

If an infinite composition of permutations is allowed to be a
"rearrangement," then tommy1729 is right, since even a
standard mathematician can't deny that there's now only
pi^2/6 dollars in the hotel.

But if only permutations are "rearrangements," then we must
consider tommy1729 to be wrong. In particular, he mentions
"Riemann's series theorem." I'm not familiar with the exact
wording of the theorem, but I bet that only permutations are
allowed as rearrangements, not infinite compostiions thereof.

than answer the simply question : at what finite position does 2 occur in the set T ?
if you answer oo ; thats not finite , and thus you admit i am correct.
else im curious what number you will come up with ...

If f:N -> T is a bijection (a permutation), then f^o-1 (the inverse
function) must exist. So in particular, f^o-1(2), the inverse of
f evaluated at 2, must exist. If tommy1729 cannot define
f^o-1(2), then he must admit that f is not a bijection.
.