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

On Thu, 14 Feb 2008 12:45:37 -0500, "Jesse F. Hughes"
<jesse@xxxxxxxxxxxxx> wrote:


Let x, y be any sets, then (the _same_ function S as above):

y is in S[x] iff there is some n such that for all m > n
y is in S_m[x]

This amounts to the _claim_

S[x] = lim S_n[x].
n

(in this case).

Proof
-----
As an exercise for the interested reader ;-)

Bull***.

Indeed.


I've already said that this is false, so how do you dare omit the
proof?

In <87d4r3r3ur.fsf@xxxxxxxxxxxxx>, I explicitly anticipated this
argument.

Here's a simple argument using the "language of limits" - why not use
this beautiful tool?

Let's consider the (special) case with x = N. Then Han's claim is

S[x] = lim S_n[N].
n

Now (S_n[N]) is a _decreasing_ sequence of sets. (Shown by Han.) Then we
have a nice theorem (already mentioned in this thread) that in this case
_
lim S_n[N] = | | S_n(N).
n n
_
And it's easy to prove/show that | | S_n(N) = {}.
n
HENCE

S[N] =/= lim S_n[N]
n

(since S[N] = {N} - also already shown by Han).

And hence -contrary to Han's claim- it's NOT the case that for every x:

S[x] = lim S_n[x].
n


F.

--

E-mail: info<at>simple-line<dot>de
.

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