Re: Request for Review/Tutorage of Amateur Proofs

On Wed, 05 Mar 2008 10:41:31 +0100, G. Frege <nomail@invalid> wrote:


We have a similar result in real analysis. Consider the
sequence (x_n) defined with

x_n = 1/(n+1) (n e N).

Now even though this sequence is decreasing, i.e.

x_n+1 <= x_n for all n e N,
and
x_n =/= 0 for all n e N,

the limit of this sequence is = 0.

Analogous:

Let X = {x_i : i e N} with x_i = {x e N : x > i} (i e N).

(x_i) is a decreasing sequence of sets (Ai e N : x_i+1 c= x_i).

Then (according to Halmos) the limit of this sequence exists, and
_ _
lim x_i = | |x_i (= | |X).
i i

Hence (see below)

lim x_i = 0
i

in this case, though Ai e N : x_i =/= 0 (Ax e X : x =/= 0).

Theorem:

Let X = {x_i : i e N} with x_i = {x e N : x > i} (i e N).
_
Then | |X = 0.

Proof:
_
Assume for an arbitrary a: a e | |X. Then Ax e X: a e x, or An e N:

a e x_n. Hence especially a e x_0. Hence a e N. Hence a !e x_a. Hence

En e N: a !e x_n, or Ex e X: a !e x. This is equivalent to ~Ax e X:
_
a e x. Hence a !e | |X. Contradiction!
_ _
Hence a !e | |X (by RAA). Hence Ax(x !e | |X) (since a was arbitrary).
_
With other words, | |X = 0. []


F.

--

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

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