Re: Definition of finite.

On May 2, 6:34 pm, zuhair <zaljo...@xxxxxxxxx> wrote:

Even if you were right about my confusions, then they are non
important things, you should concentrate on the important aspects, not
the confusional aspects.

No, it is important not to incorrectly claim that Tarski proposed a
circular definition. Also, what is important to you (e.g., whether
certain formulations are equivalent) might not be important to other
people and other people may wish to comment on other matters that you
have mentioned or are related to your posts.

So, for example, here's a proof of an equivalence that is pertinent to
this discussion, whether or not it is important to you:

Def: conv(R) = {<c b> | <b c> e R}.

Theorem: x is finite <-> ER(R is a well ordering on x & conv(R) is a
well ordering on x).
Proof:
Suppose x is finite. So let f be a bijection from x onto a natural
number n.
Let R = {<b c> | bex & cex & f(b) e f(c)}.
So R is a well ordering on x and conv(R) is a well ordering on x.
Suppose ER(R is a well ordering on x & conv(R) is a well ordering on
x).
For all p in x, there exists a natural number k such that {y | <y p> e
R} is equinumerous with k, per proof by contradiction as follows:
Let q be the R-least member of x such that there is no natural number
k such that {y | <y q> e R} is equinumerous with k.
So q is not the least member of x, lest {y | <y q> e R} = 0 (thus {y |
<y q> e R} equinumerous with a natural number).
So q has R-predecessors.
So, since conv(R) is a well ordering on x, let g be the greatest R-
predecessor of q.
So there is a natural number j such that {y | <y g> e R} equinumerous
with j.
So let h be a bijection from {y | <y g> e R} onto j.
So hu{<g j>} is a bijection from {y | <y g> e R}u{g} onto j+.
But {y | <y g> e R}u{g} = {y | <y q> e R}, so we have a contradiction.
So, for all p in x, we have that {y | <y p> e R} is finite.
And. since every nonempty subset of x has a greatest member, every
subset of x is, for some p, a subset of {y | <y p> e R}u{p}.
So every subset of x is finite, so, since x itself is a subset of x,
we have that x is finite.

MoeBlee

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