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Topic: Ordinal Proof
Replies: 5   Last Post: Jun 21, 2014 8:17 AM

 Messages: [ Previous | Next ]
 William Elliot Posts: 2,637 Registered: 1/8/12
Ordinal Proof
Posted: Jun 19, 2014 12:57 PM

Let S be a well ordered set.
Recursively define an ordinal valued function f, over S by
for all s in S, f(s) = { f(x) | x < a }

It's straight forward to show for all s, f(s) is hereditarily
transitive,
hence an ordinal;  likewise eta + { f(x) | x in S } is an ordinal.

By construction f:S -> eta is surjective.
Additionally f is increasing for if a < b, then f(a) proper subset f(b),
whence f(a) < f(b).
Thusly f:S -> eta is an order isomorphism.

eta is the unique ordinal order isomorphic to S for if
pi is order isomorphic to S, then pi and eta are order
isomorphic, hence equal.

Thus every well ordered set S, has a unique ordinal,
the order type of S, that's order isomorphic to S.

I feel uneasy about that simple ZF proof, that it's flawed.
Can I recursively construct a function into a class, ie
a function with values that satisfy a proposition, but
lacking a set for a codomain?

Hartog's lemma would provide such a set, but as it uses
order types, it's not available to demonstrate a codomain set.

It there a patch for the proof?

A surer proof would be to show if every initial segment of S
had an order type, then S would have an order type and forthwith
show if some element of S didn't have an order type, then
the least element of S without an order type, would have
an order type.  It however is three times as long.

Which proof would you prefer?

Date Subject Author
6/19/14 William Elliot
6/19/14 Herman Rubin
6/20/14 William Elliot
6/19/14 Robert E. Beaudoin
6/20/14 William Elliot
6/21/14 Robert E. Beaudoin