Re: when does induction fail? well-ordering property?

Well-ordering leads to problems, for example in well-ordering the
reals.

A well-order is a binary relation, that is in computer talk a function
with two arguments, that returns less than, or greater than or equal.
It might instead admit greater than, or less than or equal. Then, also
there exists an element for the set that is less than, for any other
element of the set. The symbol for that ordering is similar to the <
angle bracket, but the sides of the two legs of the triangle are curved
in, concave, say <{ in ASCII, it's a different symbol to describe less
than in terms besides the normal ordering, of for example numbers.

http://en.wikipedia.org/wiki/ASCII

In ZFC, the Zermelo Fraenkel, for Ernst Zermelo and A.A. Fraenkel, set
theory with the axiom of Choice or AC, ZF+C, or ZFC, a theorem of that
axiom set is that any set is well-orderable, that there exists a choice
function for a set, and for any subset of the set. The choice function
basically returns the least element in a well-ordering for the set.

Now, imagine the open interval of the reals (0,1). The endpoints would
be zero and one but that's an open interval so they're not elements of
the set of each real number r for 0 < r < 1. While that is so, the
interval has two endpoints. How can that endpoint be addressed? If
there's a well-ordering of the reals, then for some subset of R(0,1),
the choice functions returns an endpoint. If it doesn't, then it's not
a well-ordering of the reals, because for no subset would it be the
least element. The element exists, because it's a complete ordered
field.

To thus address it demands another way of looking at the reals, which I
consider to be similar to Leibniz's infinitesimals, integral
iota-multiples, as well being the complete ordered field, Tim's
pseudo-reals, R-bar-umlaut, dually complete ordered field and partially
ordered ring, in the real numbers.

I can tell you, that's _not_ a standard viewpoint, and many do not
accept it. Care, I do not.

Ross

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