Re: Non-Standard Analysis & Other Infinities

>> First, about the nested intervals, they are non-degenerate
>> closed intervals in the complete ordered field.
<snip>

You don't know that. Without the precise well ordering, you can't rule
out that your a_n's would like: 3, 3.1, 3.14, 3.141, 3.1415, etc. and
your b_n's look like 4, 3.2, 3.15, 3.142, 3.1416, etc. leading after a
countable number of steps the degenerate "closed interval" containing
just the single point of pi. (The point about using open intervals
guarantees that you do not ever have these single point "intervals".)

You also cannot rule out that there is an uncountable number less than
the size of the continuum, thanks to the proven udecidability of the
Continuum Hypothesis.

Remember that using Dedekind cuts, you can (with only a countable
number of entries on each side) uniquely determine any real number, of
which there are an uncountable number. An arbitrary real number of the
form 0.abcde... has only a countable number (Aleph_0) of decimal
places, but there are an uncountable number (10^Aleph_0 = c) of
possible reals.

Hope that helps,

Jonathan Hoyle
Eastman Kodak

.

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