Re: Logarithm of transfinite numbers

Randy Poe said:

Tony Orlow wrote:
Randy Poe said:
I think you mean all the finite substrings have a finite
value, but none of those is the string with a 1 in every
odd position.

If all those 1 bits are in finite positions, then each one has a finite value,
and is greater than the sum of the string to the right of it.

Yes, each of those substrings has a finite value.

None of those is the entire string.

TO logic:

1. Here I have a collection of things which all have
property P.

2. Here I have an object which we all agree is
not a member of the collection.

3. Therefore I conclude my object has property P.

How does not being a member of a collection lead you
to conclude that it has a property of things in the
collection? Why do you think 2 implies 3?

- Randy



Dear god, Randy, haven't I been through this with you enough already? Why do
you have to paint my logic to be so flakey? It's not. It's simple, and if you
don't get it, you're not paying attention.

If you start at 1 and go through the inductive set incrementally, at each
point, your max value n IS the size of the set, and sits in position n. Every
time you add the next value, the max increments, and the set size increments,
in tandem, so there is no point at which one is ever going to surpass the
other, and no point where the set size will become infinite, but there is no
infinite value in the set. It's that simple.

You all agree that no natural has an infinite number of predecessors, but still
insist that they have an infinite number of successors. But, if x has a
successor y an infinite number of iterations beyond it, then y has an infinite
number of predecessors between x and itself. So, no natural can have an
infinite NUMBER of successors, despite the fact that there is no last one.

What don't you get?
--
Smiles,

Tony
.

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