Re: Logarithm of transfinite numbers


Tony Orlow wrote:
Randy Poe said:

Tony Orlow wrote:
Randy Poe said:

matt271829-n...@xxxxxxxxxxx wrote:
What value, if any, can be ascribed to the logarithm of aleph_n?

If you define it, we can see if anything fits that definition.


Take log_2(aleph_0) as an example.

log_2(aleph_0) can't be finite, and it can't be bigger than aleph_0, so
it has to equal aleph_0.

You're drawing conclusions about something you haven't yet
defined.


But log_2(aleph_0) = aleph_0 implies 2^aleph_0 = aleph_0, whereas in
fact 2^aleph_0 = aleph_1.

Ah, so by "log_2(aleph_0)" you mean "the cardinality of a set
S such that |P(S)|, the cardinality of the powerset of S, is aleph_0".

No such S exists.

Where did it go wrong?

You haven't come up with a definition such that anything
meets that definition. There is no such thing as "log2(aleph_0)".

- Randy


That seems kinda weak, Randy, no offense. "Undefined". Hmph! Sorry.

We have the set of natural numbers, of size aleph_0. Let's represent each one
in binary. How many bits will we need to represent aleph_0 naturals? Well,
given N=S^L

And L is the thing that is undefined here. No L satisfies that.

, where N=aleph_0 and S=2, aleph_0=2^L, so L=log2(aleph_0), and that
is the length of binary strings you need to represent all the naturals.

No finite quantity is enough to represent all the naturals, so L is not
any finite value.

Who said L=log2(aleph_0) was finite?

There are two possibilities if it exists: Either it's finite, or it's
not finite.

The above is half of my argument, that it is not finite.

The other case is "not finite", and that is also not possible.

Therefore it doesn't exist.

I am not addressing my own theories here,
but the op's question,

If you're going to be introducing numbers that are neither finite
nor infinite, then you are indeed talking about your own screwball
theories here.

Is L infinite? Well, no, because the set of infinite-length binary
strings has more than aleph_0 elements.

The set of aleph_0-length strings contains 2^aleph_0 strings. That's simple
N=S^L combinatorics.

Except in your screwball theories, there is no "simple combinatorics"
of transfinite numbers.

There *is* however a definition of the notation 2^X as the
cardinality (by Cantor definition) of the powerset of a set of
cardinality X.

So far as I know, there is no other definition of the set of
symbols "2^X" when X is infinite.

if you say aleph_0 is the "smallest infinity", then you
may think that infinite strings can be no shorter than aleph_0.

You are unable to distinguish between "random neural firings
I had after three beers" and "rigorously proven", but it is
a rigorously established FACT of our axiomatic system that
there is no infinite set of cardinality < aleph_0.

But then the
question remains, how many bits are required to list aleph_0 naturals? Is it a
finite number, x?

As I already said, the answer to that question is no.

The 2^x=aleph_0 and aleph_0 is finite. Is it aleph_0?

Yes.

Then
you would have 2^aleph_0 strings, which standard theory says is "uncountable".

No you wouldn't. There are 2^aleph_0 strings possible with
aleph_0 bits, but that includes the strings with infinitely many
leading 1's. There are an uncountable number of those.

The natural numbers consist of that subset of the strings of length
aleph_0 in which there are infinitely many leading 0's.

I don't want to make your head hurt, so let's stick to a finite
example of what you're trying to say. Suppose I asked "how
many bits do I need to represent the binary values up to
20? The answer is 5, since 20 in binary is 10100.

Do I conclude from that, that there must be 2^5 values between
0 and 20? Well, *I* don't, but following the logic above,
apparently *you* do.

- Randy

.

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