All of Kaya Fallenstein's Comments + Replies

I only assumed the sequence was monotonic.

The second comment is fair. I think when I first read this, I ignored the bit about 7zip and understood the sentence as claiming "no Turing machine takes an input smaller than one byte and prints out a 10kb file".

"Suppose an advanced agent with a goal like, e.g., producing smiles or making paperclips."

Typo? Does not seem to be a complete sentence. Maybe "Suppose you have an..."

"Similarly, if you start with a 10 kilobyte text file, and 7zip compresses it down to 2 kilobytes, no amount of time spent trying to further compress the file using other compression programs will ever get that file down to 1 byte." (Emphasis added.)

This claim seems too strong. Do you mean that an arbitrary compression algorithm is unlikely to compress the file to one byte?

Taking what's written at face value, if I assume that every compression I apply monotonically decreases the size of the file, then I need only find a large enough sequence of compressio... (read more)

tl;dr: I did some reading on related topics, and it turns out that (1) may be sufficient to define logarithms if we take as an axiom that every set is Lebesgue measurable (which is incompatible with the axiom of choice). Otherwise, we need to add an additional condition to (1).

(1) states that $f(x\cdot y)=f\left(x\right)+f\left(y\right)$. Given a function $g$ satisfying this condition, we can generate an additional function satisfying this condition by composing $g$ with a function $h$, where $h(x+y)=h\left(x\right)+h\left(y\right)$:

$$h\left(g\right(x\cdot y\left)\right)=h\left(g\right(x\left)\right)+h\left(g\right(y\left)\right)$$

$h$, as defined, is a solution to Cauchy's functional e... (read more)

The proof of (5) only goes through for $n\in N$.

You can prove a version of (8) from (5), namely, $f\left(b\right)=1\Rightarrow f\left(b^q\right)=q$ for $q\in Q$, but this doesn't pin down $f$ completely, unless you include a continuity condition.

(8) doesn't follow from (5). The assumption in (5) was than $n$ ranged over naturals, not reals. In fact, (1) only implies (8) if you also require the function to be continuous.

(1) essentially says $f$ is a homomorphism from $(R^{>0},\cdot )$ to $(R,+)$. To generate a function satisfying (1) but not (8), we need only compose $log$ (choose a base) with an automorphism in the additive group and show that the composition is not a multiple of a logarithm. We can get such an automorphism by considering $R$ as an infinite dimensional vector space over the rationals and... (read more)