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Anonymous Tue 30 Jul 11:35:36 2019 No.10848909 View Reply Original Report

Quoted By: >>10849141 >>10849260

Can't any set of natural numbers be injected in the rationals by summing the reciprocals of the nth primes? Why is the cardinality of the Naturals' power set aleph-one and not aleph-null?

Anonymous

Anonymous Tue 30 Jul 2019 13:51:19 No.10849141 Report

Quoted By:

>>10848909
nah

Anonymous

Anonymous Tue 30 Jul 2019 15:02:51 No.10849260 Report

Quoted By: >>10849287

>>10848909
>injected
The word you’re looking for is embedded. Basically you find a way to define a subset of one set that sort of translates the original. People do this all the time with the naturals and dedekind cuts
Also the rationals and the naturals have the same cardinality. You can make an easy argument with Cantor-Berstein-Schröder:
You can map an element $n \in \mathbb{N}$ the rational $\frac{n}{1}$ . In the reverse direction, you can map every rational $\frac{a}{b} \in \mathbb{Q}, a, b \in \mathbb{Z}$ to the natural $2^m 3^a 5^b$ , where $m$ is 0 if the rational is positive and 1 if not. Even rationals that aren’t reduced fractions have a unique representation by the fundamental theorem of arithmetic.
Thus there’s a bijection between the rationals and the naturals

Anonymous

Anonymous Tue 30 Jul 2019 15:17:49 No.10849287 Report

Quoted By:

>>10849260
>The word you’re looking for is embedded.
Is it? I thought embedding was a specific type of injection that preserved properties of the original structure and not just distinctness, which isn't what I'm talking about.
>Basically you find a way to define a subset of one set that sort of translates the original. People do this all the time with the naturals and dedekind cuts
>Also the rationals and the naturals have the same cardinality. You can make an easy argument with Cantor-Berstein-Schröder:
>You can map an element n?N
>n \in \mathbb{N} the rational n1
>\frac{n}{1} . In the reverse direction, you can map every rational ab?Q,a,b?Z
>\frac{a}{b} \in \mathbb{Q}, a, b \in \mathbb{Z} to the natural 2m3a5b
>2^m 3^a 5^b , where m
>m is 0 if the rational is positive and 1 if not. Even rationals that aren’t reduced fractions have a unique representation by the fundamental theorem of arithmetic.
>Thus there’s a bijection between the rationals and the naturals
I already know all of this, I got caught up on the basics of set theory last weekend. My point was about the cardinality of the power set but I figured out since posting that the infinite subsets don't need to be constructible so it's fine.

Anonymous

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Anonymous Tue 30 Jul 2019 22:38:17 No.10850390 Report

Quoted By:

The image shows two functions. You can easily prove both are injective using the properties of decimal places in different bases. Then by using Schroeder-Bernstein theorem it's concluded that there is a bijection between the interval [0,1) and the power set of the Naturals, and from there you can choose any bijection you like between [0,1) and the Reals.
I could send you the whole detailed proof, but it's in Spanish.
Anyway if you want a more intuitive insight, it kind of feels it should be that way by applying Cantor's theorem and assuming true the continuum hypothesis.

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