# Thread: reformulate "there are infinitely many primes"

1. ## reformulate "there are infinitely many primes"

I was given this excersise as homework and to a certain extent i do understand what they are asking, but don't know where to start.

Consider "there are infinitely many primes". Find a suitable universe U of elements so that this theorem can be reformulated in the form : A (is a subset of) B

We were also told that we dont need to prove anything here.

Thank you for any suggestions.

2. ## Re: reformulate "there are infinitely many primes"

I am not sure if this is the idea of the exercise, but if you formulate "there are infinitely many primes" as $\displaystyle \forall n\in\mathbb{N}.\,P(n)$ for some property P, then it can also be stated as $\displaystyle \mathbb{N}\subseteq\{n\in\mathbb{N}\mid P(n)\}$.

3. ## Re: reformulate "there are infinitely many primes"

Originally Posted by Arturo_026
I was given this excersise as homework and to a certain extent i do understand what they are asking, but don't know where to start.

Consider "there are infinitely many primes". Find a suitable universe U of elements so that this theorem can be reformulated in the form : A (is a subset of) B

We were also told that we dont need to prove anything here.

Thank you for any suggestions.
Would this homework be for a grade?

4. ## Re: reformulate "there are infinitely many primes"

Originally Posted by Ackbeet
Would this homework be for a grade?
Not directly. We're graded more on effort, but mainly on midterms and final.

A simple set up I came up with (which was similar to his answer) was:

Let U be the Set of all Sets (even though it's understood that there's no set of all sets), and let B be the set of all infinite sets. Moreover, let A be a subset of this set but also having the condition of being the set of primes.

Therefore A (is subset of) B (is subset of) U. ... sorry for not writting the symbol.

Does this sound right?

5. ## Re: reformulate "there are infinitely many primes"

Originally Posted by Arturo_026
Let U be the Set of all Sets (even though it's understood that there's no set of all sets), and let B be the set of all infinite sets. Moreover, let A be a subset of this set but also having the condition of being the set of primes.

Therefore A (is subset of) B (is subset of) U. ... sorry for not writting the symbol.
Why not learn to post in symbols? You can use LaTeX tags
$$A\subseteq B$$ gives $\displaystyle A\subseteq B$

6. ## Re: reformulate "there are infinitely many primes"

Originally Posted by Arturo_026
Let U be the Set of all Sets (even though it's understood that there's no set of all sets), and let B be the set of all infinite sets. Moreover, let A be a subset of this set but also having the condition of being the set of primes.

Therefore A (is subset of) B (is subset of) U.
It is sufficient to take U to be the powerset of natural numbers. However, with A and B defined as above, A is infinite iff $\displaystyle A\in B$, not iff $\displaystyle A\subseteq B$.