So I've gotten this homework that I have to solve but unfortunately we have done only few tasks like this in class so I have no idea how this principle works.
I have to prove that a set of natural numbers( that divided by 7 give a rest of 3) is a countable set.
My mind has no idea how to work this through since when someone gives me an infinite set, I just figure that numbers in the set can go to infinity therefore I can not count all of them.
I would be really thank full for your help!
On
Any subset of the naturals is at most countably infinite because the naturals are. Does your definition of countable include finite sets? In that case you are done already. If not, you need to show that your set is infinite. One easy way here is to say your set consists of $\{n|n=7k+3,\ k\in \Bbb N\}$
To show a set is countable one shows that there is bijection between it and the natural numbers (or indeed some other countable set). We can construct a fairly simple bijection between $S:=\{n\in\Bbb N:n\equiv 3\pmod 7\}$ and $\Bbb N$ in the following way.
For any $a\in S$ we have $a=7b+3$ for some unique $b\in\Bbb N$. So we can construct a bijection by associating the $a$'s with their unique $b$'s as such: $$f:\Bbb N\to S\qquad f(b)=7b+3$$ I leave it to you to show that this is indeed a bijection, although this is fairly straight forward.
Note: I assume $\Bbb N$ includes $0$. If you do not use this interpretation, then there is just a slight shift in the bijection, but this is a minor correction/detail.