Tips for forming countability proofs? I'm new to writing proofs, and I'm having a really hard time enumerating sets. I'm struggling to construct functions which convert positive integers into the desired set and which is bijective. For instance, one of these sets is the list of all real numbers with decimal representations consisting of 1's.
I have no idea how to approach this question. I'm not even sure if it's countable or not.
Another example which illustrates something which I am having trouble with is whether the set of all integers divisible by 5 but not divisible by 7 is countable. Obviously it is countable, but I don't know how to prove that it is countable.
I got f: $\mathbb{N} \rightarrow \mathit{G}$
f(x) = $\frac{5x}{5}$ where x is divisble by 5
$\frac{-5(x-1)}{5}$ where (x - 1) is divisble by 5
$\frac{35(n-2)}{5} - \frac{35(n-2)}{5}$ where (x-2) is divisible by 5
At which point I felt as if I wasn't going in the right direction and stopped.
Any and all advice would be appreciated.
 A: Expanding on Elliot G's comment:
Sometimes producing an explicit bijection is not the right way to go. For example, this sketch https://www.homeschoolmath.net/teaching/rational-numbers-countable.php that the rationals are countable (which can also be seen as a reason why $\mathbb N \times \mathbb N$ is countable), as one might see in an introductory analysis class, isn't explicitly written as $f : \mathbb N \to \mathbb Q$.
Now, for this case where you have numbers with decimal expansions consisting only of $1$s, you can first note that the set of all numbers of the form $n + \dfrac{1}{10}$ where $n$ is an integer is countable. This should give you all numbers that have one decimal place with $1$. Now you can go to numbers of the form $n + \dfrac{11}{100}$, or two decimal places with $1$s. Then three decimal places, and so on.
Can you see how to finish it off using a countable union of countable sets?
Edit: Added more in response to the original question adding a new scenario.
I would hope you have a result that says something like "all subsets of a countable set are countable." An integer $n$ is divisible by $5$ if it can be written as $n = 5k$ for another integer $k$. Then the integers that are multiples of $5$ but are not divisible by $7$ is a subset of the set of all numbers divisible by $5$.
As another note, an integer $n$ is not divisible by $7$ if and only if it can be written as $n = 7k + j$ for some integer $k$ and another integer $1 \leq j \leq 6$.
Now, maybe you want to also show that it's infinite. Maybe using prime powers and the unique prime factorization theorem come in handy.
A: Broadly speaking, infinite subsets of $\mathbb{R}$ are either countable or uncountable. Set $\mathbb{N}$ of natural numbers is countable; set $\mathbb{R}$ is uncountable. All you need to check is whether an infinite subset of $\mathbb{R}$ is equinumerous with $\mathbb{N}$ or $\mathbb{R}$.
Sometimes an explicit bijection is possible between a set and $\mathbb{N}$. In that case, the set will be countable. For example, the function $f:\mathbb{N}\to \mathbb{W}$ defined by $f(x)=x-1$ is a bijection meaning that $\mathbb{W}$ is countable, $\mathbb{W}=\mathbb{N}\cup\{0\}$.
As another example, the function $g:\mathbb{R}\to (-1,1)$ given by $g(x)=\frac{x}{1+|x|}$ is a bijection meaning that the interval $(-1,1)$ is uncountable.
However, such explicit bijections are not always possible. In that case, try to see if you can use the existing literature to decide the countability. As an example, using the fact that the union of two countable sets is countable, we conclude that the set of irrational numbers $\mathbb{R}-\mathbb{Q}$ is uncountable, for otherwise, it would lead to conclude that $\mathbb{R}=\mathbb{Q}\cup(\mathbb{R}-\mathbb{Q})$ is countable, which would be a contradiction.
