Which of the sets are well ordered? Which ones are isomorphic? I'm new to StackExchange and I'd like to ask you for help. 
I have been trying to solve this exercise:
$$A = \left\{3 - \frac{1}{2n} : n \in \mathbb{N} - \left\{ 0 \right\} \right\}$$
$$ B = \left\{ \pi - \frac{2}{n} : n \in \mathbb{N} - \left\{0 \right\} \right\} \cup \left\{4 \right\} $$
$$C = \left\{ 0 \right\} \cup \left\{ \frac{1}{n} : n \in \mathbb{N} - \left\{ 0 \right\} \right\} \cup \left\{2 - \frac{1}{n} : n \in \mathbb{N} - \left\{ 0 \right\} \right\}$$
Which of sets $A, B, C, \mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{Q} - \left\{ 0 \right\}, \mathbb{R}, \mathbb{R} - \left\{ 0 \right\}$ are well ordered by $\le$? Which ones are isomorphic?
I'd love to see a detailed solution to this, as I have some problems with the Set Theory subject. I'm a first year CS student.
Thank you in advance.
 A: A set is well ordered by $\leq$ if it is totally ordered by $\leq$ (there is no pair of incomparable elements) and in every non-empty subset of that set there exists a $\leq$-least element.
All of the sets in your question are totally ordered since they are all subsets of real numbers (and the relation $\leq$ is a total ordering for that set).
With this in mind, here's how you could approach the problem of well orderedness for the first three sets in your question:


*

*Set $A$ is well ordered: for any subset generated by the positive natural numbers $n_1 < n_2 < \dots$ the least element is equal to $3 - \frac{1}{2n_1}$.

*Set $B$ is well ordered for similar reasons.

*Set $C$ is not well ordered. It is the union of three sets: the second set has no least element. We can generate positive numbers arbitrarily close to $0$ by choosing larger and larger values of $n$.
For the question of which are isomorphic, look first at the cardinality of each set. A countable set such as $\mathbb{N}$ cannot be isomorphic to an uncountable set such as $\mathbb{R}$. Also look at which are well ordered and which are not. A well ordered set such as $\mathbb{N}$ is not isomorphic to, say, $\mathbb{Z}$ (one has a least element, the other does not). Thirdly note that $B$, while countable and well-ordered, has a greatest element (but do other sets such as $A$ and $\mathbb{N}$ do not).
