Maximal element of infinite well ordered set Can an infinite well ordered set $A \subseteq \mathbb{R}$ have a greatest element? I've read it a couple of times that the answer is yes, but I just couldn't agree with that. 
If set $A$ is well ordered, then surely it has least element. Okey, but then if it have a greatest element, as well, the set will be finite. 
If I'm wrong, could you give me example of such set? Thanks!
 A: One idea is to find a well-ordered set of real numbers that fits nicely into a fixed interval, then to pick some individual value bigger than all of them and add it in as the maximum element. For example, you could pick
$$S = \{ 1 - \frac{1}{n} \ | \ n \in \mathbb{N} \} \cup \{ 137 \}.$$
Here, the first half of the definition of $S$ gives you an infinite well-ordered set, and adding in $137$ preserves the well-orderedness while giving you a maximum element.
A: Yes, an infinite well ordered set in $\mathbb{R}$, under the usual ordering, can have a greatest element. For instance consider $1\cup \{{1-1/n}, n\in \mathbb{N}^+\}$, it consists of $\{0,1/2,2/3,…,1\}$ and is clearly well ordered.

What is true on the other hand is this: If $X$ is a well ordered set, and every non empty subset of $X$,  has a greatest  element, then $X$ is finite.This can be proven with a simple recursive argument, by assuming the contrary and getting a contradiction.On the other hand the above fails for totally ordered sets, consider $\{1/n, n\in \mathbb{N}^{+}\}$ equipped with the usual ordering.
