I just started studying set theory and I'm a bit confused with some of these relation properties.
Given a set A = {8,4,2}, and a relation of order R such that aRb means "a is a multiple of b", we would have the following relations:
{8R8, 8R4, 8R2, 4R4, 4R2, 2R2}
(I think a correct notation for this set of relations would be A/R but frankly I'm not too sure so if anybody could confirm or deny that, it'd be appreciated)
If my understanding of relations is correct (which it might very well not be), the relation would be reflexive, transitive and antisymmetric, therefore we would have a partial relation of order.
Now, since in this partial order all elements are related to each other, it would be a total order relation.
And since all non empty subsets of the set A would have a least element, it's a well ordered set
If, however, instead of A = {8,4,2}, we had A = {8,4,2,3}, we would just have a relation of partial order since 3 wouldn't be related to any of the other elements in the set.
On another note, in order to have a well-ordered set, we first need a totally ordered set, right?
Also lastly, if A started out at $2^n$ where $n = \infty$ and $n$ descended until 1, the set would be totally ordered but not well ordered, right? I'm aware this notation might be terrible but I don't know how to get the idea across otherwise so yeah. Any help and advice with notation and general understanding is also appreciated
Edit: A correction on this last paragraph: after reading some of the replies and re-reading my own message I've noticed how what I've written makes basically no sense, so I'm going to re-do the question:
The set I was (very poorly) trying to describe was A = { $2^n | n \in \Bbb N $ } only backwards for some reason. So reading the replies, they say it's a totally ordered set but not a well ordered set. Seeing as $2^n | n \in \Bbb N $ has a least element (if $n = 0$), I was wondering why it wouldn't be a well ordered set.
In other words, do all of the subsets of any set with a least element also have a least element?
The source of confusion you might have comes from the sort of casual way that we often talk about "partially ordered sets". A set itself, strictly speaking, is an un-ordered entity. Two sets are equal if and only if they have the same elements. So $\{1,2,3\}=\{3,2,1\}$, etc.
When we speak of a partially ordered set, what we are really referring to is a pair $(S,R)$, where $S$ is a set and $R\subseteq S\times S$ is a relation that satisfies the axioms of a partial ordering. We sometimes will say "$S$ is a partially ordered set under the relation $R$", but that's really what could arguably be called casual abuse of terminology, since $S$ itself is just an arbitrary set. When we say something like this, we're communicating that whenever we talk about $S$ as being ordered, we are referring to the partial ordering defined by $R$.
In your final edited question, you ask if $A=\{2^n | n \in \Bbb N\}$ is well-ordered. This is a fine question to ask provided that you've already clarified what ordering relation $R$ you are talking about.
If $R$ is the usual ordering for integers (restricted to $A\times A$), then the answer is "Yes, $A$ is well-ordered".
If, on the other hand, $R$ is the relation you described in the beginning of your post, i.e., $aRb$ means $a$ is a multiple of $b$, then under this relation, $A$ is not well-ordered, since the ordering is reversed from the usual ordering - every subset will have a greatest element, but not necessarily a least one. In particular, $A$ itself does not have a least element under this ordering, as $2^0$ is the greatest element, then $2^1$ precedes $2^0$ as $2^1R2^0$, likewise $2^2$ precedes $2^1$ since $2^2R2^1$, etc.
The answer to your final question was given by Joe in the comments, but let me just re-emphasize it here, that no, just having a least element for a set does not imply every subset has a least element. The real interval $[0,1]$ under the usual ordering has a least element, but the subset $(0,1)$ does not have a least element, so $[0,1]$, though totally ordered, is not well-ordered.