A question about bounds, least and minimal elements, and partial vs strict ordered sets It's not very clear to me if the concepts of bounds, least elements and minimal elements (also, greatest elements and maximal elements, etc. ) apply only to partial orders or if the definition applies to general ordered sets. 
If I try to make an analogy between sets in general and the real numbers I would say that those definitions are explicitly made for partial orders, though I'm not sure and I don't know whether there are some sets with strict order to which the definitions apply.  
For example, there is a theorem that says that "if $b$ is the least element of $B$ in some order on $A$ and $B\subseteq A$, then $b$ is the infimum of $B$". In this case I can think of the set $A=\{1,2,3\}$ with the usual strict order $<$ on $\mathbb{R}$ and $B=\{2,3\}$. Then $\inf(B)=1$ and $\min(B)=2$. But if I take the usual partial order $\leq$ then $\min(B)=\inf(B)=2$. 
Here I'm using the next definition: Let $R$ be an ordering on $A$, and let $B\subseteq A$. Then,
1.- $a\in A$ is a lower bound of $B$ iff $\forall x\in B(a R x)$.
2.- $a\in B$ is a greatest element of $B$ iff $\forall x\in B(x R a)$.
3.- $a\in A$ is called an infimum of $B$ iff it is the greates element of the set of all lower bounds of $B$.
Note: For the example I'm assuming as proved that the greatest element an the infimum are unique.  
Edit: Esentially what I'd like to know is to what kind of orders the definitions of bounds, least and greatest elements, minimal and maximal elements, etc. apply.
 A: Ok these definitions are all for partial orders, but since total (linear) orders are partial orders as well, they work just as well for them.
Some things I will point out.  First, bounds are exactly like what they say there are.  There are bounds of the set in question (notices the s).  For example,
$A = \{1, 2, 3 \}$ as a subset of $\mathbb{N}$ and the order $<$, then $4, 5,$ and$ 1,000$ are all upper bounds, but the only lower bound is $0$.  But if instead we are thinking of $\leq$ then $3$ is also an upper bound and $1$ is also a lower bound.
Now, let $(A, \leq)$ be a partial ordered set and let $S \subset A$.
A minimal element of $S$ is an element $m$ such that there is no $n \in S$ such that $n \neq m $ and $n \leq m$.  Equivalently, for all $n \in S$ $n \leq m \implies n = m$. A maximal element of $S$ is an element $m$ such that there is no $n \in S$ such that $n \neq m$ and $m \leq n$.  Equivalently, for all $n \in S$ we have $m \leq n \implies m = n$ (see here).
A least element of $S$ is an element $l$ such that for all $s \in S$ we have $l \leq s$.  A greatest element of $S$ is an element $g$ such that for all $s \in S$ we have $s \leq g$.
I agree these two definitions (minimal/least and maximal/greatest) are a bit confusing as they are not equivalent, so let me give some examples.
So no take for example $A = \mathbb{N}$, $S = \{2, 3, 4, 5, 6, 10, 12 \}$ and $\leq$ be the divides relation i.e. $a \leq b \text{ iff } a \mid b$ (this is a true partial order).  Then $2, 3, 5$ are all minimal elements and $10, 12$ are maximal elements. (This is why it is useful to look at the Hasse diagram, because you can pick out the minimal/maximal elements fairly quickly). But notice: $S$ has no least or greatest elements.  But now consider $S = \{3, 5, 15\}$.  Then again $S$ has $3, 5$ as minimal elements (and no least element), but $15$ is a maximal element and a greatest element.
Now, for infimum and supremum, you are looking at the greatest element of the lower bounds and the least element of the upper bounds.  So in you example, the set of lower bounds is $\{1, 2\}$ and so the greatest is $2$ (notice the definition of greatest includes equality see here).
I can add more if you want and sorry if that explanation is a bit erratic.  Let me know what else you would like to know and I can expand this answer. Hope it helps a little!
A: First of all, when we say "ordered sets" we usually mean that the set is partially ordered, and often even more (i.e. totally ordered, and in some contexts even well-ordered).
Secondly, $1\neq\inf\{2,3\}$. In the real numbers if $A$ has a minimum, then $\min A=\inf A$. In fact, it is easy to see that if $\min A$ exists then $\inf A=\min A$, whenever the two notions are defined. 
Why? Well, what is $\inf A$? It is some $x$ such that for all $a\in A$ we have $x\leq a$, and whenever $y$ is such that $y\leq a$ (for all $a\in A$, of course) then we have $y\leq x$. But if $m=\min A$ exists then it is a member of $A$, so whenever $x$ is such that $x\leq a$ for all $a\in A$, in particular $x\leq m$.
If you want to prove the uniqueness of $\inf$, and $\inf$ is defined as the greatest element of lower bounds, then proving the uniqueness of "greatest element" would suffice.
And how do you prove uniqueness? Suppose that $x$ and $y$ are greatest element of $B$, and show that by definition we have to have $x=y$. Hint: You have to use antisymmetry here.
