Suppose that $A$ is an ordinal and $B\subset A$. We say that $B$ is Cofinal in $A$ iff $\sup(B) = A$.
However, elsewhere I've read definitions that make no reference to ordinals. Here is the definition I usually encounter:
For all $a\in A$ there exists $b\in B$ such that $a\le b$.
where $A$ and $B$ are just (any) ordered sets.
I would like to see how these definitions are equivalent.
Lastly, I don't understand how $\sup$ is defined for ordinals (it is used in the notes I'm reading but never defined).
The supremum of a subset $A$ of an ordered set $X$ is the minimum of the set of upper bounds of $A$ in $X$, if it exists.
So $\forall a \in A, a \leq sup(A)$, and $\forall a \in A(\forall x \in X(a\leq x) \rightarrow sup(A) \leq X)$. Equivalently, $sup(A)$ is unique in $X$ (if it exists at all) to satisfy $\forall x \in X(A\leq x \leftrightarrow sup(A) \leq x)$. (where I write $A\leq x$ to denote $\forall a \in A, a\leq x$, and I use similar denotations with the strict order)
Here the supremum of $B$ is not taken in $A$ (which does not contain itself) but in the class of ordinals, where the same definition makes sense. What's nice is that every set of ordinals has a supremum which is an ordinal.
Depending on how familiar you are with ordinals this equivalence could be quick to prove or not. First note that the equivalence does not hold in general but only if $A$ is a limit ordinal. Else $sup(A)=max(A)<A$ although $A$ is cofinal in $A$. In the general case, $sup(B)=A$ is equivalent to $B$ being strictly cofinal in $A$ in the sense that for $a\in A$, there is $b\in B$ such that $a<b$. (which is never the case if $A$ is successor).
I henceforth assume $A$ is limit. Note that $B$ is cofinal in $A$ in the second sense iff there is no $a\in A$ such that $B<a$.
-Assume $B$ is cofinal in $A$ in the first sense. Consider $a\in A$. Then $a< A$ (as ordinals), what does this tell you about the statement $B< a$?
-Assume $B$ is not cofinal in $A$ in the first sense. This implies that $sup(B)\in A$. Can you use this to find $a \in A$ such that $B<a$?