cardinal of a quotient space Let $X$ be a nonvoid set of cardinal $\alpha$. Let $\cong$ be an equivalence relation on $X$. Let $\beta$ be the cardinal of the set
(1)     $ D = \{ \, ( x, y ) \in X \times X ~|~ x \ncong y \, \} $
Let any family $( x_i )_{i \in I}$, with $I$ a nonvoid index set, and $x_i \in X$, for $i \in I$. We call such a family a chain, iff
(2)     $ x_i \ncong x_j,~~ i, j \in I,~ i \neq j $
We denote by
(3)     $ \kappa $
the smallest cardinal which is at least as large as the cardinal of any index set $I$ of a chain (2).
Clearly, we shall have
(4)     $ car ( X / {\cong} ) = \kappa $
Problem 1
Find, in terms of the cardinals $\alpha, \beta$, the cardinal $\kappa$.
Problem 2
Given the cardinal $\alpha$, and given an upper bound
(5)     $ \beta \leq \gamma $
find, in terms of the cardinals $\alpha, \beta, \gamma$, an upper bound for the cardinal $\kappa$.
Problem 3
Given the cardinal $\alpha$, and given a lower bound 
(6)     $ \beta \geq \gamma $
find, in terms of the cardinals $\alpha, \beta, \gamma$, a lower bound for the cardinal $\kappa$. 
 A: Let me first treat the case where the
underlying set is infinite. 
In the infinite case, your cardinal $\beta$ is either $0$
or equal to $\alpha$, depending on whether all points are
equivalent or not. The reason is that if the relation is
not trivial, then every point is inequivalent to some other
point, so $\alpha\leq\beta$, and conversely
$\beta\leq\alpha\cdot\alpha=\alpha$ by infinite cardinal
arithmetic.
For question 1, the answer is therefore that $\kappa$ is
not determined by $\alpha$ and $\beta$. As you observed,
$\kappa$ is the number of classes, and the same infinite
set of size $\alpha$ can be partitioned into any number
$\kappa$ of classes, provided $1\leq\kappa\leq\alpha$.
For question 2, when $\alpha$ is infinite, then since
$\beta=\alpha$ (unless there is only one class, in which
case $\beta=0$), the bound $\gamma$ is not very helpful. But the largest $\kappa$ can be is $\alpha$. (This is under AC; without AC, then it is possible that $\kappa$ could be strictly larger than $\alpha$, as I expain at the bottom.)
Similarly, for question 3, the smallest $\kappa$ can be is
$1$, when $\beta=0$, and otherwise, $\kappa=2$ is possible,
since you can divide $\alpha$ into $2$ classes, each of
size $\alpha$.
In the infinite case, there are some interesting
issues that arise with the Axiom of Choice in this
question. Your observation that the quotient has size
$\kappa$ seems to rely on AC, since the chains are
essentially choice functions. More generally, it was
observed in a previous MO answer by Dr. Strangechoice that
$\kappa$ can actually be strictly larger than $\alpha$! That is, one can partition a set into strictly more classes than there are points! For
example, consider the relation $E$ on the reals, where
$xEy$ if $x=y$ or if both $x$ and $y$ code a well order on
the natural numbers having the same order type. This is an
equivalence relation on the reals, but it is consistent
with ZF that there is no $\omega_1$-sequence of reals, and
in this case there can be no injection from the $E$-classes
into the reals, since this would provide such an
$\omega_1$-sequence. But there is a converse injection,
since we can injectively map reals to reals that don't code
well-orders. So this is a situation where the number of
equivalence classes is a strictly larger cardinality than
the underlying set.

Update. In the finite case, I happened to observe that again $\kappa$ is not a function of $\alpha$ and $\beta$. To see this, let $\sim_1$ and $\sim_2$ be two relations on 6 points, the first partitioning it as $2+2+2$, with three classes, and the second partitiioning it as $3+1+1+1$, with four classes. In each case, we have $\alpha=6$. But unless I have made a counting mistake, it seems we also have $\beta=24$ in each case, since each equivalence relation adds 3 equivalent (unordered) pairs beyond the identity pairs, making for 12 equivalent pairs (a,b), and hence $\beta=36-12=24$ inequivalent pairs in each case. But the first relation has $\kappa=3$ and the second has $\kappa=4$; so $\kappa$ is not determined by $\alpha$ and $\beta$.
