An index set for a limit has very few morphisms I'm just learning about limits. My book (Kashiwara/Schapira) defines projective limits on sets in the following way: Suppose $F \in [I^{\text{op}} , \text{Sets}]$. Then 
$$\varprojlim F:= \text{Hom}_{[I^{\text{op}},\text{Sets}]}(\Delta_{\text{pt}}, F),$$
where $\Delta_{\text{pt}}$ is the constant functor at the set $\{*\}$. Thus $\varprojlim F$ can be viewed as an subset of $\prod\limits_{i}F(i)$. (I believe this is somewhat different from the way it is usually defined, though equivalent.)
My question is: it seems that the elements of $\varprojlim F \subseteq \prod\limits_iF(i)$ that are present must be very few, unless $I$ has hardly any morphisms at all. For instance, if $s \in F(i)$ is such that there exist $\omega, \eta \in \text{Hom}_{I^{\text{op}}}(j,i)$ such that $F(\omega)(s) \neq F(\eta)(s)$, then $\alpha_i \neq s$ for any $\alpha \in \varprojlim F$. In particular, if there exists an $i$ such that all the elements of $F(i)$ are like $s$ above, then $\varprojlim F$ must be the empty set.
Is this right? So we mostly have to choose $I$ to have few morphisms (like in the case of the product where $I$ is discrete)?

What you are saying mathematically is correct, but the conclusion is wrong (or at least, hard to judge since there is no precise meaning of "few morphisms"). Better look at some classical examples first ($p$-adics $\mathbb{Z}_p = \varprojlim_n ~ \mathbb{Z}/p^n$) before diving into general category theory.
