How come an element $s\in (\varprojlim S_i)^c$ does not obey $\pi_{ji}(s_j)=s_i$? Notation:
1. $S_i$ is a hausdorff space for every $i\in \mathbb N$
2. $(S_i, \pi_{ji})$ is an inverse system where $\pi_{ji}:S_j\rightarrow S_i$
3. $S=\varprojlim S_i $ 
I wish to show $S$ is a closed subset of $\Pi S_i$ (the Cartesian product).
As a part of the proof, I'm trying to show that $S^c$ (the complement of $S$) is open. So I grab $s\in S^c$ and show it has an open neighborhood disjoint to $S$.
Because $s\notin S$, we get $\pi_{ji}(s_j)\neq s_i$.  
My question is: How come the $j^{th}$ component of an element from $\Pi S_i$ does not obey $\pi_{ji}(s_j)=s_i$? That's the definition of $\pi_{ji}$: for every element $s\in \Pi S_i$ it takes the $j^{th}$ component and returns an $i^{th}$ component.
Can you give a simple nontrivial example of such a case?
EDIT: 
With the assistance of a community member, I found out $\pi_{ji}$ can be any map, not necessarily a kind of projection. As such, it doesn't have to map all $S_j$'s elements to $S_i$, so there might be an element $s_j\in S_j$ such that $\pi_{ji}(s_j)\notin S_i$.
It brings up the question: If not all elements of $S_j$ are mapped to $S_i$ where are they're mapped to?
 A: Note that $S$ is the following subspace of $\prod_{i \in I} S_i$:
$$S=\{(x_i)_{i \in S}: \forall i,j \in I: (j \ge i) \implies \pi_{ji}(x_j)=x_i\}$$
the set of all "threads": a point with coordinates $x_i, i \in I$ is in $S$ precisely when all of its coordinate points are mapped to each other via the bonding/connecting maps $\pi_{ji}$ whenever this map is defined. The whole product itself is a mass of all points $(x_i)$ with the only condition being $x_i \in S_i$ for all $i$, disregarding the bonding maps. So $(x_i) \notin S$ means that there is at least one pair $i\le j$ of indices where the thread condition fails, so that $\pi_{ji}(x_j) \neq x_i$ for those specific $i$, $j$ and $x_i$ and $x_j$ that are in that point. We do know that as $(x_i)$ is in the full product that $x_i \in S_i$ and $x_j \in S_j$, so $\pi_{ji}$ is defined on $x_j$ but happens to be not equal to this specific $x_i$ (which is in fact very likely to happen, were you to choose points randomly in the product: the thread condition from the definition is very strict). Note that we work with a specific, concrete $(x_i) \notin S$. Hopefully now you can continue to work trying to understand the proof of why $S$ is closed in $\prod_{i \in I} X_i$.
