Every coalgebra is the sum of its finite-dimensional subcoalgebras In the article about coalgebra in Wikipedia, it says that

Every coalgebra is the sum of its finite-dimensional subcoalgebras

I want to know how to prove this but I have no idea where to start.
I appreciate any help.
 A: $\bullet$ First of all, note that if $C$ is a coalgebra and $(C_{α})_{α}$ a family of subcoalgebras, then their sum $\sum_{α}C_{α}$ is also a subcoalgebra, because: 
$$
\Delta(\sum_{α}C_{α}) = \sum_{α}\Delta(C_{α})
\subseteq \sum_{α} C_{α} \otimes C_{α} \subseteq (\sum_{α}C_{α})
\otimes (\sum_{α}C_{α})
$$
$\bullet$ Second, note that every coalgebra is locally finite (the terminology is due to Sweedler). This is frequently called the fundamental theorem of coalgebras, stating that: 

Every element of a coalgebra $C$ is contained in a finite dimensional subcoalgebra.  

(the proof of the fundamental theorem on coalgebras is standard and can be found in almost all introductory texts in coalgebras and hopf algebras, see for example the book of S.Montgomery, "Hopf algebras and their actions on rings", p.56 or the book of Dascalescu-Nastasescu-Raianu, "Hopf algebras: an introduction", p.25). 
Now, combining the above two, it is easy to conclude the statement of your question. 
P.S.: Actually note that, the fundamental theorem of coalgebras is implying  something even more general than your statement: "every coalgebra is the union of its finite dimensional subcoalgeb-ras". (see also the book of Sweedler on Hopf algebras, p.170). 
A: I am unfamiliar with coalgebras, in particular whether they have finite-dimensional -subcoalgebras, so here is a very vague but hopefully motivating answer.
Suppose we have an infinite-dimensional coalgebra $\mathcal C$ with a sequence of finite-dimensional subcoalgebras $C_1 \subset C_2 \subset \ldots$ such that dim$(C_n)$ increases to infinity and such that $\mathcal C = \bigcup C_n$. For simplicity suppose there are no other subcoalgebras. What might it mean to say $\mathcal C$ is the sum of its finite-dimensional coalgebras?
Naievely we might try $\mathcal C = C_1 \oplus C_2 \oplus \ldots$ but this seems a rather strange thing to do. Because the incarnations of $C_1$ and $C_2$ inside $\mathcal C$ share elements. In particular select any element of $c \in C_1$. The sum contains both distinct elements $(c,0,0, \ldots)$ and $(0,c,0,\ldots)$ which do not seem to correspond to anything in particular in the original $\mathcal C$.
The problem here is that $C_n$ intersect and our direct sum ignores that fact. The way to acknowledge it is using direct limits. What we do is we consider the collection $\mathcal \{G(i): i \in I \}$ of all finite-dimensional subcoalgebras and the inclusion maps $\{F^i_j \colon G(i) \to G(j)\}$. The direct limit is a way to 'glue together' all the coalgebras.
You should look up the specific construction of direct limits of coalgebras. But I would suspect it is similar to te usual for algebraic structures. . . .
Consider the set $D$ of all elements of $\bigcup_i G(i)$ under the weakest equivalence relation generated by all relations $x \sim ^i_j y$ iff $x \in G(i), y \in G(j)$ are such that $F^i_j(x)=y$. Define the comultiplication and counit seperately on each $G(i)$. It is an exercise to show the extend consistently to the set $D$. Then $D$ is our direct limit.
Consider how the above construction simplifies for $\mathcal C$. In that case $\bigcup_i G(i)$ is just a chain.
Edit: This page of nLab claims a proof of an assertion similar to the above, called the fundamental theorem of coalgebras, can be found in Walter Michaelis, Coassociative Coalgebras, Handbook of Algebra Volume 3, Elsevier (2003).
