Lie coalgebra with no finite-dimensional subcoalgebras In Walter Michaelis' paper Lie Coalgebras, he gives on page 9 an explicit example of a Lie coalgebra which is not the union of its finite-dimensional Lie subcoalgebras. In fact, Michaelis' example has exactly two finite-dimensional Lie subcoalgebras: the zero coalgebra and a certain one-dimensional subcoalgebra.
What I want to know is whether one can do even better: is there a Lie coalgebra $C\neq0$ (over any field) whose only finite-dimensional Lie subcoalgebra is $0$? Ideally, I'd like an explicit example.
 A: Consider the Lie algebra of vector fields on the formal disk, $\mathfrak g=k[[t]]d/dt$, where $k$ is a field of characteristic zero and $k[[t]]$ is the $k$-algebra of formal Taylor power series in the variable $t$.  Then $k[[t]]$ is naturally a topological $k$-algebra with a pro-finite-dimensional ( = pseudocompact = linearly compact) topology, and $\mathfrak g$ is a topological Lie algebra with a pro-finite-dimensional topology.
Consequently, there exists a unique coassociative coalgebra $\mathcal C$ such that $k[[t]]\simeq \mathcal C^*$ as a topological associative algebra, and there exists a unique Lie coalgebra $\mathcal L$ such that $\mathfrak g = \mathcal L^*$ as a topological Lie coalgebra.  Now, of course, $\mathcal C$ is the union of its finite-dimensional subcoalgebras.  However, $\mathcal L$ has no nonzero finite-dimensional Lie subcoalgebras.  In fact, $\mathcal L$ has no nonzero proper Lie subcoalgebras at all.
To prove these assertions about $\mathcal L$, one can translate them back into the topological Lie algebra language, where they become more intuitively clear and can be checked explicitly by hand.  The claims to prove are that the topological Lie algebra $\mathcal g$ has no proper open Lie ideals, and in fact, it does not even have any nonzero proper closed Lie ideals.
The topological Lie algebra $\mathfrak g$ has a topological basis $L_{-1}$, $L_0$, $L_1$, $L_2$, $\dots$, where $L_n = t^{n+1}d/dt$.  The commutation relations are $[L_i,L_j]=(j-i)L_{i+j}$.  Suppose $\mathfrak h$ is a nonzero closed Lie ideal in $\mathfrak g$.  Considering the action of $L_0$ in $\mathfrak h$, one shows that $\mathfrak h$ is homogeneous, i.e., if it contains an infinite linear combination of $L_n$ with nonzero coefficient at some $L_j$, then it also contains the element $L_j$ itself.  Now one can use the action of $L_{-1}$ and $L_n$ with $n\gg0$ in $\mathfrak h$ to prove that $\mathfrak h=\mathfrak g$ if $\mathfrak h\ne0$.
Exercises: pinpoint a three-dimensional closed Lie subalgebra in $\mathfrak g$ isomorphic to $\mathfrak{sl}_2(k)$, and also an infinite family of pro-nilpotent open Lie subalgebras in $\mathfrak g$ forming a base of the topology of $\mathfrak g$. Realize that these Lie subalgebras are not Lie ideals in $\mathfrak g$.  Conclude that the Lie coalgebra $\mathcal L$ is a union of its finite-dimensional Lie coideals, but these Lie coideals are not Lie subcoalgebras.
