Example of a topological vector space such that $E = M \oplus N$ algebraically, but not topologically I have the following question:

Give an example of a topological vector space $E$ with subspace $M$ and $N$, such that $E = M \oplus N$ algebraically, but not topologically (so $E \ncong M \sqcup N$).

I suspect that it must be infinite-dimensional, but I have no clue how to construct an example.
Any help would be appreciated.
 A: Let $M$ be a proper dense subspace of an infinite dimensional $E$, and pick $N$ to be any algebraic complement. Then $E=M\oplus N$ algebraically by construction, but the direct sum is not of topological vector spaces, as the projection $E=M\oplus N\to N$ is not continuous: its kernel is $M$, which is not closed.
A: I have found an answer (I think) thanks to the previous answer, but I'm not sure it's correct.
Take $\mathbb{R}$ as a $\mathbb{Q}$-vectorspace, with the euclidean topology. Suppose we have a $\mathbb{Q}$-basis $(e_i)_{i \in I}$, with some $e_j = 1$. Take $\mathbb{Q}$ as a subspace and take the quotient space $\mathbb{R}/\mathbb{Q}$, so we have $\mathbb{R} = \mathbb{Q} \oplus \mathbb{R}/\mathbb{Q}$ and a quotient map $\phi: \mathbb{R} \longrightarrow\mathbb{R}/\mathbb{Q}$.
The topology on $\mathbb{R}/\mathbb{Q}$ is the trivial topology, since it is produced by sets $\phi(V)$ with $V \subseteq \mathbb{R}$ such that $V \oplus \mathbb{Q}$ is open in $\mathbb{R}$. But the only open set in $\mathbb{R}$ containing $\mathbb{Q}$ is $\mathbb{R}$, so $V$ contains a representative of each class in $\mathbb{R}/\mathbb{Q}$, so $\phi(V) = \mathbb{R}/\mathbb{Q}$. We naturally have
$$\mathbb{R}/\mathbb{Q} \cong \bigoplus_{i\in I, i \neq j} \mathbb{Q}e_i$$
But then we're done, because for $\mathbb{R}$ with the euclidean topology to be the topological direct sum of $\mathbb{Q}$ and $\mathbb{R}/\mathbb{Q}$, $\psi: \mathbb{R}/\mathbb{Q}\rightarrow\mathbb{R}$ has to be continuous, but then $\psi^{-1}(]-1,1[)$ should be open in $\mathbb{R}/\mathbb{Q}$ and not empty (every $e_i$ can be reduced with a $q \in \mathbb{Q}$ to an element in $]-1,1[$). So $\psi^{-1}(]-1,1[) = \mathbb{R}/\mathbb{Q}$, but every $e_i$ can be enlarged with a $q \in \mathbb{Q}$, so it has a norm greater than 1. So it is not open and we're done.
Is this correct ?
A: Construction
In Addition to:
M. Suárez-Álvarez
Hilbert Space: $\mathcal{H}$
Ortonormal Basis: $\mathcal{E}$

Infinite Dimension:
  $$\dim\mathcal{H}=\infty:\quad\mathcal{S}:=\langle\mathcal{E}\rangle<\overline{\langle\mathcal{E}\rangle}=\mathcal{H}$$
  Desired Subspace!

