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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.

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2 Answers 2

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.

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Can you give a concrete example ? I thought of $\mathbb{R}$ as $\mathbb{Q}$-vectorspace, but I don't know how to construct the algebraic complement of $\mathbb{Q}$. –  KevinDL Nov 4 '11 at 12:59
    
I can, but it will be much more useful to you if you keep looking. In any case, there is no need to explicitly exhibit a complement! –  Mariano Suárez-Alvarez Nov 4 '11 at 13:23
    
Kevin: if you can prove that there is a linear functional $f$ on $E$ that is not continuous, you can take $M = \ker E$ and $N$ the span of any $v$ in $E$ satisfying $f(v) \neq 0$. (Sorry if this is giving away too much; I don't think it is, if only for the reason that depending on what $E$ you start with, this approach can be far from "concrete". For many $E$, it is possible to prove that such $f$ exist, but not possible to "construct" them in any reasonable sense.) –  leslie townes Nov 4 '11 at 18:59
    
For example, in $C[0,1]$ (the continuous real-valued functions on $[0,1]$ with topology of uniform convergence), let $M$ be the polynomials. You can't exhibit an algebraic complement of this in $C[0,1]$, but you can pick $N$ to be the span of any non-polynomial continuous function and let $E$ be the algebraic direct sum of $M$ and $N$, with the topology it inherits from $C[0,1]$. –  Robert Israel May 1 '12 at 7:28
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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 ?

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The example works. However, you write "But the only open set in $\mathbb{R}$ containing $\mathbb{Q}$ is $\mathbb{R}$". This is not true: To see this, enumerate the rationals as $\{q_{n}\}_{n=1}^{\infty}$, put $U = \bigcup_{n=1}^{\infty} (q_n - 2^{-n+1},q_n+2^{-n+1})$ and observe that $U$ has measure $\leq 4$ (the sum of lengths of the intervals defining $U$), so it can't be all of $\mathbb{R}$. The point here really is that the pre-image of an open set in $\mathbb{R}/\mathbb{Q}$ under the quotient map is open in $\mathbb{R}$ and invariant under translations by rational numbers. –  t.b. Nov 7 '11 at 12:05
    
Maybe a better (more natural) example would be the space of continuous functions inside the space $L^p[0,1]$. –  t.b. Nov 7 '11 at 12:11
    
Thank you, my mistake. So basically, you can take for $V$ to be an open interval $]a-\varepsilon,a+\varepsilon[$ and then you have $V + \mathbb{Q} = \mathbb{R}$, right? –  KevinDL Nov 7 '11 at 12:16
    
Yes. That's right. Just to be sure: If $V \subset \mathbb{R/Q}$ is open and non-empty then so is $U = \phi^{-1}(V) \subset \mathbb{R}$. In particular, $U$ contains an interval $I = (a-\varepsilon,a+\varepsilon)$, and because $U + q = U$ for all $q \in \mathbb{Q}$ we have that $I + q = (a+q-\varepsilon, a+q+\varepsilon) \subset U$. But every real number is contained in an interval $I+q$, so $U = \mathbb{R}$ and as $\phi$ is surjective we have $V = \phi(\phi^{-1}(U)) = \phi(\mathbb{R}) = \mathbb{R/Q}$. –  t.b. Nov 7 '11 at 12:25
    
Ok, thanks for the help –  KevinDL Nov 7 '11 at 12:44
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