On a quasi-complete dense subspace of a complete locally convex space

Question

Let $X$ be a complete locally convex (topological vector) space, and let $M$ be a dense subspace of $X$. If we suppose that $M$ is quasi-complete (i.e., every bounded closed subset of $M$ is complete), is it true that (necessarily) $M=X$ ? I think that this should be true, but I am unable to verify it. Can anybody give a hint ? Thanks in advance.

Answer 1

I think this is false by considering quasi-closure. In constructing the example below the idea behind it is the standard one that $(1,1,1,\ldots )$ lies in $l_\infty$ but not in $c_0$.

The quasi-closure $\widehat M$ of a subset $M$ of a locally convex space in the intersection of all quasi-closed sets $Q \supseteq M$ and a set $Q$ is quasi-closed if it contains all the closure points in $X$ of its bounded subsets. Each point of $\widehat M$ is called a strict closure point and any strict closure point is a closure point. However, a strict closure point does not have to be the closure point of a bounded subset of $M$.

So, let $X$ be the locally convex sum of countably many copies of $l_\infty$, i.e. $(\bigoplus l_\infty )_0 $. For any $x\in X$ we have: $$ x = (x_{11}, x_{12}, \ldots ; x_{21}, x_{22}, \ldots ;\ \ldots \ ;x_{i_01},x_{i_02},\ldots;0,0,\ldots;\ \ldots ) $$ where $x_{ik} = 0$ for all $k$ whenever $i \geq i_0$. We take basis vectors $e_{ik}$ where the k$^{th}$ element of the i$^{th}$ component is $1$ and everything else is $0$, so $$ e_{ik} = (0, 0, \ldots ; 0, 0, \ldots ;\ \ldots \ ;0, 0, \ldots ,0,1, 0, 0, \ldots;0,0,\ldots;\ \ldots ) $$ and so we can write $$ x = \sum_{i,k \in {\mathbb N}} x_{ik}e_{ik} $$ as a generic element of the space.

Now let $M = \mathop{span}\langle e_{1n}+e_{nk}\rangle_{n,k \in {\mathbb N}}$. Note first that $e_{12} \not\in M$ but that $\lim_k e_{12}+e_{2k} = e_{12}$ so $e_{12} \in \widehat M$. It follows that $e_{1k} \in \widehat M \setminus M$ for all $k$. So all finite sums $\sum_{k=2}^N e_{1k}$ lie in $\widehat M$ but not in $M$, and letting $N\rightarrow \infty$ yields that the limit of the sums $x_0 := \sum_{k \geq 2} e_{1k}$ also lies in $\widehat M$.

However, $x_0$ cannot be the closure point of a bounded subset of $M$ by construction because any such bounded subset must contain finite linear combinations of the $\{e_{1n}+e_{nk}\}_{n,k\in{\mathbb N}}$. This means $x_0$ is a strict closure point of $M$ but not a closure point of any bounded subset of it.

Finally, let $M'$ be the collection of such $x_0$ and consider $M\setminus M'$. This is dense and quasi-complete in $M$ by construction, but $M \setminus M' \not= M$.