Open linear subspace of a Hilbert space. Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.
Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = L^[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of  to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.
 A: If $M \leq \mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 \in B_r(0) \subset M$, but for every (none-zero) vector $v \in \mathcal{H}$, we have
 $$ \frac{r}{2\Vert v \Vert} v \in B_r(0) \subset M $$
But M is a linear subspace so $ v \in M $. Thus the only open subspaces of $ \mathcal{H} $ are $ \mathcal{H} $ itself.
A: No.
Let $N$ be a normed space and
$M \subsetneq N \tag 1$
a proper subspace.  Then $M$ contains no nonempty open set.  For if
$\emptyset \ne U \subset M \tag 2$
were open, with
$M \ni m \in U, \tag 3$
we could find $\rho > 0$ such that the open ball
$B(m, \rho) \subset U; \tag 4$
then picking any 
$0 \ne v \in N \setminus M \tag 5$
the vector 
$m + \alpha (v - m) \in B(m, \rho) \tag 6$
if $0 \ne \alpha \in \Bbb R$ is sufficiently small, since
$\Vert (m + \alpha (v - m)) - m \Vert = \Vert \alpha (v - m) \Vert = \vert \alpha \vert \Vert v - m \Vert < \rho \tag 7$
for
$\vert \alpha \vert < \dfrac{\rho}{\Vert v - m \Vert}; \tag 8$
but then
$m + \alpha(v - m) \in M, \tag 9$
whence
$\alpha(v - m) = m + \alpha(v - m) - m \in M, \tag{10}$
whence
$v - m \in M, \tag{11}$
whence
$v = v - m + m \in M, \tag{12}$
in contradiction to (5); therefore no $B(m, \rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.
