Example for a Hilbert space and a subset $M \subset H$ with $M^{\bot}= \{0\}$ In the lecture we had the following theorem:

For a Hilbert space $(H, (\cdot,\cdot))$ and a subspace $M \subset H$ we have:
$$M^\perp=\{0\} \implies M \text{ is dense in } H$$
with $M^{\bot}$ defined as $M^{\bot}:=\{v \in H | (w,v)=0 , \forall w \in M\}$

I have proven  the above already. Now, I wanted to try using it on an example, but I can't think of a Hilbert space $H$ and a subspace $M$ so that the above is true. Could someone give me one?
Thanks in advance.
 A: For a non-trivial example, let $H=\ell^2(\mathbb{N})$ with the usual inner product, and let $M$ be the subspace of sequences which are eventually zero.
A: Consider the Hilbert space $L^2[0,1]$ and the subspace of all polynomials $\mathcal{P}[0,1]$.
By Stone-Weierstrass we know that $\mathcal{P}[0,1]$ is dense in $C[0,1]$ in the supremum norm, but since $\|\cdot\|_2 \le \|\cdot\|_\infty$, we get that $\mathcal{P}[0,1]$ is dense in $C[0,1]$ in $\|\cdot\|_2$.
Now, since $C[0,1]$ is dense in $L^2[0,1]$, we conclude that $\mathcal{P}[0,1]$ is a dense subspace of $L^2[0,1]$.
This gives:
$$\mathcal{P}[0,1]^\perp = \{0\}$$
but of course $\mathcal{P}[0,1] \ne L^2[0,1]$.
You even reduce $\mathcal{P}[0,1]$ to the space of all polynomials with rational coefficients to obtain a countable set whose orthogonal complement is trivial (it's no longer a subspace though).
A: A slightly boring example: take any Hilbert space $H$, and chose $M=H.$
A: One nontrivial example that I especially like is the following. Consider the Hilbert space $H=L^2(\mathbb R; e^{-x^2}\,dx)$ of all measurable functions $f\colon \mathbb R\to\mathbb C$  that satisfy 
$$
\|f\|_H^2:=\int_{-\infty}^\infty |f(x)|^2e^{-x^2}\, dx <\infty.$$ 
The set $P$ of all polynomial functions on $\mathbb R$ is contained in this space. It is nontrivial to prove that $P^\bot=\{0\}$, and this is typically done in treatments of the Hermite polynomials, such as the one on the Wikipedia page.
