Is every subspace of an incomplete inner product space closed?

Question

Let ($V$,$\langle \cdot , \cdot \rangle$) be an infinte-dimensional incomplete inner product space. Is it possible to find in it some subspace that is is not closed ? or is every subspace of this space closed ? consider finite and infinite dimensional subspaces.

Answer 1

A finite-dimensional subspace $U$ of $V$ is always closed: $U$ is complete with the induced norm (since all norms on $U$ are equivalent), so $\|u_n - x\| \to 0$ with $u_n \in U, x \in V$ implies that $u_n$ is a Cauchy sequence and hence converges to some $u \in U$. From this it is easy to see that $x = u \in U$ (because metric spaces are Hausdorff), i.e. $U$ is closed.

Infinite-dimensional subspaces may not be closed: Take for example $$c_{00}(\mathbb{C}) = \{(x_n)_{n \in \mathbb{N}} \in \mathbb{C}^{\mathbb{N}}: \exists\, m \in \mathbb{N} \text{ such that } x_n = 0 \, \forall {n \geq m}\}$$ and consider the inherited inner product of $\ell^2(\mathbb{C})$ on $V := c_{00}(\mathbb{C}) + \mathbb{C} \cdot z$ with $$z = (z_n)_{n \in \mathbb{N}} \in \ell^2(\mathbb{C}), \, z_n := \frac{1}{n}.$$

Then $V$ is not complete since it is not closed as a subset of $\ell^2(\mathbb{C})$. Furthermore $c_{00}(\mathbb{C})$ is dense in $\ell^2(\mathbb{C})$ and with this it is easy to see that $c_{00}(\mathbb{C})$ also is dense in $V$ with respect to the induced inner product. In particular, $c_{00}(\mathbb{C})$ is not closed in $V$ since $z \notin c_{00}(\mathbb{C})$.