By the continuity of the inner product, a subspace of a Hilbert space is closed?

Question

Let $(e_n)$ be an orthonormal sequence in a Hilbert space $H$. The following are equivalent:

1. $(e_n)$ is complete.
2. $\text{clin}\{e_n \ | \ n\in \mathbb{N}\} = H$,
3. $||x||^2 = \sum_{n=1}^\infty |(x,e_n)|^2, \quad \text{for all} \ x\in H$.

For showing $(2)=>(1)$, the following text is used in the proof

Let $x\in H$ such that $(x,e_n)=0$ for all $n$. Define $$ E = \{y\in H : (x,y) = 0\}. $$ $E$ is a subspace of $H$ and by the continuity of the inner product it is closed.

I don't understand how the continuity of the inner product is used to state $E$ is closed?

Answer 1

Let $(y_n)$ be a convergent sequence in $E$ and denote by $y_0$ its limit.

We have $(x,y_n)=0$ for all $n$ and, by continuity of the inner product:

$(x,y_n) \to (x,y_0)$.

This shows $ (x,y_0) =0$, hence $y_0 \in E$.

Answer 2

Fix $x \in H$, and consider the function $$ H \to \mathbb C$$ sending $$ y \mapsto (x, y).$$ As you say, this is a continuous function.

Being continuous, the preimage under this function of every closed subset of $\mathbb C$ is a closed subset of $H$. In particular, the preimage of $\{ 0 \} \subset \mathbb C$ is $$ E = \{ y \in H \ : \ (x, y) = 0 \},$$ and this is a closed subset of $H$.