Is ultraproduct of separable Hilbert space is separable? Let $(H_n)_n$ be a sequence of separable Hilbert spaces and $\omega$ be a free ultrafilter. Is it true that $\Pi_{\omega}H_n$ is also separable? Note that $\Pi_{\omega}H_n$ is the quotient space $\mathcal{l}^{\infty}(\mathbb{N},H_n)/I$, where $I=\{(x_n)_n\in\mathcal{l}^{\infty}(\mathbb{N},H_n):\ \lim_{n\rightarrow\omega}\|x_n\|=0 \}.$ If we take each $H_n=\mathbb{C}$, Then $\mathcal{l}^{\infty}(\mathbb{N},H_n)=\mathcal{l}^{\infty}(\mathbb{N})$, which is non separable. What about $I= \{(x_n)_n\in\mathcal{l}^{\infty}(\mathbb{N}):\ \lim_{n\rightarrow\omega}|x_n|=0 \}$. Is it separable?
 A: I will write $(\xi_n)^\bullet$ for the equivalence class of the sequence $(\xi_n)$ in the ultraproduct. I will also write $H^\omega$ for the ultraproduct $\prod_{n\to\omega}H$.
Since we can embed every separable Hilbert space into $\ell_2$, the question is really if $\ell_2^\omega$ is separable. To show this, let $(e_k)$ denote the canonical ONB of $\ell_2$ and let
$$
T\colon \ell_\infty(\ell_2)\to\bigoplus_{k=1}^\infty\mathbb C^\omega,\,(\xi_n)\mapsto ((\langle \xi_n,e_1\rangle)^\bullet,(\langle \xi_n,e_2\rangle)^\bullet,\dots).
$$
By a Fatou-type argument,
$$
\lVert T((\xi_n))\rVert^2=\sum_{k=1}^\infty\lim_{n\to\omega}\lvert\langle \xi_n,e_k\rangle\rvert^2\leq \lim_{n\to\omega}\sum_{k=1}^\infty \lvert \langle \xi_n,e_k\rangle\rvert^2=\lim_{n\to\omega}\lVert \xi_n\rVert^2=\lVert (\xi_n)^\bullet\rVert^2.
$$
Thus $T$ can be decomposed as
$$
\ell_\infty(\ell_2)\overset{q}{\to}\ell_2^\omega\overset{\iota}\to \bigoplus_{k=1}^\infty\mathbb C^\omega,
$$
where $q$ is the canonical projection and $\iota$ is a contraction.
As discussed in the comments, $\ell_\infty=C_b(\mathbb N)\cong C(\beta \mathbb N)$ and $I$ is a maximal ideal in $C(\beta \mathbb N)$ under this identification, so that $\mathbb C^\omega=\ell_\infty/I\cong \mathbb C$. Thus $\ell_2^\omega$ embeds into a separable Hilbert space, which means that it itself must be separable. In fact, since $\ell_2\hookrightarrow \ell_2^\omega$ as constant sequences, it follows that $\ell_2^\omega\cong\ell_2$.
