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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?

David C. Ullrich
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Rajeev ranjan
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  • Yes, it is a Hilbert space. The inner product on $\Pi_{\omega}H_n$ is given by $\left\langle (x_n)n,(y_n)_n \right\rangle=\lim{n\rightarrow \omega}\left\langle x_n,y_n \right\rangle$. – Rajeev ranjan Aug 06 '21 at 05:21
  • My bad. $ \ \ $ – Martin Argerami Aug 06 '21 at 05:34
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    $\ell^\infty(\mathbb N)\cong C(\beta\mathbb N)$ and $\omega$ corresponds to a point $x\in\beta\mathbb N\setminus \mathbb N$ so that $I\cong {f\in C(\beta\mathbb N\mid f(x)=0}$ under the isomorphism above. But this is a maximal ideal, hence $\ell^\infty(\mathbb N)/I\cong \mathbb C$. – MaoWao Aug 06 '21 at 09:32

1 Answers1

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Unfortunately, $\Pi_\omega H_n$ is not separable if the $H_n$ are infinite dimensional (separable or not). There are probably issues as long as they have more than 1 (real) dimension, but I don't have a clean argument for that case.

We can do this as a proof by contradiction, suppose that $(\xi_n^m)_n \in \Pi_\omega H_n$ was a countable dense subset, here I denote an element of the indices of the ultra product with subscripts whereas superscripts denote the indexing of the countable collection of elements.

For each $m\in \mathbb{N}$ we can find $e_m\in H_m$ such that $e_m\cdot \xi^j_m =0$ for all $1\leq j\leq m$ and $||e_m||=1$, this is where we use the fact that each $H_m$ in infinitely dimensional so that we can find a unit vector perpendicular to a given finite collection of vectors, $\{\xi_{m}^j\}_{j=1}^m$. We can now see that $(e_n)\in \Pi_\omega H_n$ with $||(e_n)|| = \lim_{n\rightarrow \omega} ||e_n|| = \lim_{n\rightarrow \omega}1 =1$ and that $\langle (e_n), (\xi^m_n)\rangle = \lim_{n\rightarrow \omega} \langle e_n, \xi^m_n\rangle$ but this sequence $\langle e_n, \xi^m_n\rangle$ is zero for each $n>m$ and thus we know that $\lim_{n\rightarrow \omega} \langle e_n, \xi^m_n\rangle =0$. Hence, we have found a unit vector orthogonal to the countable dense set which is a contradiction.

Sorry, I don't have the rep to comment, so I will point out here why the previous response doesn't work. They have not shown the map is an embedding (i.e. injective). Using a construction similar to the above you find such an element $(\xi_n)\neq 0$ where $T(\xi_n)=0$.

jhop
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    Presumably you meant $|e_m| = 1$ in the beginning of your third paragraph? As for your speculation at the beginning, it is not true that $H_n$ having more than $1$ dimension implies the ultraproduct is nonseparable - $(\mathbb{C}^n)^\omega = \mathbb{C}^n$ by compactness of the unit ball of $\mathbb{C}^n$, for example. However, $\lim_\omega \dim(H_n) = \infty$ is sufficient (also necessary, in fact) to imply the ultraproduct is nonseparable, by modifying your construction to $e_m \cdot \xi_m^j = 0$ for $1 \leq j \leq \dim(H_m) - 1$. – David Gao Apr 26 '25 at 01:07
  • Three additional remarks: 1. The Hilbert dimension of $\ell_2^\omega$ can be explicitly computed - it is exactly the continuum. The lower bound can be established by noting that there is a continuum sized collection of infinite subsets of $\mathbb{N}$ which are pairwise almost disjoint and using this to construct an orthonormal set in $\ell_2^\omega$ of cardinality continuum. The upper bound follows by noting that $\ell_2^\omega$ has cardinality continuum. – David Gao Apr 26 '25 at 01:54
  • (The exact same argument also works for $\prod_\omega H_n$ whenever all $H_n$ have Hilbert dimensions less than or equal to continuum and $\lim_\omega \dim(H_n) = \infty$.) – David Gao Apr 26 '25 at 01:56
  • This also works for Banach spaces in general. Here, instead of asking $e_m$ to be orthogonal to $\xi^j_m$, one asks they have distance at least $1/2$. One can then show that, for a sequence of Banach spaces $E_n$, $\prod_\omega E_n$ is nonseparable iff $\lim_\omega \dim(E_n) = \infty$. And, when $\lim_\omega \dim(E_n)$ is finite instead, we have $\prod_\omega E_n$ has dimension $\lim_\omega \dim(E_n)$. (One also needs the result that all norms on an $n$-dimensional space are equivalent to inner product norms in a uniform way to show the second half.)
    • – David Gao Apr 26 '25 at 02:00
  • (The dimension computation also works for general Banach spaces: if $E_n$ all have density character at most continuum, then $\prod_\omega E_n$ has density character continuum whenever $\lim_\omega \dim(E_n) = \infty$.) – David Gao Apr 26 '25 at 02:02
  • This is more or less the analogue of the fact in model theory that an ultraproduct is either uncountable, when the ultralimit of the cardinalities of the structures is infinite (and when all the structures have cardinalities at most continuum, the ultraproduct has cardinality exactly continuum); or finite, in which case the cardinality equals the ultralimit of the cardinalities of the structures. That the results match is not a surprise - Banach space ultraproducts are model-theoretic ultraproducts, just in continuous model theory.
    • – David Gao Apr 26 '25 at 02:07