Invariant subspace problem: non-separable Hilbert space I was reading an article, about the invariant subspace problem. The statement of the problem is as follows:
Given an $n$-dimensional Hilbert Space (or complex Banach Space) $\mathcal{H}^n$, does every bounded linear operator $T \in L(\mathcal{H}^n, \mathcal{H}^n)$, that is, $T : \mathcal{H}^n \to \mathcal{H}^n$, have a non-trivial closed $T$-invariant subspace? i.e. a closed linear subspace $\Omega \subset \mathcal{H}^n$ such that $\Omega$ is not $\{0\}$ or $\mathcal{H}^n$ with $T(\Omega) \subseteq \Omega$.
Various special cases have been proven, such as the trivial case of non-zero finite-dimensional vector spaces, as every linear operator admits an eigenspace; the generalized case, however, has yet to be proven.
The author notes that a non-separable Hilbert Space $\mathcal{H}$ has an invariant subspace for $T \in L(\mathcal{H},\mathcal{H})$ if one considers $x \in \mathcal{H}$, the supspace generated by the orbit, $\mathcal{M}_x = \operatorname{span}\{T^nx \}_{n \in \mathbb{N}}$. Where, however, does this use the assumption that $\mathcal{H}$ has an uncountable basis?
 A: The space $\mathcal{M}_x$ is always $T$-invariant, whether $\mathcal{H}$ is separable or not.  However, if $\mathcal{H}$ is separable then we may have $\mathcal{M}_x = \mathcal{H}$, so it isn't a nontrivial invariant subspace.
For example, take $\mathcal{H} = \ell^2(\mathbb{N})$ with its usual orthonormal basis $\{e_1, e_2, \dots\}$, and let $T$ be defined by $T e_n = e_1 + e_{n+1}$.  Then if $x = e_1$, we have $T^n x = e_{n+1}$, so $\mathcal{M}_x = \operatorname{cl}(\operatorname{span}\{e_1, e_2, \dots\}) = \mathcal{H}$.
(Edit: In that example, I was thinking of $\mathcal{M}_x$ including $x$ itself, i.e. starting with $T^0 x$, but I just noticed that was not assumed.  This $T$ can't work if $x$ is not included because its range is not dense.  If I think of a better example I'll add it.  Perhaps something like the left shift $T^*$ with $x$ something like $x = \sum \frac{1}{n} e_n$.)
Since $\mathcal{M}_x$ is always separable, if $\mathcal{H}$ is not separable then we indeed have $\mathcal{M}_x \subsetneq \mathcal{H}$ which is what we want.
