In the characterization of purely nondeterministic processes, more precisely at proposition 2.2.7 of "Topics in Stochastic Processes", Ash, Gardner et al., after having defined, given $\{X(t), t \in T\}$ a stationary $L^2$ process, $$ \begin{aligned} H(X) & =L^2\{X(t), t \in T\} \\ H_t(X) & =L^2\{X(s), s \leq t\}, \quad t \in T \\ H_{-\infty}(X) & =\bigcap_{t \in T} H_t(X) . \end{aligned} $$ and having said that if $H_{-\infty}(X)=H(X)$ or, equivalently, $H_t(X)$ is the same for all $t$, we call $\{X(t), t \in T\}$ deterministic, nondeterministic if $H_{-\infty}(X) \varsubsetneqq H(X)$ and purely nondeterministic if $H_{-\infty}(X)=\{0\}$, the following is claimed:
Assume $\{X(n)\}$ purely nondeterministic. Write $H_n(X)=$ $H_{n-1}(X) \oplus D_n$; since $H_{n-1}(X) \cup\{a X(n): a \in C\}$ spans $H_n(X), \operatorname{dim} D_n \leq 1$. If $\operatorname{dim} D_n=0$, then $H_n(X)=H_{n-1}(X)$, so that $H_n(X)=H(X)$ for all $n$, and thus $\{X(n)\}$ is deterministic, a contradiction. Therefore $\operatorname{dim} D_n=1$ for all $n$.
It seems to me, however, that from this it is only possible to conclude that $\exists n: \text{dim }D_n=1$, and that we could indeed have $\text{dim }D_m=0$ for some m even if the process is nondeterministic. I figured that I am probably supposed to assume that $\{X(n)\}$ is stationary, so that I could say that $D_n=<X_{n+1}-P_{n}X_{n+1}>$ and $||X_{n+1}-P_{n}X_{n+1}||$ does not depend on $n$ by stationarity and is greater than $0$, otherwise the process would be deterministic, hence $\text{dim }D_n=1 \,\,\forall n$; I have however no idea on how it could be done without this assumption. Any hint would be greatly appreciated.