seratend said:
{0,1}^n for any finite n is countable however {0,1}^|N is uncountable (same cardinality as |R) what do you conclude?
I conclude that this example is not relevant for the construct of Fock space.
For V a separable Hilbert space, set
<br />
T = \sum_{n=0}^\infty \otimes^n V,<br />
with \otimes^0 V = \mathbb{C}. Just as an operational definition has to be given to the infinite sum
<br />
S = \sum_{n=0}^\infty a_n,<br />
an operational definition has to be given to the expression for T. The infinite sum of real numbers is defined by
<br />
S = \lim_{n\rightarrow\infty}s_n,<br />
where
<br />
s_n = \sum_{i=0}^n a_i.<br />
Note that an infinite sum does directly add up an infinite number of terms; infinity only appears in the limit process.
T is the set of sequences \left\{ \left( w_0 , w_1 , ... , w_i , ... \right) \right\} such that:
1) w_i \in \otimes^i V;
2)
<br />
\sum_{i=0}^\infty \|w_i\|^2 <\infty.<br />
Condition 1) gives T as a countably infinite direct sum of vector spaces, while condition 2) allows an inner product to be defined on T.
Let \left\{\psi_{ij} | j \in \mathbb{N} \right\} be an orthonormal basis for \otimes^i V. Then B_T = \left\{\psi_{ij} | \left(i ,j \right) \in \mathbb{N}_0 \times \mathbb{N} \right\} is an orthnormal basis for T. Since \mathbb{N}_0 \times \mathbb{N} is countable, B_T is countable, and T is separable.
In order for your example to be relevant, i would have to equal \mathbb{N}, which is not possible since \mathbb{N} is not an element of \mathbb{N}.
It is well-known that the Fock space based on a separable Hilbert space is itself separable.
Regards,
George
