Inductive limit of separable Banach spaces

740 Views Asked by At

Let $\mathcal{C} = \{A_\alpha\}_\alpha$ be a chain of nesting separable Banach subspaces of $C_b(X)$, the space of all bounded continuous functions, on a locally compact space $X$. I am wondering if the inductive limit, denoted by $\lim_\alpha A_\alpha$, forms a separable Banach space.

PS I would like to know more about direct limit of Banach spaces, please advise me on a good reference as well.

1

There are 1 best solutions below

0
On BEST ANSWER

I found my answer. All we need is looking at the Gelfand spectrum of the chain of algebras. Due to a corollary in [Takeda, Zirô, Inductive limit and infinite direct product of operator algebras. Tôhoku Math. J. (2) 7 (1955), 67–86], if $\{A_\alpha\}$ is as mentioned in the question, there is a maximal element for the chain called $C^*$-inductive limit of $\{A_\alpha\}$ which is another $C^*$-algebra and its Gelfand spectrum is the projective limit of the Gelfand spectrum of $\{A_\alpha\}$.

Now recall that a commutative $C^*$-algebra is separable if and only if its Gelfand spectrum is metrizable. On the other hand, the projective limit of a (non-countable) net of metrizable metric spaces is not metrizable necessarily. Hence, we may find a (non-countable) chain of separable commutative $C^*$-algebra admitting an non-separable $C^*$-inductive limit.