Let $\Omega$ be an open set in a topological space and $C(\Omega)$ be the vector space of continuous complex valued functions with the topology given by the following family of seminorms:

$$\rho_n(f)=\sup\{|f(x)|:x\in K_n\}$$

Where $K_n$ are nested compacts whose union is $\Omega$. The book then proceeds to define a metric that coincides with this topology given by:

$$d(f,g) = \sum_n \cfrac{\rho_n(f-g)}{1+\rho_n(f-g)}$$

And that got me thinking. Is $C(\Omega) \cong\prod_{n \in \mathbb{N}} C(K_n)$? (by which I mean homeomorphicc).

If so, is this a particular case of a more general construction? (The product is given the universal topology of the product)

  • $\begingroup$ Can you precise what meaning you give to $\cong$? $\endgroup$ Jul 12 '15 at 14:03
  • $\begingroup$ Sorry about that. I really am looking only for a homeomorphism $\endgroup$ Jul 12 '15 at 14:30

First, even though you don't say which book it is, I bet the book doesn't just say the $K_n$ are nested, I bet it assumes the stronger condition $$K_{n}\subset K_{n+1}^o,$$where $A^o$ is the interior of $A$. People sometimes says that compact sets satisfying this condition form an "exhaustion" of $\Omega$.

If you assume that stronger condition then $C(\Omega)$ is homeomorphic to a certain closed subspace of $\prod C(K_n)$, not to the whole product space.

We'll write elements of that product as $$(f_n)=(f_1,f_2,\dots),$$where $f_n\in C(K_n)$. Let $$S=\{(f_n)\in\prod C(K_n)\,:\,f_{n+1}|_{K_n}=f_n\}.$$ Let $\Phi:C(\Omega)\to S$ be the natural map $$\Phi(f)=(f|_{K_n}).$$It's easy to see that $\Phi$ is injective. To show it's surjective, start with $(f_n)\in S$. There's a unique $f:\Omega\to\mathbb C$ such that $$f_n=f|_{K_n}$$for all $n$. That says $\Phi(f)=(f_n)$, right? Well no. That says $\Phi(f)=f_n$ if we can show that $f$ is continuous. Showing that $f$ is continuous uses the fact that we actually have an exhaustion as defined above.

So $\Phi$ is a bijection. I'll let you show that $\Phi$ and its inverse are continuous.

  • $\begingroup$ Thanks, that's what i suspected! Do you know of another example where one can use infinite products to describe a construction not coming necessarily from a product. $\endgroup$ Jul 12 '15 at 16:10
  • $\begingroup$ Every complete locally convex space is a closed subspace of a product of Banach spaces. For Frechet spaces like $C(\Omega)$ (where $\Omega$ is locally compact and $\sigma$-compact) you can have a countable product. $\endgroup$
    – Jochen
    Jul 13 '15 at 6:28

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