# Are there topological properties that are finitely productive but not countably productive?

Let $X_i, i \in I$ be a set of topological spaces with property $P$. We would like to know whether $P$ holds for $\prod_{i \in I} X_i$ for $I$ of different cardinalities.

I have looked over a list of topological properties, it seems only "finiteness" is finitely productive (finite products of finite set is finite) but not countably productive.

In the literature, it seems there is a tendency to divide topological properties to finitely productive, countably productive, and arbitrarily productive. For example, in Munkres, compactness is first proved to be finitely productive, then immediately we jump to Tychonoff theorem.

But I have rarely seen instances where a property $P$ is held under finitely product, but fails hold under countable product.

On the other hand, a lot of properties seems to fail to cross the line between countably productive and arbitrarily productive. This includes separability, first countable, second countable, suslin, metrizability.

Are there any interesting or well known topological properties that holds under finite product but not countable products?

## 1 Answer

• You could argue that the very definition of the product topology gives a kind of answer to your question. Let $(X_i)_{i \in I}$ be a family of topological space and let $U_i \subset X_i$ be an open set for each $i \in I$. If the family is finite, then $\prod_{i \in I} U_i$ is an open subset of $\prod_{i \in I} X_i$ in the product topology. This is not (typically) the case when the family is infinite.
• A finite product of discrete spaces is discrete, but this isn't so for infinite products.
• A finite product of manifolds is a manifold, but not so for infinite products (according to the most conventional definition of a manifold).
• See also the answer of Nate Elredge here. So, for example, $\mathbb{R}^n$ can be given a compatible Banach space norm for each finite $n$, but the product topology on $\mathbb{R}^\mathbb{N}$ is not induced by any complete norm. (Added: actually this is only really true if we think of $\mathbb{R}^\mathbb{N}$ as having a preferred vector space structure.)
• Here's a pretty good one. Let's restrict attention to Hausdorff spaces so that the right definition of "locally compact" is not a point of contention. The product of two (hence of finitely many) locally compact spaces is locally compact. However, an infinite product of locally compact spaces is locally compact if and only if all but finitely many terms of the product are compact. Thus, for example, $\mathbb{R}^n$ is locally compact for each finite $n$, but $\mathbb{R}^\mathbb{N}$ is not locally compact.
• I thought about Banach spaces, but is the impossibility of assigning a complete norm to $\mathbb{R}^{\mathbb{N}}$ a topological property? – AJY Aug 12 '16 at 1:59
• No, I guess it is not. The argument of Nate Elredge involves the built-in vector space structure of $\mathbb{R}^\mathbb{N}$. In fact some googling turns up papers by R.D. Anderson asserting that the countable infinite product of lines is homeomorphic to Hilbert space! Surprising, to me.... link, another link. – Mike F Aug 12 '16 at 2:20
• Fascinating stuff. – AJY Aug 12 '16 at 5:28