A product space is never first countable if there are uncountably many factors. For example, is not first countable. In fact any dense subspace of is not first countable. In particular, the subspace of consisting of points which have at most countably many nonzero coordinates is not first countable. This subspace is called the product of many copies of the real line and is denoted by . However, this product is a Frechet space (or a FrechetUrysohn space). In this post, we show that the product of first countable spaces is a Frechet space.
Consider the product space . Fix a point . Consider the following subspace of :
The above subspace of is called the product of the spaces about the base point . When the base point is understood, we simply say the product of the spaces and use the notation to denote the space.
For each , define to be the set of all such that , i.e., the support of the point . Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set where each is an open subset of . The support of , denoted by is the finite set of all such that .
A space is said to be first countable if there exists a countable local base at each point in . A space is said to be a Frechet space if for each and for each , if , then there exists a sequence of points of such that the sequence converges to . Frechet spaces also go by the name of FrechetUrysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.
Theorem 1

Suppose each factor is a first countable space. Then the product is a Frechet space.
Proof of Theorem 1
Let . Let and let . We proceed to define a sequence of points such that the sequence converges to . For each , choose a countable local base at the point . Assume that . Then enumerate the countable set by . Let . The following set is an open subset of .
Note that is an open set containing . Choose . Enumerate the support by . Form the finite set by picking the first two points of and the first two points of , i.e., . Then form the following open subset of .
Choose . Enumerate the support by . Then let , i.e., picking the first three points of , the first three points of and the first three points of . Now, form the following open subset of .
Choose . Let this inductive process continue and we would obtain a sequence of points of . We claim that the sequence converges to . Before we prove the claim, let’s make a few observations about the inductive process of defining . Let .
 Each is the support of the open set .
 The sequence of open sets is decreasing, i.e., . Thus for each integer , we have for all .
 The support of the point is contained in , i.e., .
 The support of the each is contained in , i.e., .
 In fact, .
 The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets .
 Consequently, for each , for each . In other words, and each agree (and agree with the base point ) on the coordinates outside of the countable set .
Let be a standard open set in the product space such that . Let . We show that for some , for all .
Let be the support of . Let and . Consider the following open set:
Note that . For each , choose . Let be the maximum of all where . Then for each . Choose a positive integer such that:
Let . It follows that there exists some such that . Then for all . It is also the case that for all . This is because on the coordinates not in .
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