Coordinate spaces inherit certain properties of a product spaces. There is a theorem Kelley's General Topology says that:

Let $X_{a}$ be a topological space satisfying the first axiom of countability for each member $a$ of an index set $A$. Then the product $\prod\limits_{a\in A}X_{a}$ satisfies the first axiom of countability if and only if all but a countable number of the spaces $X_{a}$ are indiscrete.

Then he says in the end of the section that:

If the product space has a countable local base at each point, then so does each coordinate space.

My questions are:

*

*Does it means if the product space is $A_{1}$ space, then so is each coordinate space; but the product of $A_{1}$ spaces may not be a $A_{1}$ space?


*How to prove the second propsition?
Thanks for help in advance!

*

*Yes, exactly. The product of uncountably many first countable spaces is seldom first countable (you mentioned the exact condition).

*Let $f\colon X\to Y$ continuous, open and onto (as a projection) and suppose $x\in X$ has a countable local base of open neighbourhoods $(U_n)$. Let $y = f(x)$. Let $V_n = f[U_n]$, then (as $f$ is open) is a neighbourhood of $y$. Now let $V$ an arbitrary neighbourhood of $y$. Set $U = f^{-1}[V]$. $U$ is an open neighbourhood of $x$, so there is an $n$ such that $U_n \subseteq U$, hence (as $f$ is onto) $$ V_n = f[U_n] \subseteq f[U] = f[f^{-1}[V]] = V. $$
Therefore $(V_n)$ is a local base at $y$.

*Note that 2. only applies to projections if each factor is non-empty. But in the case of empty factors the proposition is wrong anyway: Let $X$ be a space without any countable local base. Then $X\times\emptyset$ has countable local bases at each of its (zero) points.

