Product of two first countable sets is first countable Prove that the product of two first countable spaces $X$ and $Y$ is first countable.
Attempt:Let $X$ and $Y$ be two first countable spaces. Let $(x,y) \in X \times Y$. Then $x \in X$ and $y \in Y$. Then $x$ has a countable neighborhood basis, $\beta_x$.Also $y$ has a countable neighborhood basis $\beta_y$. Then the set $\beta_x \times \beta_y=\{B_x \times B_y|B_x \in \beta_x,B_y \in \beta_y\}$ is a countable neighborhood basis at $(x,y)$. The set is countable since the cardinality is $|\beta_x \times \beta_y|$ and the cartesian product of two countable sets is countable. Each basic open set $U \times V$ containing $(x,y)$ contains some $B_x \times B_y$, since if $x \in U$ and $y \in V$, there is some $B_x$ with $x \in B_x\subset U$ and some $B_y$ with $y \in B_y \subset V$.So $(x,y) \in B_x \times B_y \subset U \times V$.
Comment: I looked at the proof of this for countable products and noticed the proof to be much more demanding, which makes me unsure if this attempt suffices.However I noticed this is generally the case for proofs involving infinite/finite products. Is this ok, if not how could I change/improve it?Also I am unsure if it appropriate to use the notation $\beta_x \times \beta_y$ for the set $\{B_x \times B_y|B_x \in \beta_x,B_y \in \beta_y\}$.
 A: Yes, it's correct. You rightly realised that you only need to show that a basic open neighbourhood of $(x,y)$ contains a set from $\beta_x \times \beta_y$ (this is quite a standard notation BTW) to show that the latter forms a local base at $(x,y)$.
Finite products now follow by induction of course, and we can get the countable index case as well (we cannot go higher), but we need a bit more set theory to count the base where we use basic open sets but with all non-trivial components from the correct countable component local bases. It's a bit more notation but not really a lot harder.
A: The issue is that an open set in $X \times Y$ doesn't necessarily have the form $U \times V$ for open $U \subset X$ and open $V \subset Y$. Consider an open ball in $\mathbb{R} \times \mathbb{R}$. This set is certainly open, but it cannot be expressed in the form $U \times V$.
The key here is realizing that $\{U \times V: U \subset X \textrm{ open}, V \subset Y \textrm{ open}\}$ is a basis for the topology on $X \times Y$.
