1-st countability of a uncountable topological product Let $X = \displaystyle\prod_{\lambda \in \Lambda} X_{\lambda}$, where each $X_\lambda$ is $T_2$ and has at least two points. Prove that if $\Lambda$ is uncountable then $X$ is not 1st-countable.
I dont even know how to start proving. Maybe in the opposite direction, $X$ 1st-countable $\implies$ $\Lambda$ countable...
I know I have to use the $T_2$ assumption, but I dont know wheter its as simple as noticing that X is  also $T_2$, or do I have to do something with the projections to ceratin $\lambda$, and taking into account that the projections are continious.
Thanks
 A: Hint: Membership in a countable family of open sets depends only on countably many coordinates, as is being a superset of a member of such a family. Being a neighbourhood of a point in the product depends on $\lvert \Lambda\rvert$ many coordinates (due to each $X_\lambda$ being nontrivial).
In fact, if you choose your point carefully, all you need about $X_\lambda$ is for each to be nontrivial (as in nontrivial topology), so that there's an $x_\lambda\in X_\lambda$ with a neighbourhood distinct from $X_\lambda$ itself.
A: Let $x=(x_\lambda)_λ$ denote a point in $X$. Assume that $N^1, N^2,...$ is a countable family of open basic neighborhoods of $x$. That means each $N^k$ has the form $\prod_l N^k_l$ where for a fixed $k$ all but finitely many of the sets $N^k_l$ are equal to $X_l$. Let $\{λ^k_1,λ^k_2,...,λ^k_{n_k}\}$ be the set such that $N^k_{λ^k_i}$ is a proper subset of $X_{λ^k_i}$. By $\Lambda$ denote the union $\{λ^k_i\mid k\in\Bbb N\}$. What can you say about the cardinality of $\Lambda$? Can you find a neighborhood of $x$ which does not contains any of the $N^k$ ?
Note how similar this is to the proof that the cofinite topology on an uncountable space is not first-countable. The idea is the same: For a countable family of neighborhoods you can only make countably many choices as to where the neighborhoods are "small" (or "non-existent").
