# First countability and Hausdorff uncountable product

Let $$I$$ be uncountable. Suppose each $$X_i$$ is Hausdorff with atleast two distinct points, $$p_i,q_i$$. Put $$X=\prod_{i\in I}X_i$$ (product topology). Then $$X$$ cannot be first countable.

Proof: Suppose for a contradiction that $$X$$ is first countable. Fix $$f\in \prod_iX_i$$. Suppose it has a nested local basis, $$(U_n)_{n\in \mathbb{N}}$$.

Claim: There exists an open set $$U$$ such that $$f\in U$$, but for all $$n$$, $$U_n$$ is not contained in $$U$$.

The proof I read was something like:

Step 1: Recall that the product topology has a basis. Observe that, for each $$n$$ there exists a basis element, $$B_n$$, such that $$f\in B_n \subseteq U_n$$. Note that $$B_n$$ is the finite intersection of sets of the form $$\pi_i^{-1}(U_i)$$ where $$\pi_i$$ denotes the projection map and $$U_i$$ is open in $$X_i$$. So, define $$D$$ to be the the set of indices in the finite intersection for each $$n$$. Therefore, $$D$$ is countable, being the countable union of finite sets.

Step 2: Therefore, we may choose some $$P\in I$$ not in $$D$$. Consider $$X_P$$. Since $$X_P$$ has two distinct points, we may suppose $$q_P$$ is different from $$\pi_P(f)=f(P)$$. Since $$X_P$$ is hausdorff, let $$M_1$$ be a neighborhood of $$f(P)$$ and $$M_2$$ a neighborhood of $$q_P$$, such that $$M_1\cap M_2=\varnothing$$. Put $$U=\prod_{i\in I}O_i$$ where $$O_i= M_1$$ if $$i=P$$ and $$X_i$$ if $$i\neq P$$. This set is open and contains $$f$$.

But... I can't see why $$U_n$$ is not contained in $$U$$?

Suppose there was: So $$U_m \subseteq U$$ for some $$m$$. Then also $$f \in B_m \subseteq U_m$$ and we can write $$B_m = \bigcap_{j=1}^{N(m)} \pi^{-1}_{i_j}[U_{i_j}]$$ for some finite set of indices $$\{i_1, \ldots, i_{N(m)}\}$$ and open sets $$U_{i_j} \subseteq X_{i_j}$$ (this notation should have been introduced at step 1; it was described in words; here I make it more formal). By design we have that $$P$$ is not equal to any of the $$i_j$$ (it was chosen to be not in any of the countably many finite sets we need (the set $$D$$), one for each $$U_n ,B_n$$ pair).
Define a point $$f' \in \prod_i X_i$$ by $$f'(j)= f(j)$$ for $$j \neq P$$ and $$f'(P)= q_P$$; this point is in (even all) $$B_m$$ because it obeys the finitely many conditions trivially, but is by design not in $$U$$ (its value at $$P$$ must lie in $$M_1$$, which does not contain $$q_P$$).
This contradiction (we thought we had $$B_m \subseteq U$$ but $$f'$$ contradicts that) concludes the proof.
Note: we do not need Hausdorff, but $$T_1$$ is enough (and all spaces having at least two points): we could then pick $$X_P\setminus \{q_P\}$$ as the non-trivial open set at coordinate $$P$$, for any $$q_p \neq f(P)$$ again. We also don't use the nestedness, just the fact that open sets contain basic open sets (that depend on only finitely many coordinates), plus that a countable union of finite sets is countable.
• @topologicalorientablesurface the finitely many conditions to be in $B_m$. It's a finite intersection; the conditions are $f(i_j) \in U_{i_j}$ for $j=1, \ldots N(m)$. And for $f$ it holds by assumption, and $f'=f$ on these coordinates. Apr 28 '20 at 22:31
• thanks, that really helped. The notation was really confusing me, so I switched $f$ to $(y_i)_{i \in I}$ and that part was immediately obvious. So interesting how a change of notation could do such a thing. But thank you so much Henno. Your help is always appreciated. Apr 28 '20 at 23:26