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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 $U$?

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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.

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  • $\begingroup$ what do you mean by "it obeys the finitely many conditions trivially"? $\endgroup$
    – user643073
    Apr 28, 2020 at 22:29
  • $\begingroup$ @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. $\endgroup$ Apr 28, 2020 at 22:31
  • $\begingroup$ 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. $\endgroup$
    – user643073
    Apr 28, 2020 at 23:26

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