1
$\begingroup$

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 $U$?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ what do you mean by "it obeys the finitely many conditions trivially"? $\endgroup$ Apr 28 '20 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 '20 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. $\endgroup$ Apr 28 '20 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.