# The product of basic neighbourhood systems is a basic neighbourhood system for the product topology.

Lemma

If $$\mathfrak{X}=\{X_i:i\in I\}$$ is a collection of topological spaces and if $$\mathfrak{B}=\{\mathcal{B_i}: i\in I\}$$ is a collection of basic neighbourhood system for $$\pi_i(x)$$ for any $$i\in I$$ and for $$x\in\prod_{i\in I}X_i$$ then the collection $$\mathcal{B}=\{B\subseteq\prod_{i\in I}X_i:\pi_i[B]\in\mathcal{B}_i,\forall i\in I\}$$ is a basic neighbourhood system for $$x$$.

Proof. Since the proiections are open then if $$V$$ is a neighbourhood of $$x\in\prod_{i\in I}X_i$$ then $$\pi_i[V]$$ is a neighbourhood of $$\pi_i(x)$$ for each $$i\in I$$ and so there exist $$B_i\in\mathcal{B}_i$$ such that $$\pi_i(x)\in B_i\subseteq\pi_i[V]$$ so that $$x\in\bigcap_{i\in I}\pi^{-1}_i[B_i]\subseteq\bigcap_{i\in I}\pi^{-1}_i\big[\pi_i[V]\big]=V$$ but $$\bigcap_{i\in I}\pi^{-1}_i[B_i]\in\mathcal{B}$$ because $$\pi_j\big[\bigcap_{i\in I}\pi^{-1}_i[B_i]\big]=B_j$$ for any $$j\in J$$.

So I ask if the statement of the lemma is true and if not I ask to take a counterexample. Furthermore if the proof is correct: in particular I suspect that the equilities $$\bigcap_{i\in I}\pi^{-1}_i\big[\pi_i[V]\big]=V$$ and $$\pi_j\big[\bigcap_{i\in I}\pi^{-1}_i[B_i]\big]=B_j$$ are false so I ask to prove them. So could someone help me, please?

• Which book(s) are you using for learning this stuff? I ask this so as to be clear on the usage. May 28 '20 at 10:34
• @SaaqibMahmood My text book is ""Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa May 28 '20 at 11:04
• OK. Please give the relevant definitions such as that of "neighborhood" and "neighborhood system". That will enable one to be clear on the usage of the terms. May 28 '20 at 11:06
• @SaaqibMahmood Okay. So if $X$ a topologycal space then for $x\in X$ a neighbourhood $V$ of $x$ is a set that contains an open set $U$ such that $x\in U\subseteq V$. Furthermore a collection of neighbourhoods $\mathcal{B}(x)$ of some $x\in X$ is a basic neighbourhood system if for any neighbourhood $V$ of $x$ there exist $B\in\mathcal{B}(x)$ such that $B\subseteq V$. May 28 '20 at 11:12
• @SaaqibMahmood First to answer read the last part of the question: now I have edited it because there was a mistake. May 28 '20 at 11:22

The sets in $$\mathcal{B}$$ are not even neighbourhoods in the product topology, so they certainly do not form a neighbourhood system at all, if $$I$$ is infinite.

A better choice (that does work)

$$\mathcal{B}(x) = \{\bigcap_{i \in F} \pi^{-1}[B_i]: F \subseteq I \text{ finite and } \forall i \in F: B_i \in \mathcal{B}_i\}$$

Using this we can e.g. show that a countable product of first countable spaces is first countable.

• Okay, so the statement is only true for finit product? and if $|I|<\omega$ what can you say about my proof? May 28 '20 at 11:36
• @AntonioMariaDiMauro The proof doesn't work: $\pi_i[V] \subseteq \pi_i[W]$ for all $i$ does not imply $V \subseteq W$. The formualtion of $\mathcal{B}$ is too lax: if includes "diagonal sets" too. You really need sets $\prod_i B_i$ where almost all $B_i = X_i$, i.e. sets as in my base sets May 28 '20 at 11:38
• Okay, now I remember what you showed to me here. Then is true that using this result I can prove that the (close) rectangle of $\Bbb{R}^n$ is a basic neighbourhood system? May 28 '20 at 12:21
• However It seems to me that this is a simple consequence of the regularity of $\Bbb{R}^n$, right? May 28 '20 at 12:51
• @AntonioMariaDiMauro that as well. May 28 '20 at 12:52