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?

  • $\begingroup$ Which book(s) are you using for learning this stuff? I ask this so as to be clear on the usage. $\endgroup$ May 28 '20 at 10:34
  • $\begingroup$ @SaaqibMahmood My text book is ""Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa $\endgroup$ May 28 '20 at 11:04
  • $\begingroup$ 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. $\endgroup$ May 28 '20 at 11:06
  • $\begingroup$ @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 $B\subseteq V$. $\endgroup$ May 28 '20 at 11:12
  • $\begingroup$ @SaaqibMahmood First to answer read the last part of the question: now I have edited it because there was a mistake. $\endgroup$ 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.

  • $\begingroup$ Okay, so the statement is only true for finit product? and if $|I|<\omega$ what can you say about my proof? $\endgroup$ May 28 '20 at 11:36
  • $\begingroup$ @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 $\endgroup$ May 28 '20 at 11:38
  • $\begingroup$ 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? $\endgroup$ May 28 '20 at 12:21
  • $\begingroup$ However It seems to me that this is a simple consequence of the regularity of $\Bbb{R}^n$, right? $\endgroup$ May 28 '20 at 12:51
  • $\begingroup$ @AntonioMariaDiMauro that as well. $\endgroup$ May 28 '20 at 12:52

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