Prove that $\prod X_j$ is first-countable if and only if $X_j$ is first-countable, $\forall j\in J.$

At the 4th chapter on the "Elementos de Topología general" by Angel Tamariz and Fidel Casarrubias there is the following theorem

Theorem 4.19

Let be $$\mathfrak{X}=\{(X_j\ ,\mathcal{T_j}): j\in J\wedge|J|\le\aleph_0\}$$ numerable collection of topological spaces: then $$\Pi_{j\in J}X_j$$ is second-countable if and only if each $$X_j$$ is second-countable.

Well the authors ask to the student to prove the following statement

Let be $$\mathfrak{X}=\{(X_j\ ,\mathcal{T_j}): j\in J\wedge|J|\le\aleph_0\}$$ numerable collection of topological spaces: then $$\Pi_{j\in J}X_j$$ is first-countable if and only if each $$X_j$$ is first-countable.

Well I did it imitating the proof that the authors show for the case of the second-countable spaces.

Proof. Well first we suppose that $$\Pi_{j\in J}X_j$$ is first countable: then since each projection $$\pi_j$$ is a continuous, open and surjective function from previous theorem we can claim that its image $$X_j=\pi_j(\Pi_{j\in J}X_j)$$ is a first-countable topological space.

Now we suppose that each $$X_j$$ is a first-countable. Well we distinguish two cases: first we will suppose that $$|J|<\aleph_0$$ and then that $$|J|=\aleph_0$$.

• Well for starters we remember that the product of two first-countable topological spaces is first-countable (I proved this here ) so in the case that $$|J|<\aleph_0$$ through induction and through the associativity property of the cartesian product we can claim that $$\Pi_{j\in J}X_j$$ is first countable.
• So now we suppose that $$|J|=\aleph_0$$.Well first of all we consider that for any point $$x$$ of some topological space $$X$$ it result that for each neighborhood $$V_x$$ of $$x$$ there exist a open set $$U$$ such that $$x\in U\subseteq V_x$$ so, since any open set is a neighborhood of its points, we can prove the statement using local and open basis: indeed for what we observe it result that any (numerable) local basis contains a open (numerable) local basis. So now we consider the collection $$\mathcal{B}=\{\pi^{-1}_{i_1}(U_1)\cap...\cap\pi^{-1}_{i_n}(U_n): n\in\mathbb{N}\wedge i_1,...,i_n\in J\wedge U_j\in\mathcal{T_j},\forall j\in\{i_1,...,i_n\}\}$$ that is a basis for $$\Pi_{j\in J}X_j$$; then remembering the previous observation let be $$x\in\Pi_{j\in J}X_j$$ and we consider the collection of open and numerable local basis $$\mathfrak{B}=\{\mathcal{B }(\pi_1(x)),....,\mathcal{B}(\pi_j(x)),...\}_{j\in J}$$ for each projection $$\pi_j(x)$$ of $$x$$. Now we consider the collecion $$\mathcal{A}=\{\pi^{-1}_{i_1}(A_1)\cap...\cap\pi^{-1}_{i_n}(A_n): n\in\mathbb{N}\wedge i_1,...,i_n\in J\wedge A_j\in\mathcal{B}(\pi_j(x)),\forall j\in\{i_1,...,i_n\}\}$$ and we prove that it is numerable and that it is a open local basis for $$x$$. $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ Well we define the function $$\phi:\mathcal{A}\rightarrow\bigcup_{k\in\mathbb{N}}\mathbb{N}^k$$ such that $$\phi(\pi^{-1}_{i_1}(A_1)\cap...\cap\pi^{-1}_{i_n}(A_n))=(i_1...,i_n,l_1,...,l_n)$$ for $$A_j=B_{l_j}\in\mathcal{B}(\pi_{i_j}(x))=\{B^{i_j}_1,...,B^{i_j}_n,...\}$$ and $$j\in\{i_1,...i_n\}$$: it isn't difficult to demonstrate that $$\phi$$ is a injection and so, since $$|\bigcup_{k\in\mathbb{N}}\mathbb{N}^k|=\aleph_0$$, we proved that $$\mathcal{A}$$ is numerable. $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ Well now first of all we observe that for the continuity of the projections $$\pi_j$$ each element of $$\mathcal{A}$$ is a neighborhood of $$x$$. So let be $$V_x$$ an open neighborhood of $$x$$: in particular since for each point $$z\in\Pi_{j\in J}X_j$$ of some open set $$U$$ there exist $$B\in\mathcal{B}$$ such that $$z\in B\subseteq U$$, we suppose that $$V_x\in\mathcal{B}$$, that is $$V_x=\pi^{-1}_{i_1}(U_1)\cap...\cap\pi^{-1}_{i_n}(U_n)$$ where $$n\in\mathbb{N}$$ and $$i_1,...i_n\in J$$ and expecially $$U_j\in\mathcal{T_{i_j}}$$. Well if $$x\in V_x\pi^{-1}_{i_1}(U_1)\cap...\cap\pi^{-1}_{i_n}(U_n)$$ it result that $$\pi_k(x)\in\pi_k(V_x)=\begin{cases}U_k,\quad\mathscr{if\quad} k=i_j,\mathscr{\quad for\quad some\quad}i_j\in\{i_1,...,i_n\}\\X_k,\quad\mathscr{otherwise}\end{cases}$$ and so, since $$\pi_k(V_x)$$ is an open set for each $$k\in J$$ and since each open set is a neighborhood of its points, for each $$k\in J$$ there exist $$B_k\in\mathcal{B}(\pi_k(x))$$ such that $$x\in B_k\subseteq \pi_k(V)$$ from which we it result that $$x\in\pi^{-1}_{i_1}(B_1)\cap...\cap\pi^{-1}_{i_n}(B_n)\subseteq\pi^{-1}_{i_1}(U_1)\cap...\cap\pi^{-1}_{i_n}(U_n)=V_x$$ and so since $$\pi^{-1}_{i_1}(B_1)\cap...\cap\pi^{-1}_{i_n}(B_n)\in\mathcal{A}$$ we conclude that $$\Pi_{j\in J}X_j$$ is first-countable.

Well I ask if the proof is correct and in this case I ask also if the funcion $$\phi$$ is really an injection and if really $$\pi_k(V_x)=\begin{cases}U_k,\quad\mathscr{if\quad} k=i_j,\mathscr{\quad for\quad some\quad}i_j\in\{i_1,...,i_n\}\\X_k,\quad\mathscr{otherwise}\end{cases}$$ indeed since the projection function are not surjective it is $$E=\pi_{i_j}(\pi^{-1}_{i_j}(E))$$ for each $$E$$. Instead if the proof is uncorrect how prove the statement?

Could someone help me, please?

There are two things at play here: first a purely set theoretical fact:

The set of all finite subsets of $$J$$ is countable when $$J$$ is countable. This can be shown by showing that all finite subsets of $$\Bbb N$$ is a countable set (e.g. by enumerating all primes $$p_n, n \in \Bbb N$$ and sending a finite set $$F=\{n_1, n_2,\ldots, n_k\}$$ to $$p_{n_1}p_{n_2}\ldots p_{n_k} \in \Bbb N$$ and noting that unique prime factorisation proves this is an injection), or by noting all powers $$J^n$$ for $$n \in \Bbb N$$ are also countable (by induction starting from $$\Bbb N^2$$ being countable etc.) and then using that a countable union of countable sets is countable. Etc. No need for excluding finite index sets: all of those finite subsets still form a finite subset, and we can still catch them under the name countable. All we need is a countable local base in the end.

The second fact we use is topological: the standard base $$\mathcal{B}$$ for the topology on $$X=\prod_{j \in J} X_j$$ is all sets of the form $$\bigcap_{j \in F} \pi_j^{-1}[O_j]$$ where $$F \subseteq J$$ is finite and for all $$j \in F$$, $$O_j \in \mathcal{T}_j$$ (using the fact that inverse images of open sets under all projections form a subbase).

This gives us (remembering the case of two spaces here) the basic idea for showing first countability for $$X$$. Let $$p=(p_j)_{j \in J}$$ be a fixed point in $$X$$ and for each $$j$$ we have a countable local base $$\mathcal{B}(p_j)$$ for $$p_j \in X_j$$ by assumption. Now define

$$\mathcal{B}(p)=\{\bigcap_{j \in F} \pi_j^{-1}[B_j], F \subseteq J \text{ finite and } \forall j \in F: B_j \in \mathcal{B}(p_j)\}$$

and note that this is countable: we have countably (possibly even finitely) many choices for the finite subset $$F$$ and then for each of those finitely many sets we have a finite product of countable choices to make from the relevant $$\mathcal{B}(p_j)$$. So we can make all such sets in countably many ways as well (countable union of countable sets...).

Now if $$p \in O$$ and $$O$$ is product open, we use the standard base $$\mathcal{B}$$ from earlier to see that we have a finite subset $$F$$ again and finitely many open $$O_j$$ from $$X_j$$ for $$j \in F$$ such that

$$p \in \bigcap_{j\in F} \pi_j^{-1}[O_j] \subseteq O$$

and then apply the fact that we have local bases in the $$X_j$$ to pick for each $$j \in F$$ a $$B_j \in \mathcal{B}(p_j)$$ such that $$p_j \in B_j \subseteq O_j$$ and again note (as in the two spaces case):

$$p \in \bigcap_{j\in F} \pi_j^{-1}[B_j] \subseteq \bigcap_{j\in F} \pi_j^{-1}[O_j] \subseteq O$$ and the first set is in $$\mathcal{B}(p)$$ and we're done.

• Hi professor Brandsma, here I asked a question to which unfortunately I didn't receive a clear answer: could you help me, please? Feb 24 '20 at 20:03