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.

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

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