A family such that separates points of closed sets.

Question

I have a some troubles with the next exercise.

Let X be a topological spaces and $\{X_j:j\in J\}$ be a family of topological spaces. For all $j\in J$ let $f_j:X\rightarrow X_j$ be a continuous function. We say that the family $\mathcal{C}=\{f_j:j\in J\}$ separates points of closed sets in $X$ if for all $F\subseteq X$ closed set and for all $x\in X\setminus F$, exist $j\in J$ such that $f_j(x)\notin\text{cl}_{X_j}(f_j[F])$ where $\text{cl}_{X_j}$ is the closure of a set in the $X_j$ space.

1) Prove that $\mathcal{C}$ separates points of closed sets in $X$ if and only if the family $A=\{f_j^{-1}[U]|j\in J, U\in\tau_j\}$ is a basis for $X$

2) Prove that if $\mathcal{C}$ separates points of closed sets in $X$, then $X$ has the weak topology induced by $\mathcal{C}$

My attempt:

1)

$\Rightarrow)$

Clearly, $\bigcup A=X$, Then, if we consider $B,C\in A$, we need find some $D\in A$ such that $D\subseteq B\cap C$

Let $f^{-1}_j[Uj],f^{-1}_i[U_i]\in A$. Then, if $x\in f^{-1}_j[Uj]\cap f^{-1}_i[U_i]$ , so, $x\notin X\setminus(f^{-1}_j[Uj]\cap f^{-1}_i[U_i])$ and clearly this set is closed, because $f^{-1}_j[Uj]\cap f^{-1}_i[U_i]$ is open because $f_{i,j}$ is continuous and $U_{i,j}$ is open. Then, by hypothesis there exist $k\in J$ such that $$f_k(x)\notin \text{cl}_{X_k}\left(f_k\left[X\setminus(f^{-1}_j[Uj]\cap f^{-1}_i[U_i]) \right] \right)$$Then, $f_k(x)\in X_k\setminus \text{cl}_{X_k}\left(f_k\left[X\setminus(f^{-1}_j[Uj]\cap f^{-1}_i[U_i]) \right] \right)$

In this way, $x\in f^{-1}_k\left[ X_k\setminus \text{cl}_{X_k}\left(f_k\left[X\setminus(f^{-1}_j[Uj]\cap f^{-1}_i[U_i]) \right] \right)\right]=X\setminus\text{cl}_{X_k}(f_k[X\setminus f^{-1}_j[U_j]\cap f^{1}_i[U_i]])$ and this set is open, again, because $f$ is continuos. I think that this set can be $D$, but, how can I prove it?

$\Leftarrow)$

Let $F\subseteq X$ be a closed set and $x\in X\setminus F$. We know that $X\setminus F$ is an open set, then, like $A$ is a basis, $X\setminus F=\displaystyle\bigcup_{i\in I}^{}f^{-1}_i[U]$.

Like $x\in X\setminus F$, then, there exist $i\in I$ such that $x\in f^{-1}_i[U]$ and $f^{-1}[U]$ is open. Then, $f_i(x)\in U$ and $U\subseteq f_i[X\setminus F]$. From here, I don't know how can I conclude.

2)

If we take another topology $\tau$ such that for all $i\in J$ $f_i$ is continuos, then, the topology generated by $A$, denoted by $\tau_A$ holds $\tau_A\subseteq\tau$. Then, $\tau_A$ is the weak topology. This fact is because if $\mathcal{C}$ separetes points of closed sets, then, $A$ is a basis for some topology.

I really appreciate any help you can provide me.

Answer 1

Suppose $\mathcal{B} = \{f_i^{-1}[U] : i \in I, U \text {open in } X_i \}$ is a base for $X$.

Then the family separates points from closed sets: let $x \notin C$ ,$C \subseteq X$ closed. Then $x \in X\setminus C$, which is open, and as $\mathcal{B}$ forms a base, there is some $i \in I$ and $U$ open in $X_i$, such that

$$x \in f_i^{-1}[U] \subseteq X \setminus C$$ and the latter means 1. that $f_i(x) \in U$ and 2. that $U \cap f_i[C] = \emptyset$, Because if there existed a $y\in U \cap f_i[C]$, then $f_i(c) \in U$ for some $c \in C$, but then $c \in f_i^{-1}[U] \cap C$ contradicting that $f_i^{-1}[U]$ should be disjoint from $C$. This $U$ shows that $f_i(x)$ has a neighbourhood missing $f_i[C]$, so $f_i(x) \notin \overline{f_i[C]}$

For the reverse implication, assume that the family $\{f_i : i \in I\}$ separates point and closed sets, and to now see that $\mathcal{B}$ is a base: let $U$ be open with $x \in U$. Then $C = X\setminus U$ is closed, and $x$ is not in it. So there is some $i \in I$ such that $f_i(x) \notin \overline{f_i[C]}$. So define $$O = f_i^{-1}[X_i \setminus \overline{f_i[C]}] \in \mathcal{B}$$ and clearly $x \in O$. But also $$O = f_i^{-1}[X_i \setminus \overline{f_i[C]}] \subseteq f_i^{-1}[X_i \setminus f_i[C]] \subseteq X\setminus C = X \setminus (X \setminus U) = U$$ which shows that $\mathcal{B}$ is a base for $X$

As to the second point , for any such family $\mathcal{C}$ the weak topology on $X$ has subbase $\mathcal{B}$ as defined above, in the sense that it generates the weak (initial) topology. So it always generates the topology, but in this case it's even a base, not a mere subbase.