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Royden's (Real analysis, 4th edition, p.231) definition:

Let $X$ be a nonempty set and consider a collection of mapping $F=\{f_\alpha:X\rightarrow X_\alpha\}_{\alpha \in \Delta}$, where each $X_\alpha $ is a topological space. The weakest (coarsest) topology for $X$ that contains the collections of sets $\mathbb{F}=\{f_\alpha^{-1}(A_\alpha):f_\alpha \in F, A_\alpha \ open \ in \ X_\alpha\}$ is called the weak topology for $X$ induced by $F$.

Let $\tau_w$ be the weak topology for $X$ induced be $F$. I want to show that $\tau_w$ is a topology. So I thought to proceed in the following way.

(1) I start by showing that $\mathbb{F}=\{f_\alpha^{-1}(A_\alpha):f_\alpha \in F, A_\alpha \ open \ in \ X_\alpha\}$ is a subbasis for some topology $\tau$ on $X$ $^{[*1]}$. For this I show that the collection of all finite intersections of elements of $\mathbb{F}$, say $\mathbb{F}'$, is a basis for $\tau$ on $X$. So far, I assumed that $\tau$ is a topology on $X$.

(2) Now I invoke the result: let $\mathcal{B}$ be a basis for a topology $\tau$ on $X$. Then $\tau$ equals the collection of all unions of elements of $\mathcal{B}$. Now, I gave a form for the topology generated by the subbasis $\mathbb{F}$ (which is a topology).

(3) Next, I show that this $\tau$ is actually the coarsest topology which contains $\mathbb{F}$. That is, $\tau = \tau_w$.

  • Do you think it is correct?
  • Do you think it is too much? I feel like it is much more straighforward than that.
  • How would you stab this question?
  • Is there a way to show this by directly showing that $\tau_w$ (the topology generated by the subbasis $\mathbb{F}$) satisfies the axioms of topological spaces?

Thanks in advance!

$^{[*1]}$Do I need to put $\emptyset$ and $X$ together with $\mathbb{F}$ to obtain a subbasis? I didn't require $f_\alpha$ to be surjective.

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    $\begingroup$ What are you trying to do? By the very definition that you wrote, the weak topology is a topology. This would be like trying to prove that the "smallest interval in $\mathbb{R}$ containing $0$ and $1$" is indeed an interval. $\endgroup$ – Luiz Cordeiro Jul 23 at 2:08
  • $\begingroup$ @LuizCordeiro I added the last question for you, in the post. $\endgroup$ – Celine Harumi Jul 23 at 2:14
  • $\begingroup$ Pick your favourite $\alpha$, then $\emptyset = f^{-1}_\alpha (\emptyset)$ and $X = f^{-1}_\alpha(X_\alpha)$ $\endgroup$ – Calvin Khor Jul 23 at 2:18
  • $\begingroup$ In (1), there is literally nothing to do, as any collection of subsets is a subbase (for some topology). I think you are confused about the construction of the coarsest topology containing a subbase $\endgroup$ – Calvin Khor Jul 23 at 2:21
  • $\begingroup$ @CalvinKhor I already saw others saying the same, $X=f_\alpha^{-1} (X_\alpha)$. But why this is necessarily true? Why not $X \supseteq f_\alpha^{-1} (X_\alpha)$ $\endgroup$ – Celine Harumi Jul 23 at 2:23
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I believe you have a few concepts mixed up. Let $X$ be a set.

  • A basis for a topology $\rho$ on $X$ is a collection $\mathcal{B}$ of open sets (i.e., elements of $\rho$) such that every element of $\rho$ is a union of elements of $\mathcal{B}$.
  • A subbasis for $\rho$ is a collection $\mathcal{F}$ of open sets such that the set $\mathscr{B}(\mathcal{F})$ of finite intersections of elements of $\mathcal{F}$ (i.e., those of the form $F_1\cap\cdots\cap F_n$, where $F_i\in\mathcal{F}$) is a basis for $\rho$.
  • Any collection $\mathcal{F}$ of subsets of $X$ such that $X=\bigcup\mathcal{F}$ (i.e., any element of $X$ belongs to some element of $\mathcal{F}$) is a subbasis for a unique topology on $X$. Namely, if we define $\mathscr{B}(\mathcal{F})$ as above, and let $\rho(\mathcal{F})$ be the set of all unions (possibly empty or infinite) of elements of $\mathscr{B}(\mathcal{F})$, then $\rho(\mathcal{F})$ is a topology and has $\mathcal{F}$ as a subbasis. By the very definition of subbasis, this is the only possible topology which has $\mathcal{F}$ as a subbasis.
  • (Note that the empty set is the union of the empty subcollection of $\mathscr{B}(\mathcal{F})$; More precisely, if we take the empty collection $\mathcal{A}:=\varnothing\subseteq\mathscr{B}(\mathcal{F})$, then $\varnothing=\bigcup A$, so $\varnothing\in\rho(\mathcal{F})$.)
  • Actually, $\rho(\mathcal{F})$ is the coarsest (weakest, smallest) topology on $X$ which makes all sets of $\mathcal{F}$ open, in the sense that any topology $\eta$ for which $\mathcal{F}\subseteq\eta$ also satisfies $\rho(\mathcal{F})\subseteq\eta$ (prove this).

Now to your problem, as a particular case of the last statement above, note that the weak topology is, by definition, $\rho(\mathbb{F})$, the collection of unions of finite intersections of elements of $\mathbb{F}$. I think this is what you were trying to prove.

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    $\begingroup$ "every element of $\rho$ is a union of elements of $\rho$ or$ \mathcal{B}$"? in the first point. $\endgroup$ – Celine Harumi Jul 23 at 2:34
  • $\begingroup$ @Danmat Corrected. Thank you $\endgroup$ – Luiz Cordeiro Jul 23 at 16:51

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