# Coarsest topology is a topology

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?

$$^{[*1]}$$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.

• 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. – Luiz Cordeiro Jul 23 at 2:08
• @LuizCordeiro I added the last question for you, in the post. – Celine Harumi Jul 23 at 2:14
• Pick your favourite $\alpha$, then $\emptyset = f^{-1}_\alpha (\emptyset)$ and $X = f^{-1}_\alpha(X_\alpha)$ – Calvin Khor Jul 23 at 2:18
• 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 – Calvin Khor Jul 23 at 2:21
• @CalvinKhor I already saw others saying the same, $X=f_\alpha^{-1} (X_\alpha)$. But why this is necessarily true? Why not $X \supseteq f_\alpha^{-1} (X_\alpha)$ – Celine Harumi Jul 23 at 2:23

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.
• "every element of $\rho$ is a union of elements of $\rho$ or$\mathcal{B}$"? in the first point. – Celine Harumi Jul 23 at 2:34