Weak topology generated by functions is a topology.

Question

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 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 $F$.

I was asked to show that $F$ is formed by continuous functions in the weak topology (for $X$ induced by $F$), and to show that the weak topology is indeed a topology.

For the first question, let $f_\alpha \in F$ be an arbitrary element and $A_\alpha\in \tau_\alpha$ an open set. Since the weak topology induced by $F$ contains $\mathbb{F}$, the preimage $f_\alpha^{-1}(A_\alpha)$ is open. So the continuity follows (where I supposed that the weak topology is indeed a topology).

I am a bit confused with the second question. To show that the weak topology is in fact a topology, I suppose that I have to show the axioms of a topological space. As far as I understand, I cannot observe the elements of the weak topology directly, which is a problem for who chooses to inspect the topological axioms. Instead, I could try to show that the weak topology induced by $F$ is the topology generated by $\mathbb{F}$. Does it automatically implies that the weak topology is a topology?

I am a beginner in Topology, so please, try to give a detailed answer if possible.

Thanks!

Answer 1

First note that the topology is indeed well-defined: if we have any collection $\mathcal{S}$ of subsets of some set $X$ (could even be an empty collection, it does not matter), we can always talk about the set of all topologies $\mathcal{T}$ on $X$ that contain $\mathcal{S}$ as a subcollection. The discrete topology ($\mathcal{P}(X)$, really) will be one of those, but maybe many more. And it’s a simple exercise to check that if $\{\mathcal{T}_i, i \in I\}$ is a set of topologies on $X$, then $\bigcap_{i \in I} \mathcal{T}_i$ is also a topology on $X$, and when we apply this to the set of all topologies that contain $\mathcal{S}$, this topology is the minimal topology (wrt inclusion) that contains $\mathcal{S}$.

So once you’ve verified the fact that the intersection of topologies is again a topology you have shown that the weak topology defined by $F$ is well-defined.

You can actually describe this topology pretty well: your “generating collection” (subbase is the usual name) is $$\mathcal{S}=\{f^{-1}[O]: f=f_\alpha \in F, O \subseteq X_\alpha \text{ open }\}$$ and the weak topology has as a base the set of all finite intersections from $\mathcal{S}$, so all sets of the form

$$f_{\alpha_1}^{-1}[O_1] \cap f^{-1}_{\alpha_2}[O_2] \cap \ldots \cap f^{-1}_{\alpha_n}[O_n]$$ where $n$ is a natural number, $f_{\alpha_1},\ldots, f_{\alpha_n} \in F$ and each $O_i$ open in its respective $X_{\alpha_i}$ of course. All open sets of the weak topology thus are unions of such basic open sets (as many as we like, I’ll spare you a notation for it).

Indeed all $f_\alpha \in F$ are trivially continuous in the weak topology as that topology contains $\mathcal{S}$, exactly the sets it needs to make all $f_\alpha$ continuous (and no more).

Answer 2

Royden defines the weak topology as the weakest topology containing the collection $\mathbb F$. This does not mean that $\mathbb F$ is a topology on $X$. As Lord Shark the Unknown said in his comment, it is a subbasis for the weak toplogy which we shall denote by $\mathbb T _w$. In fact, $\mathbb T _w$ is generated by $\mathbb F$ as follows:

1. Let $\mathbb F'$ denote the set of all finite intersections of elements of $\mathbb F$. It is clear that any topology containing $\mathbb F$ must also contain $\mathbb F'$.

2. Let $\mathbb T _w$ denote the set of all unions of elements of $\mathbb F'$. It is clear that any topology containing $\mathbb F'$ must also contain $\mathbb T _w$ .

It is an easy exercise to show that $\mathbb T _w$ is a topology. Thus it is the weakest topology containing $\mathbb F$.

Note that it is more common to use the phrase coarsest topology instead of weakest topology. Moreover, $\mathbb T _w$ is also called the initial topology generated by the family of functions $F$.

Edited on request:

Let us show that $\mathbb T _w$ is a topology.

1. $\emptyset, X \in \mathbb T _w$: Both sets are already contained in $\mathbb F$. Picking any $\alpha$, we see that $\emptyset = f^{-1}_\alpha(\emptyset)$ and $X = f^{-1}_\alpha(X_\alpha)$.

2. All unions of elements of $\mathbb T _w$ are contained in $\mathbb T _w$: Each element of $\mathbb T _w$ is a union of elements of $\mathbb F'$ and a union of unions of elements of $\mathbb F'$ is again a union of elements of $\mathbb F'$.

3. The intersection $U \cap V$ contained in $\mathbb T _w$ if $U, V \in \mathbb T _w$: Write $U = \bigcup_{a \in A} D_a , V = \bigcup_{b \in B} E_b$ with $D_a, E_b \in \mathbb F'$. Then $$U \cap V = (\bigcup_{a \in A} D_a) \cap (\bigcup_{b \in B} E_b) = \bigcup_{(a,b) \in A \times B} D_a \cap E_b$$ But each $D_a \cap E_b$ is an element of $\mathbb F'$ (intersection of two finite intersections of elements of $\mathbb F$). Thus $U \cap V \in \mathbb T_w'$.