# It is possible to prove that these two collections generate the same topology on $\mathbb{X}$?

Let $\mathbb{Y}$ a topological space whose topology $\tau$ is generated by a collection of subsets $\mathcal{B} \subset 2^\mathbb{Y}$.

In other words the topology $\tau$ of $\mathbb {Y}$ is the smallest topology ("Smaller" in partial order on $\subset$ sub-collections in the class of $2^\mathbb{Y}$) containing the collection $\mathcal{B}$ of subsets of $\mathbb{Y}$.

Let $\{\varphi_i\}_{i \in \Lambda}$ a family of functions defined in a set $\mathbb{X}$ and taking values ​​in topological space $\mathbb{Y}$. Put $\mathcal{A}_{\tau}$ as the collection of all subsets of $\mathbb{X}$ in the form $$\{\varphi_i^{-1}(V): V \in \tau \mbox{ and } i\in \Lambda \}.$$ and $\mathcal{A}_\mathcal{B}$ as $$\{\varphi_i^{-1}(B): B\in\mathcal{B}\mbox{ and } i \in \Lambda \}.$$ It is possible to prove that these two collections generate the same topology on $\mathbb{X}$? If not what assumptions are sufficient for this to be true? I am particularly interested in the case where $\mathbb{Y} = \mathbb{R}$.

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If $(S,\mathcal T)$ is a topological space, and $\mathcal C\subset 2^S$, then the topology generated by $\mathcal C$ can be given in the following way: add $S$ and $\emptyset$ to $\mathcal C$ in order to assume WLOG they are in $\mathcal C$, and let $\mathcal F$ the collection of finite intersections of elements of $\mathcal C$. Then the topology generated by $\mathcal C$ consists of arbitrary unions of elements of $\mathcal F$.

Back to the problem. As $\mathcal A_{\tau}$ contains $\mathcal A_{\mathcal B}$, we have to show that each element of $\mathcal A_{\mathcal B}$ is in the topology generated by $\mathcal A_{\tau}$. We just have to show it for elements of the form $\varphi_i^{-1}(V)$ for $V\in\mathcal T$ and $i\in I$. Indeed, if we have done it for these elements, we can do it for arbitrary unions of finite intersections of such elements. Write $V:=\bigcup_{i\in J}\bigcap_{j\in F_i}B_j$, where $J\subset I$ is arbitrary and $F_i$ is finite, $B_j\in\mathcal B.$

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Suppose that $F\subseteq\Lambda$ is finite, $V_i\in\tau$ for each $i\in F$, and $x\in\bigcap_{i\in F}\varphi_i^{-1}[V_i]$. For each $i\in F$ let $y_i=\varphi_i(x)\in V_i$; there is a $B_i\in\mathcal{B}$ such that $y_i\in B_i\subseteq V_i$, so $x\in\bigcap_{i\in F}\varphi_i^{-1}[B_i]\subseteq\bigcap_{i\in F}\varphi_i^{-1}[V_i]$. It follows immediately that $\mathcal{A}_\tau$ and $\mathcal{A}_{\mathcal{B}}$ generate the same topology on $\Bbb X$.

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Sorry this question was a mistake. Please disregard. – MathOverview Oct 12 '12 at 14:45

If $\mathcal{B}$ generates $\tau$ then for every $V \in \tau$ there are $(B_{i,j})_{i \in I,1\leq j \leq n_i}$, $B_{i,j} \in \mathcal{B}$ such that $V = \bigcup_{i \in I} \bigcap_{j=1}^{n_i} B_{i,j}$. Since $\varphi^{-1}(\bigcup_{i \in I} \bigcap_{j=1}^{n_i} B_{i,j})$ = $\bigcup_{i \in I} \bigcap_{j=1}^{n_i} \varphi^{-1}(B_{i,j})$ every open set generated by $A_\tau$ is also generated by $A_\mathcal{B}$. The converse is obivous because $\mathcal{B} \subset V$.

$A_\tau$ btw is called the initial topology on $\mathbb{X}$ and is the coarsest (smallest wrt $\subset$) topology on $\mathbb{X}$ which makes all the $\varphi_i$ continuos for a given topology on $\mathbb{Y}$.

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Note that it was not assumed that $\mathcal{B}$ is a base for a topology. For instance $\mathcal{B}$ could be the family of all unbounded open intervals $( - \infty , b )$, $( a , + \infty )$ for real numbers $a,b$, Then $\mathcal{B}$ generates the usual Euclidean topology on $\mathbb{R}$, but is not a base for it. (It is a sub-base, however.) – inactive... for now Oct 12 '12 at 14:39
@ArthurFischer Oh, right, I missed that. The proof still works though, because $\varphi^{-1}$ commutes with $\bigcup$ as well as $\bigcap$. – fgp Oct 12 '12 at 14:49

(I know the OP said to disregard this question, but I discovered some nice things that I'd like to share in this answer.)

Notations. Let me start by introducing a slightly different notation, writing $D^\#$ instead of $\mathcal{A}_D$: $$D^\# = \{ V,f : V \in D : f^{-1}[V] \}$$ Read this as "the set which, for each $V$ and $f$ such that $V \in D$, contains $f^{-1}[V]$". Here, and in the rest of the post, we let $f$ range implicitly over $\{\varphi_i\}_{i \in \Lambda}$.

We use $D^\cup$ and $D^\cap$ for the set of all arbitrary unions and of finite intersections of elements of $D$, respectively: \begin{align} & D^\cup = \left\{ \mathscr{F} : \mathscr{F} \subseteq D : \bigcup \mathscr{F} \right\} \\ & D^\cap = \left\{ F,G : F,G \in D : F \cap G \right\} \\ \end{align} (See also another answer of mine.) This means that we can write "the topology generated by $B$" as $B^{\cap\cup}$.

Problem statement. Using these notations, we are asked whether, for all $B$ and $\tau$ with $\tau = B^{\cap\cup}$, we have $\tau^{\#\cap\cup} = B^{\#\cap\cup}$; or equivalently, whether for all $B$ $$(0) \;\;\; B^{\cap\cup\#\cap\cup} = B^{\#\cap\cup}$$ Proof. The only thing we know about $^\#$, $^\cap$, and $^\cup$ are their definitions: these apply $f^{-1}[\cdot]$, $\cap$, and $\cup$ many times. And looking at the latter operators, we know that the order of $f^{-1}[\cdot]$ and $\cap$ doesn't matter, and the same for $f^{-1}[\cdot]$ and $\cup$. (I will not prove those distribution properties here.) Therefore it seems likely that for all $D$ $$(1) \;\;\; D^{\cup\#} = D^{\#\cup} \\ (2) \;\;\; D^{\cap\#} = D^{\#\cap} \\$$ If we would be able to prove these, then $(0)$ directly follows from a simple calculation: \begin{align} & B^{\cap\cup\#\cap\cup} \\ = & \;\;\;\;\;\text{"using (1) with D := B^\cap"} \\ & B^{\cap\#\cup\cap\cup} \\ = & \;\;\;\;\;\text{"using (2) with D := B"} \\ & B^{\#\cap\cup\cap\cup} \\ = & \;\;\;\;\;\text{"^{\cap\cup} is idempotent"} \\ & B^{\#\cap\cup} \\ \end{align} (I won't prove here that $^{\cap\cup}$ is idempotent.)

It turns out we can prove $(1)$, as follows: \begin{align} & D^{\cup\#} \\ = & \;\;\;\;\;\text{"definition of ^\#"} \\ & \{ V,f : V \in D^\cup : f^{-1}[V] \} \\ = & \;\;\;\;\;\text{"write V as a union using the definition of ^\cup"} \\ & \{ \mathscr{F},f : \mathscr{F} \subseteq D : f^{-1}\left[\bigcup \mathscr{F} \right] \} \\ = & \;\;\;\;\;\text{"distribution property of ^{-1}"} \\ & \{ \mathscr{F},f : \mathscr{F} \subseteq D : \bigcup \{ V : V \in \mathscr{F} : f^{-1}[V]\} \} \\ = & \;\;\;\;\;\text{"rewrite using the definition of ^{\cup}"} \\ & \{ V,f : V \in D : f^{-1}[V] \}^{\cup} \\ = & \;\;\;\;\;\text{"definition of ^\#"} \\ & D^{\#\cup} \\ \end{align} The proof for $(2)$ is very similar.

Observations. As a first observation, I quite like how the $^\#$ notation works out, and interacts with the $^\cap$ and $^\cup$ notations I'd invented earlier: $(0)$ brings out the structure of the problem in a way that is much clearer than writing $(\mathcal{A}_{B^{\cap\cup}})^{\cap\cup} = (\mathcal{A}_B)^{\cap\cup}$: after making the latter explicit, I suddenly realized that $\mathcal{A}$ is essentially a unary operator.

Also, I like how we've been able to avoid $\subseteq$, and instead have an equality proof. That is not what I expected when I started to work on this proof.

Finally, I dislike the set comprehension manipulation: the two steps above "using the definition of $^\cup$" are not really easy to follow. I don't yet see how to fix that.

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+1 For you. Excelent answer! Very nice notation. – MathOverview May 14 '13 at 11:26