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Given a set $X$ and $\mathcal{A} \subseteq \mathcal{P}(X)$, prove that there exists a unique smallest topology on $X$ that contains $\mathcal{A}$

My attempt:

Define $$\mathcal{B}:= \mathcal{A} \cup \{\emptyset, X\}$$ $$\mathcal{C} := \{\bigcap \mathcal{E}\mid \mathcal{E} \subseteq \mathcal{B}, \mathcal{E} \ \mathrm{finite}\}$$ $$\mathcal{D} := \{\bigcup \mathcal{F}\mid\mathcal{F} \subseteq \mathcal{C}\}$$

We now prove that $\mathcal{D}$ is the topology we look for.

Take $\mathcal{E}:= \{\emptyset\} \in \mathcal{B}$. Then $\bigcap \mathcal{E} = \emptyset \in \mathcal{C}$ and if we put $\mathcal{F}:= \{\emptyset\} \subseteq \mathcal{C}$, it follows that $\bigcup \mathcal{F} = \emptyset \in \mathcal{D}$

In the same way, we can prove that $X \in \mathcal{D}$.

Now, let $A,B \in \mathcal{D}$. Then, there exists families $\mathcal{F_1}:= (F_i^1)_{i \in I}, \mathcal{F_2}:= (F_j^2)_{j \in J} \subseteq \mathcal{C}$ such that $A = \bigcup \mathcal{F_1}, B = \bigcup \mathcal{F_2}$

Hence, $A \cap B = (\bigcup_{i \in I} F_i^1) \cap (\bigcup_{j \in J}F_j^2) = \bigcup_{i \in I, j \in J}F_i^1 \cap F_j^2$

and hence it suffices to prove that $F_i^1 \cap F_j^2 \in C$, for every $i$ and $j$.

To do this, we note that $F_i^1, F_j^2 \in \mathcal{C}$. Hence, there are finite collections $ \mathcal{E_1}:=(E_k^1)_{k \in K}, \mathcal{E_2}:=(E_l^2)_{l \in L} \subseteq \mathcal{B}$ with $F_i^1 := \bigcap \mathcal{E_1}, F_j^2 := \bigcap \mathcal{E_2}$

and hence $$F_i^1 \cap F_j^2 = \bigcap_{k \in K} E_k^1 \cap \bigcap_{l \in L} E_l^2$$

and this is a finite intersection of sets out of $\mathcal{B}$, so $F_i^1 \cap F_j^2 \in \mathcal{C}$

Now, let $\mathcal{G}:= (G_n)_{n \in N} \subseteq \mathcal{D}$. Then for each $n \in N$, there is a collection $\mathcal{F_n} := (F_m^n)_{m \in M} \subseteq \mathcal{C}$ with $G_n = \bigcup\mathcal{F_n}$

So, $$\bigcup \mathcal{G} = \bigcup_{n \in N} G_n =\bigcup_{n \in N}(\bigcup \mathcal{F_n}) =\bigcup_{n \in N}\bigcup_{m \in M} F_m^n = \bigcup_{(n,m) \in N\times M} F_m^n$$

which is a union of elements in $\mathcal{C}$, so it follows that $\bigcup \mathcal{G} \in \mathcal{D}$

and this shows that $\mathcal{D}$ is a topology.

It is clear that $\mathcal{A} \subseteq \mathcal{D}$, and if the topology $\mathcal{T}$ contains $\mathcal{A}$, then, because $\mathcal{T}$ contains $\emptyset, X$ as well (it's a topology), it follows that $\mathcal{B} \subseteq \mathcal{T}$ and because $\mathcal{T}$ is a topology, $\mathcal{C} \subseteq \mathcal{T}$ and also $\mathcal{D} \subseteq \mathcal{T}$

This shows that $\mathcal{D}$ is the smallest topology that contains $\mathcal{A}$ (and the uniqueness is evident).

Is this correct? Do you have any suggestions? I'm not entirely sure if the equality $\bigcup_{n \in N}\bigcup_{m \in M} F_m^n = \bigcup_{(n,m) \in N\times M} F_m^n$ is valid. In my proof, I also use that I can index collections of sets. Is this always possible? Thanks.

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Your idea is correct (I haven't digested your whole answer yet) which is definitely the main thing here.

Actually you can start with $\mathcal A$ itself instead of $\mathcal B=\mathcal A\cup\{\varnothing,X\}$.

This under the convention that the empty (hence finite) intersection of elements of $\mathcal A$ is identified with $X$. Further $\varnothing$ can be identified with empty union so that $\varnothing\in\mathcal D$.

This setup provides a nice inductive definition of the topology that is generated by subbasis $\mathcal A$.

The collection $\{\bigcap\mathcal E\mid\mathcal E\subseteq\mathcal A,\mathcal E\text{ finite}\}$ is a basis for a topology according to this link.

The arbitrary unions of such a basis always form a topology.

Mostly the topology generated by $\mathcal A\subseteq\wp(X)$ is presented as the intersection of all topologies on $X$ that contain $\mathcal A$ as a subcollection. That is a correct route, but your route (inductive definition) gives a far better insight on the way the topology is constructed.


edit (my proof).

Let $\mathcal C:=\{\bigcap\mathcal E\mid\mathcal E\subseteq\mathcal A,\mathcal E\text{ finite}\}$. Then (by convention) $X=\bigcap\varnothing\in\mathcal C$ and secondly it is immediate that $\mathcal C$ is closed under finite intersections. This makes $\mathcal C$ a basis of a topology according to this link.

Let $\mathcal D:=\{\bigcup\mathcal F\mid\mathcal F\subseteq\mathcal C\}$ so that $\varnothing=\bigcup\varnothing\in\mathcal D$ and $X\in\mathcal C\subseteq\mathcal D$. Also it is immediate that $\mathcal D$ is closed under unions (a union of unions of elements of $\mathcal C$ is a union of elements of $\mathcal C$). Further for $\mathcal F_i\subseteq\mathcal C$ with $i=1,2$ we find: $(\bigcup\mathcal F_1)\cap(\bigcup\mathcal F_2)=\bigcup\{F_1\cap F_2\mid F_1\in\mathcal F_1, F_2\in\mathcal F_2\}$. From $F_i\in\mathcal F_i\subseteq\mathcal C$ for $i=1,2$ it follows that $F_1\cap F_2\in\mathcal C$ since $\mathcal C$ is closed under finite intersections. So $(\bigcup\mathcal F_1)\cap(\bigcup\mathcal F_2)\in\mathcal D$ and proved is now that $\mathcal D$ is closed under finite intersections.

Proved is now that $\mathcal D$ is a topology, and this with $\mathcal A\subseteq\mathcal D$. If conversely $\mathcal E$ is a topology with $\mathcal A\subseteq\mathcal E$ then it is evident that also $\mathcal C\subseteq\mathcal E$ and secondly that also $\mathcal D\subseteq\mathcal E$. So it is clear that $\mathcal D$ is the "smallest" topology that contains $\mathcal A$ as a subcollection. In this context $\mathcal A$ is a subbase of $\mathcal D$ and $\mathcal C$ is a base of $\mathcal A$.

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  • $\begingroup$ Thanks for your answer. I don't mind that you didn't read the entire answer, but can you verify that the union identity that I wrote down in my proof is correct? $\endgroup$ – user370967 Feb 13 '18 at 11:28
  • $\begingroup$ Yes, I think its okay. $\endgroup$ – drhab Feb 13 '18 at 11:33
  • $\begingroup$ I added a proof so that you can compare. Actually the fact that $\mathcal D$ is closed under unions is immediate by its definition. $\endgroup$ – drhab Feb 13 '18 at 12:13
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    $\begingroup$ @emblindened actually I was a bit too eager to justify myself. Sorry, and thank you for sharing your alternative look on this matter. $\endgroup$ – drhab Feb 13 '18 at 18:19
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    $\begingroup$ I'm used to my own definition as well, that's why I was prepared to take a stance. I see now that (using the common definition) the subbase need not be a cover of $X$. Thank you as well. $\endgroup$ – suhogrozdje Feb 13 '18 at 18:27

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