Topology generated by a collection of subsets of ##X##

[mahler1]

Homework Statement .

Let $X$ be a set and $\mathcal A \subset \mathcal P(X)$. Prove that there is a topology $σ(A)$ on $X$ that satisfies

(i) every element of $A$ is open for $σ(A)$

(ii) if $\tau$ is a topology on $X$ such that every element of $\mathcal A$ is open for $\tau$, then $σ(A) \subset \tau$

The attempt at a solution.

I suppose the idea is to construct the smallest possible topology which contains $\mathcal A$

As the elements of $\mathcal A$ have to be open, then each $U \in A$ has to be in $σ(A)$. Also, $X,\emptyset$ have to be in $σ(A)$.

Since $σ(A)$ has to be closed under finite intersections and arbitrary unions, I'll have to add all sets that come from finite intersections of $X$ with elements of $\mathcal A$.

So my idea was to define $σ(A)=\{S \subset X : S=\bigcap_{i \in I} U_i, U_i \in \mathcal A, I \text{finite}\} \cup \{X\}$ but when I've tried to show that $σ(A)$ was a topology I got stuck, so I don't know if that set works.

It is clear that $X \in σ(A)$, and that if $S_j \in σ(A)$ for $j \in J$ with $J$ finite, then since $S_j=\bigcap_{k=1}^{s_j} U_k$, $\bigcap_{j \in J} S_j=\bigcap_{j \in J} (\bigcap_{k=1}^{s_j} U_k)$, which is clearly a finite intersection of elements in $σ(A)$. The problem is with arbitrary unions, I couldn't show that arbitrary unions remain in $σ(A)$. Where is my mistake?

 

[jbunniii]

There's a much slicker way to do this. Hint: the intersection of an arbitrary number of topologies is a topology.

 

[gopher_p]

There's a much slicker way to do this. Hint: the intersection of an arbitrary number of topologies is a topology.

... as long as that arbitrary number isn't 0.

 

[jbunniii]

... as long as that arbitrary number isn't 0.

Yes! Good catch. The OP will need to show that there is at least one suitable topology being intersected to form his desired topology.

 

[mahler1]

There's a much slicker way to do this. Hint: the intersection of an arbitrary number of topologies is a topology.

Oh, $\bigcap_{i \in I, \mathcal A \subset \tau_i} T_i$, the intersection of all topologies which contain $\mathcal A$ is a topology (I've verified that intersection of an arbitrary number different from $0$ of topologies is a topology, and if $\tau$ is a topology such that every element in $\mathcal A$ is open in $\tau$, this implies $\mathcal A \subset \tau$, but then $\bigcap_{i \in I, \mathcal A \subset \tau_i} T_i \subset \tau$. Thanks very much for the super hint, is this correct now?

 

[jbunniii]

Oh, $\bigcap_{i \in I, \mathcal A \subset \tau_i} T_i$, the intersection of all topologies which contain $\mathcal A$ is a topology (I've verified that intersection of an arbitrary number different from $0$ of topologies is a topology, and if $\tau$ is a topology such that every element in $\mathcal A$ is open in $\tau$, this implies $\mathcal A \subset \tau$, but then $\bigcap_{i \in I, \mathcal A \subset \tau_i} T_i \subset \tau$. Thanks very much for the super hint, is this correct now?

Yes. The intersection is not empty because $\mathcal{P}(X)$ is a topology containing $A$. Any topology containing $A$ must be one of the $T_i$, so it contains the intersection.

This kind of argument is used all the time. "The intersection of all _____ with property P (assuming there is at least one) is a _____ with property P, and therefore it is the smallest such _____." It's an elegant and standard argument, but completely nonconstructive and so it may not give much insight into what the elements of the intersection actually look like.