# Homework Help: Topology generated by a collection of subsets of $X$

1. Aug 14, 2014

### mahler1

The problem statement, all variables and given/known data.

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?

2. Aug 14, 2014

### jbunniii

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

3. Aug 14, 2014

### gopher_p

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

4. Aug 14, 2014

### jbunniii

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

5. Aug 14, 2014

### mahler1

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?

6. Aug 14, 2014

### jbunniii

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.

Last edited: Aug 14, 2014