Slides of a talk I gave on this topic

Recall the definition of the subgroup generated by a set and the closure of a set in a topological space. They look curiously similar, involving intersections of subgroups and closed sets respectively.

It turns out that mathematical structures that are closed under countable or uncountable intersections give rise to a minimal structure generated by some set. We would go through examples in algebra, topology and measure theory to explore this idea. This is closely related to structures closed under unions and similar ideas in topology.

## Subgroups generated by a set

From group theory, arbitrary intersection of subgroups are subgroups.

As such for any subset $$S$$ of a group $$G$$, we can define the subgroup generated by $$S$$ to be

$\braket{S}=\bigcap_{\substack{H \text{ subgroup of G} \\ \mathstrut \mathstrut S \subset H} } H$

and it satisfies the following properties

1. $$S \subset \braket{S}$$

2. $$\braket{S}$$ is a subgroup of $$G$$

3. If $$H$$ is a subgroup and $$S \subset H$$, then $$H$$ is a subgroup of $$\braket{S}$$

I.e. $$\braket{S}$$ is the smallest subgroup that contains $$S$$

Very similar ideas can be formulated for other algebraic structures such as normal subgroups, subrings, ideals and vector subspaces. For example, given subspaces $$U, W$$ of a vector space $$V$$, we have

$\braket{U \cup W} = U+W$

## Closures

From topology, arbitrary intersection of closed sets are closed.

As such for any subset $$A$$ of a topological space $$X$$ we can define the closure of $$A$$ to be

$\overline{A}=\bigcap_{\substack{F \text{ closed in } X \\ A \subset F}} F$

and it satisfies the following properties

1. $$A \subset \overline{A}$$

2. $$\overline{A}$$ closed

3. If $$F$$ closed and $$A \subset F$$, then $$F \subset \overline{A}$$

I.e. $$\overline{A}$$ is the smallest closed set that contains $$A$$

In light of this, I wonder if it would be more intuitive to think of the closure of $$A$$ as the closed set generated by $$A$$.

## Sigma algebras generated by a set

In measure theory, intersection of a family of $$\sigma-$$algebras is a $$\sigma-$$algebra. This motivates us to do the following.

Let $$\Omega$$ be some set. Let $$\beta$$ some subset of $$\sigma$$. Then we can define the $$\sigma-$$algebra generated by $$\beta$$ to be

$\mathcal{F}_\beta = \bigcap_{\substack{\mathcal{F}\text{ }\sigma-\text{algebra in } \Omega \\ \beta \subset \mathcal{F}}} \mathcal{F}$

and it satisfies the following properties

1. $$\beta \subset \mathcal{F}_\beta$$

2. $$\mathcal{F}_\beta$$ is a $$\sigma-$$algebra

3. If $$\mathcal{F}$$ is a $$\sigma-$$algebra in $$\Omega$$ and $$\beta \subset \mathcal{F}$$, then $$\mathcal{F}_\beta \subset \mathcal{F}$$

I.e. $$\mathcal{F}_\beta$$ is the smallest $$\sigma-$$algebra that contains $$\beta$$

## Generalisation

Let X be a set and $$F$$ be a family of subsets of X,

with the property that it’s closed under intersections $$F_i \in F \; \forall i \in I \implies \bigcap_{i \in I} F_i \in F$$

Then for some subset A of X, we can define $$\braket{A} = \bigcap_{S \in F, A \subset S } S$$ which has the following properties

1. $$\braket{A} \in F$$

2. $$A \subset \braket{A}$$

3. If $$S \in F$$ and $$A \subset S$$, then $$\braket{A} \subset S$$

## Closure Operator

It turns out that we can further generalise the above using the language of closure operators. A closure operator on a set $$S$$ is a function $$\text{cl}: P(X) \to P(X)$$ from the power set of $$X$$ to itself such that for all sets $$A, B \subset S$$,

1. $$A \subset \text{cl}(A)$$

2. $$A \subset B \implies \text{cl}(A) \subset \text{cl}(B)$$

3. $$\text{cl}(\text{cl}(A)) = \text{cl}(A)$$

With this generalisation in mind, let’s prove that the braket we’ve defined earlier is a closure operator. Define $$\braket{\phantom{}} : P(X) \to P(X)$$ as

$\braket{A} = \bigcap_{\substack{S \in F \\ A \subset S}} S$

with $$F$$ being closed under intersections.

Claim: $$A \subset \braket{A}$$

Proof: We have shown this as property 2 of the angle braket. $$\qquad \square$$

Claim: $$A \subset B \implies \braket{A} \subset \braket{B}$$

Proof: Suppose $$A \subset B$$. If $$S \in F$$ and $$B \subset S$$, then $$A \subset S$$. So $$\{S \vert S \in F, B \subset S\} \subset \{S \vert S \in F, A \subset S\}$$. As such $$\bigcap_{S \in F, A \subset S } S \subset \bigcap_{S \in F, B \subset S } S$$ as we’re intersecting over less sets. $$\qquad \square$$

Claim: $$\braket{\braket{A}} = \braket{A}$$

Proof: It is sufficient to show that $$\braket{A} = A \; \forall A \in F$$.

Let $$A \in F$$, then by definition $$A \subset \braket{A}$$.

As $$A \in F$$ and $$A \subset A$$, we have

\begin{align*} \braket{A} &= \bigcap_{\substack{S \in F \\ A \subset S }} S \\ &= A \cap \bigcap_{\substack{S \in F \\ A \subset S }} S \\ &\subset A\\ &\phantom{\subset A}\qquad \square \end{align*}

Other exmamples of closure operators include the convex closure.

## Conclusion

I find this parallel across algebra, topology and measure theory to be rather interesting. Thanks to my lecturer in integration, Stuart White, for pointing this idea out. You can read more about closure operators here and Kuratowski closure axioms, which could be used to define a topological space, here.

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