Topologies and Sigma-Algebras

# Topologies and Sigma-Algebras

Both topologies and $\sigma-$algebras are collections of subsets of a set $X$. What exactly is the difference between the two, and is there a relationship? We explore these notions by noting the definitions first. Let $X$ be any set.

### Topology

A topology $\tau$ is a collection of subsets of a set $X$ (also called a topology in $X$) that satisfies the following properties:

(1) $\emptyset \in \tau$, and $X \in \tau$

(2) for any finite collection of sets in $\tau$, $\{V_{i}\}_{i=1}^{n}$, $\cap_{i=1}^{n}V_{i} \in \tau$

(3) for any arbitrary collection of sets $\{V_{\alpha}: \alpha \in I\}$ in $\tau$ (countable or uncountable index set $I$), $\cup_{\alpha}V_{\alpha} \in \tau$

A topology is therefore a collection of subsets of a set $X$ that contains the empty set, the set $X$ itself, all possible finite intersections of the subsets in the topology, and all possible unions of subsets in the topology.

The simplest topology is called the trivial topology, where for a set $X$, $\tau = \{\emptyset, X\}$. Notice that (1) above is satisfied by design. All intersections we can make with the sets in $\tau$ are finite ones. (There’s just one, $X \cap \emptyset = \emptyset$.) Thus, (2) is satisfied. Any union here gives $X$, which is in $\tau$, so this is a topology.

This isn’t a very interesting topology, so let’s create another one.

Let’s take $X = \{1,2,3,4\}$ as the set. Let’s give a collection of subsets $\tau = \{\{1\},\{2\}, \{1,3\}, \{2,4\}, \{1,2,3\}, \{1,2,4\}, \emptyset, X\}.$ Notice that I didn’t include every single possible subset of $X$. There are two singleton sets, 4 pairs, and 1 set of triples missing. This example will illustrate that you can leave out subsets of a set and still have a topology. Notice that (1) is met. You can check all possible finite intersections of sets inside $\tau$, and notice that you either end up with $\emptyset$ or another of the sets in $\tau$. For example, $\{1,3\} \cap \{1,2,3\} = \{1,3\}$, $\{2\} \cap \{1\} = \emptyset$, etc. Lastly, we can only have finite unions here, since $\tau$ only has a finite number of things. You can check all possible unions, and notice that all of them result in a set already in $\tau$. For example, $\{1,3\} \cup \{2,4\} = X$, $\emptyset \cup \{1,2,4\} = \{1,2,4\}$, $\{1\} \cup \{1,3\} \cup \{2,4\} = X$, etc. Thus, $\tau$ is a topology on this set $X$.

We’ll look at one final example that’s a bit more abstract. Let’s take a totally ordered set $X$ (like the real line $\mathbb{R}$). Then the order topology on $X$ is the collection of subsets that look like one of the following:

• $\{x : a < x\}$ for all $a$ in $X$
• $\{x : b > x \}$ for all $b \in X$
• $\{x : a < b < x\}$ for all $a,b \in X$
• any union of sets that look like the above

To put something concrete to this, let $X = \{1,2,3,4\}$, the same set as above. This is a totally ordered set, since we can write these numbers in increasing order. Then

• The sets that have the structure $\{x : a < x\}$ for all $a \in X$ are
• $\{x : 1 < x\} = \{2,3,4\}$
• $\{x : 2 < x\} = \{3,4\}$
• $\{x : 3 < x\} = \{4\}$
• $\{x : 4 < x \} = \emptyset$
• The sets that have the structure $\{x : b > x\}$ for all $b \in X$ are
• $\{x : 1 > x\} = \emptyset$ (which we already handled)
• $\{x : 2 > x\} = \{1\}$
• $\{x : 3 > x \} = \{1,2\}$
• $\{x : 4 > x\} = \{1,2,3\}$
• The sets that have the structure $\{x : a < x < b\}$ for all $a,b \in X$ are
• $\{x : 1 < x < 3\} = \{2\}$
• $\{x : 1 < x < 4\} = \{2,3\}$
• $\{x : 2 < x < 4\} = \{3\}$
• The remaining combinations yield $\emptyset$
• The sets that are a union of the above sets (that aren’t already listed) are
• $X = \{x : 1 < x\} \cup \{x : 2 > x\}$
• $\{1,2,4\} = \{x : 3 > x\} \cup \{x : 3< x \}$
• $\{1,3,4\} = \{x : 2 < x\} \cup \{x : 2 > x\}$
• $\{1,3\} = \{x : 2 > x\} \cup \{x : 2< x < 4\}$
• $\{1,4\} = \{x : 2 > x\} \cup \{x : 3 < x\}$
• $\{2,4\} = \{x : 1 < x < 3\} \cup \{x : 3 < x \}$

The astute reader will note that in this case, the order topology on $X = \{1,2,3,4\}$ ends up being the collections of all subsets of $X$, called the power set.

### Sigma-Algebra

Let $X$ be a set. Then we define a $\sigma$-algebra.

A $\sigma$-algebra is a collection $\mathfrak{M}$ of subsets of a set $X$ such that the following properties hold:

(1) $X \in \mathfrak{M}$

(2) If $A \in \mathfrak{M}$, then $A^{c} \in \mathfrak{M}$, where $A^{c}$ is the complement taken relative to the set $X$.

(3) For a countable collection $\{A_{i}\}_{i=1}^{\infty}$ of sets that’s in $\mathfrak{M}$, $\cup_{i=1}^{\infty}A_{i} \in \mathfrak{M}$.

Let’s look at some explicit examples:

Again, take $X = \{1,2,3,4\}$. Let $$\mathfrak{M} = \{\emptyset, X, \{1,2\}, \{3,4\}\}.$$ We’ll verify that this is a $\sigma-$algebra. First, $X \in \mathfrak{M}$. Then, for each set in $\mathfrak{M}$, the complement is also present. (Remember that $X^{c} = \emptyset$.) Finally, any countable union will yield $X$, which is present in $\mathfrak{M}$, so we indeed have a $\sigma-$algebra.

Taking another example, we’ll generate a $\sigma$-algebra over a set from a single subset. Keep $X = \{1,2,3,4\}$. Let’s generate a $\sigma-$algebra from the set $\{2\}$. $$\mathfrak{M}(\{2\}) = \{X, \emptyset, \{2\},\{1,3,4\}\}$$ The singleton $\{2\}$ and its complement must be in $\mathfrak{M}(\{2\})$, and we also require $X$ and its complement $\emptyset$ to be present. Any countable union here results in the entire set $X$.

### What’s the difference between a topology and a $\sigma-$algebra?

Looking carefully at the definitions for each of a topology and a $\sigma-$algebra, we notice some similarities:

1. Both are collections of subsets of a given set $X$.
2. Both require the entire set $X$ and the empty set $\emptyset$ to be inside the collection. (The topology explicitly requires it, and the $\sigma-$algebra requires it implicitly by requiring the presence of $X$, and the presence of all complements.)
3. Both will hold all possible finite intersections. The topology explicitly requires this, and the $\sigma-$algebra requires this implicitly by requiring countable unions to be present (which includes finite ones), and their complements. (The complement of a finite union is a finite intersection.)
4. Both require countable unions. Here, the $\sigma-$algebra requires this explicitly, and the topology requires it implicitly, since all arbitrary unions–countable and uncountable–must be in the topology.

That seems to be a lot of similarities. Let’s look at the differences.

1. A $\sigma-$algebra requires only countable unions of elements of the collection be present. The topology puts a stricter requirement —all unions, even uncountable ones.
2. The $\sigma-$algebra requires that the complement of a set in the collection be present. The topology doesn’t require anything about complements.
3. The topology only requires the presence of all finite intersections of sets in the collection, whereas the $\sigma-$algebra requires all countable intersections (by combining the complement axiom and the countable union axiom).

It is with these differences we’ll exhibit examples of a topology that is not a $\sigma-$algebra, a $\sigma-$algebra that is not a topology, and a collection of subsets that is both a $\sigma-$algebra and a topology.

#### A topology that is not a $\sigma-$algebra

Let $X = \{1,2,3\}$, and $\tau = \{\emptyset, X, \{1,2\},\{2\}, \{2,3\}\}$. $\tau$ is a topology because

1. $\emptyset, X \in \tau$
2. Any finite intersection of elements in $\tau$ either yields the singleton $\{2\}$ or $\emptyset$.
3. Any union generates $X$, $\{1,2\}$, or $\{2,3\}$, all of which are already in $\tau$

However, because $\{2\}^{c} = \{1,3\} \not\in \tau$, we have a set in $\tau$ whose complement is not present, so $\tau$ cannot also be a $\sigma-$algebra. We used (2) in the list of differences to construct this example.

#### A $\sigma-$algebra that is not a topology

This example is a little trickier to construct. We need a $\sigma$-algebra, but not a topology, so we need to find a difference between the $\sigma-$algebra and the topology where the topology requirement is more strict than the $\sigma-$algebra’s version. We focus on difference (1) here.

Let $X = [0,1]$. Let $\mathfrak{M}$ be the collection of subsets of $X$ that are either themselves countable, or whose complements are countable. Some examples of things in $\mathfrak\{M\}:$

• all rational numbers between 0 and 1, represented as singleton sets. (countable)
• the entire collection of rational numbers between 0 and 1, represented as a set itself (countable)
• $\left\{\frac{1}{2^{x}}, x \in \mathbb{N}\right\}$. (countable)
• $[0,1] \setminus \{1/2, 1/4, 1/8\}$ (not countable, but its complement is $\{1/2, 1/4, 1/8\}$, which is countable)
• $\emptyset$ (countable)
• $X = [0,1]$ (not countable, but its complement $\emptyset$ is countable)
$\mathfrak{M}$ is a $\sigma-$algebra because:

• $X \in \mathfrak{M}$
• All complements of sets in $\mathfrak{M}$ are present, since we’ve designed the collection to be all pairs of countable sets with countable complements, and all uncountable sets with countable complements.
• Finally, all countable unions of countable sets are countable, so those are present. The countable union of uncountable sets with countable complements will have a countable complement1, and thus all countable unions of elements of $\mathfrak{M}$ are also in $\mathfrak{M}$, so we have a $\sigma-$algebra.

n particular, every single point of $[0,1]$ is in $\mathfrak{M}$ as a singleton set. To be a topology, any arbitrary union of elements of $\mathfrak{M}$ must also be in $\mathfrak{M}$. Take every real number between 0 and 1/2, inclusive. Then the union of all these singleton points is the interval $[0,1/2]$. However, $[0,1/2]^{c} = (1/2,1]$, which is uncountable. Thus, we have an uncountable set with an uncountable complement, so $[0,1/2] \notin \mathfrak{M}$. Since it can be represented as the arbitrary union of sets in $\mathfrak{M}$, $\mathfrak{M}$ is not a topology.

#### A collection that is both a $\sigma-$algebra and a topology

Take any set $X$ that is countable, and let $2^{X}$ be the power set on $X$ (the collection of all subsets of $X$). Then all subsets of $X$ are countable. We have that $\emptyset$ and $X$ are present, since both are subsets of $X$. Since the finite intersection of some subcollection of subsets of $X$ is a subset of $X$, it is in the collection. The arbitrary union of subsets of $X$ is either a proper subset of $X$ or $X$ itself. Thus $2^{X}$ is a topology (called the discrete topology).

The complement of a subset of $X$ is still a subset, and if all arbitrary unions are in $2^{X}$, then certainly countable unions are. Thus $2^{X}$ is also a $\sigma-$algebra.

To be explicit, return to the above where $X = \{1,2,3,4\}$. Write out all possible subsets of $X$, including singletons, $\emptyset$, and $X$ itself, and notice that all axioms in both the $\sigma-$algebra and the topology definitions are satisfied.

#### Footnotes

1. This is a weird statement to parse. I recommend trying to prove this