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.

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.*

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

A

$\sigma$-algebrais 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 $X$.

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

- Both are collections of subsets of a given set $X$.
- 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.)
- 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.)
- 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.

- 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. - The $\sigma-$algebra requires that the complement of a set in the collection be present. The topology doesn't require anything about complements.
- 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.

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

- $\emptyset, X \in \tau$
- Any finite intersection of elements in $\tau$ either yields the singleton $\{2\}$ or $\emptyset$.
- 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.

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 complement. (This is a weird statement to parse. I recommend trying to prove this.) 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.

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.