### Browsed byCategory: Topology

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.

Sequences & Tendency: Topology Basics Pt. 2

## Introduction

In my previous post I presented abstract topological spaces by way of two special characteristics. These properties are enough to endow a given set with vast possibilities for analysis. Fundamental to mathematical analysis of all kinds (real, complex, functional, etc.) is the sequence.

We have covered the concept of sequences in some of our other posts here at The Math Citadel. As Rachel pointed out in her post on Cauchy sequences, one of the most important aspects of the character of a given sequence is convergence.

In spaces like the real numbers, there is convenient framework available to quantify closeness and proximity, and which allows naturally for a definition of limit or tendency for sequences. In a general topological space missing this skeletal feature, convergence must be defined.

This post will assume only some familiarity with sequences as mathematical objects and, of course, the concepts mentioned in Part 1. For a thorough treatment of sequences, I recommend Mathematical Analysis by Tom M. Apostol.

## Neighborhoods

Suppose $(X,\mathscr{T})$ is a given topological space, and nothing more is known. At our disposal so far are only open sets (elements of $\mathscr{T}$), and so it is on these a concept of vicinity relies.

Definition. Given a topological space $(X,\mathscr{T})$, a neighborhood of a point $x\in X$ is an open set which contains $x$.

That is, we say an element $T\in\mathscr{T}$ such that $x\in T$ is a neighborhood1 of $x$. To illustrate, take the examples from my previous post.

#### The Trivial Topology

When the topology in question is the trivial one: $\{\emptyset,X\}$, the only nonempty open set is $X$ itself, hence it is the only neighborhood of any point $x\in X$.

#### The Discrete Topology

Take $X=\{2,3,5\}$ and $\mathscr{T}$ to be the collection of all subsets of $X$:

 $\emptyset$ $\{2\}$ $\{3\}$ $\{5\}$ $\{2,3\}$ $\{2,5\}$ $\{3,5\}$ $\{2,3,5\}$

Then, for, say $x=5$, neighborhoods include $\{5\}$, $\{2,5\}$, $\{3,5\}$, and $\{2,3,5\}$.

#### The Standard Topology on $\mathbb{R}$

The standard topology on $\mathbb{R}$ is defined to be the family of all sets of real numbers containing an open interval around each of its points. In this case, there are infinitely2 many neighborhoods of every real number. Taking $x=\pi$ for instance, then $(3,4)$, $(-2\pi,2\pi)$, and even

$$\bigcup_{n=1}^{\infty}\left(\pi-\frac{1}{n},\pi+\frac{1}{n}\right)$$

are all neighborhoods of $\pi$.

Remark. A special type of neighborhood in the standard topology is the symmetric open interval. Given a point $x$ and a radius $r>0$, the set

$$(x-r,x+r)=\{y\in\mathbb{R}\mathrel{:}|x-y|

is a neighborhood of $x$. These sets form what is referred to as a basis for the standard topology and are important to definition of convergence in $\mathbb{R}$ as a metric space.

## Convergence

“…the topology of a space can be described completely in terms of convergence.” —John L. Kelley, General Topology

At this point in our discussion of topological spaces, the only objects available for use are open sets and neighborhoods, and so it is with these that convergence of a sequence are built3.

Definition. A sequence $(\alpha_n)$ in a topological space $(X,\mathscr{T})$ converges to a point $L\in X$ if for every neighborhood $U$ of $L$, there exists an index $N\in\mathbb{N}$ such that $\alpha_n\in U$ whenever $n\geq N$. The point $L$ is referred to as the limit of the sequence $(\alpha_n)$.

Visually, this definition can be thought of as demanding the points of the sequence cluster around the limit point $L$. In order for the sequence $(\alpha_n)$ to converge to $L$, it must be the case that after finitely many terms, every one that follows is contained in the arbitrarily posed neighborhood $U$.

As you might expect, the class of neighborhoods available has a dramatic effect on which sequences converge, as well as where they tend. Just how close to $L$ are the neighborhoods built around it in the topology?

We will use the example topologies brought up so far to exhibit the key characteristics of this definition, and what these parameters permit of sequences.

#### The Trivial Topology

In case it was to this point hazy just how useful the trivial topology is, sequences illustrate the issue nicely. For the sake of this presentation, take the trivial topology on $\mathbb{R}$. There is precisely one neighborhood of any point, namely $\mathbb{R}$ itself. As a result, any sequence of real numbers converges, since every term belongs to $\mathbb{R}$. Moreover, every real number is a limit of any sequence. So, yes, the sequence $(5,5,5,\ldots)$ of all $5$‘s converges to $\sqrt{2}$ here.

#### The Discrete Topology

Whereas with the trivial topology a single neighborhood exists, the discrete topology is as packed with neighborhoods as can be. So, as the trivial topology allows every sequence to converge to everything, we can expect the discrete topology to be comparatively restrictive. Taking the set $\{2,3,5\}$ with the discrete topology as mentioned above, we can pinpoint the new limitation: every set containing exactly one point is a neighborhood of that point. Notice the sets4 $\{2\}$, $\{3\}$, and $\{5\}$ are all open sets.

What does this mean? Any sequence that converges to one of these points, say $3$, must eventually have all its terms in the neighborhood $\{3\}$. But that requires all convergent sequences to be eventually constant! This seems to be a minor issue with the finite set $\{2,3,5\}$, but it presents an undesirable, counter-intuitive problem in other sets.

Take $\mathbb{R}$ with the discrete topology, for example. Under these rules, the sequence

$$(\alpha_n)=\left(\frac{1}{n}\right)=\left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right),$$

though expected to converge to $0$, does not converge at all.

So, the discrete topology is too restrictive, and the trivial topology lets us get away with anything. Fortunately, a happy middle ground exists by being a little more selective with neighborhoods.

#### The Standard Topology

By requiring an open set to contain an open interval around each of its points, it is impossible that a singleton be an open set. Therefore a singleton cannot be a neighborhood, and we eliminate the trouble posed by the discrete topology. Yet every open interval around a real number $L$ contains a smaller one, and each of these is a neighborhood.

This effectively corrals the points of any convergent sequence, requiring the distance between the terms and the limit to vanish as $n$ increases. Take again the sequence

$$(\alpha_n)=\left(\frac{1}{n}\right)=\left(1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right).$$

We suspect $(\alpha_n)$ converges to $0$, but this requires proof. Therefore, we must consider an arbitrary neighborhood of $0$, and expose the index $N\in\mathbb{N}$ such that all terms, from the $N$th onward, exist in that neighborhood.

Suppose $U$ is a given neighborhood of $0$, so that $U$ contains an open interval surrounding $0$. Without loss of generality, we may assume this interval is symmetric; that is, the interval has the form $(-r,r)$ for some radius $r>0$. Take $N$ to be any integer greater than $\tfrac{1}{r}$. Then, whenever $n\geq N$,

$$\alpha_n = \frac{1}{n} \leq \frac{1}{N} < \frac{1}{1/r} = r.$$

But this means $\alpha_n\in(-r,r)\subset U$ so long as $n\geq N$. Since we chose $U$ arbitrarily, it follows $(\alpha_n)$ converges to $0$.

## Conclusion

The behavior of a sequence in a given set can change rather drastically depending on the network of neighborhoods the topology allows. However, with careful construction, it is possible to have all the sequential comforts of metric spaces covered under a briefly put definition.

My next post in this series will push the generalization of these concepts much further, by relaxing a single requirement. In order to define convergence in the preceding discussion, the set of indices $\mathbb{N}$ was important not for guaranteeing infinitely many terms, but instead for providing order. This allows us to speak of all terms subsequent to one in particular. It turns out that if we simply hold on to order, we can loosen the nature of the set on which it is defined. That is the key to Moore-Smith Convergence, to be visited next.

Building a Ground Floor: Topology Basics Pt. 1

## Building a Ground Floor: Topology Basics Pt. 1

Like some other terms in mathematics (“algebra” comes to mind), topology is both a discipline and a mathematical object. Moreover like algebra, topology as a subject of study is at heart an artful mathematical branch devoted to generalizing existing structures like the field of real numbers for their most convenient properties. It is also a favorite subject of mine, ever since my first introduction to it. This is due in large part to its exceedingly simple first principles, which make the wealth of expansion they allow all the more impressive.

It is my intent to discuss some of these starting points here, in the first of a short series of posts toward the goal of presenting one of my favorite results arising in topology: Moore-Smith convergence, a vast extension of the notion of the limit of a sequence. My representation here follows the explanation given by John L. Kelley in his classic text General Topology, which I recommend to all curious readers.

## What is a topology?

Definition. By topology is meant any collection $\mathscr{T}$ of sets satisfying two conditions:

$$\begin{array}{lrcl}\text{(1)}&A,B\in\mathscr{T}&\Rightarrow&A\cap B\in\mathscr{T};\\\text{(2)}&\mathscr{C}\subset\mathscr{T}&\Rightarrow&\bigcup\{C\in\mathscr{C}\}\in\mathscr{T}\end{array}$$

It is worthwhile to break this definition down. Condition $(1)$ requires that the intersection of any two elements of the collection $\mathscr{T}$ must itself be a member of $\mathscr{T}$. Condition $(2)$ states that the union of any subcollection of $\mathscr{T}$ must also belong to $\mathscr{T}$. These are referred to as closure to finite intersection and closure to arbitrary union, respectively, in some texts.

Notably, the definition speaks only of a collection of sets with no specification beyond the two conditions. Yet, even with these, one can deduce some further characteristic properties.

Corollary. If $\mathscr{T}$ is a topology, then

$$\begin{array}{ll}\text{(i)}&\emptyset\in\mathscr{T};\\\text{(ii)}&\bigcup\{T\in\mathscr{T}\}\in\mathscr{T}.\end{array}$$

Since $\emptyset\subset S$ for every set $S$, and $\mathscr{T}\subset\mathscr{T}$, it is enough to apply $(2)$ to both of these cases to prove the corollary. In fact, many texts make the definition $X\mathrel{:=}\bigcup\{T\in\mathscr{T}\}$, and refer to the pair $(X,\mathscr{T})$ as the topological space defined by $\mathscr{T}$.

This way, the space is given its character by way of the scheme that builds $\mathscr{T}$, rather than the set $X$. It is an important distinction, for many topologies are possible on a given set. With that, we can look at some examples.

## From Trivial to Complicated

### 1. The Trivial Topology

Based on the corollary just presented, it is enough to gather a given set $X$ and the empty set $\emptyset$ into a collection $\{\emptyset,X\}$ and have created a topology on $X$. Because $X$ and $\emptyset$ are its only members, the collection is easily closed to arbitrary union and finite intersection of its elements. This is known as the trivial or indiscrete topology, and it is somewhat uninteresting, as its name suggests, but it is important as an instance of how simple a topology may be. As per the corollary, every topology on $X$ must contain $\emptyset$ and $X$, and so will feature the trivial topology as a subcollection.

### 2. The Discrete Topology

For this example, one can start with an arbitrary set, but in order to better illustrate, take the set of the first three primes: $\{2,3,5\}$. Suppose we consider the collection of all possible subsets of $\{2,3,5\}$. This is also referred to as the power set of $\{2,3,5\}$, and denoted $\wp(\{2,3,5\})$. Fortunately, the set is small enough to list exhaustively1. Here they are listed from top-to-bottom in order of increasing inclusion:

 $\emptyset$ $\{2\}$ $\{3\}$ $\{5\}$ $\{2,3\}$ $\{2,5\}$ $\{3,5\}$ $\{2,3,5\}$

Note these are all possible subsets of $\{2,3,5\}$. It is clear any union or intersection of the pieces in the table above exists as an entry, and so this meets criteria $(1)$ and $(2)$. This is a special example, known as the discrete topology. Because the discrete topology collects every existing subset, any topology on $\{2,3,5\}$ is a subcollection of this one.

For example, taking the sets

$$\emptyset,\quad\{5\},\quad\{2,3\},\quad\{2,3,5\}$$

from the collection in the table is enough to produce a topology2.

Remark. Given a topological space $(X,\mathscr{T})$, the elements of $\mathscr{T}$ are referred to as open sets. This nomenclature is motivated in the next example.

### 3. $\mathbb{R}$ and Open Intervals

This example will be more constructive than the previous ones. Consider the set of real numbers, $\mathbb{R}$. Let us define a special collection $\mathscr{T}$ of subsets of real numbers the following way: a set $T$ belongs to $\mathscr{T}$ if, and only if, for every $x\in T$, there exist real numbers $a$ and $b$ such that $x\in(a,b)$ and $(a,b)\subset T.$ That is, we say $T\in\mathscr{T}$ to mean $T$ contains an open interval around each of its elements.

It is good practice to take the time to prove this collection defines a topology on $\mathbb{R}$. To do so, it must be shown that $\bigcup\{T\in\mathscr{T}\}=\mathbb{R}$, and that $\mathscr{T}$ meets conditions $(1)$ and $(2)$.

Proof. To show $\bigcup\{T\in\mathscr{T}\}=\mathbb{R}$, it must be verified that $\bigcup\{T\in\mathscr{T}\}\subset\mathbb{R}$ and $\mathbb{R}\subset\bigcup\{T\in\mathscr{T}\}$. The first containment follows by defining every $T\in\mathscr{T}$ as a subset of $\mathbb{R}$ to begin with, so the reverse containment is all that is left. Let $x\in\mathbb{R}$ be given. Then certainly $x\in(x-1,x+1)$, and surely $(x-1,x+1)\in\mathscr{T}$, as it contains an open interval around all its points by its very design. Thus $x\in\bigcup\{T\in\mathscr{T}\}$.

On to proving $\mathscr{T}$ satisfies $(1)$ and $(2)$. For $(1)$, let $A,B\in\mathscr{T}$ be given and suppose3 $x\in A\cap B$. This holds if, and only if, $x\in A$ and $x\in B$. Since $A$ and $B$ both belong to $\mathscr{T}$, there exist real numbers $a$, $b$, $c$, and $d$ such that $x\in(a,b)\subset A$, and $x\in(c,d)\subset B$. But this means $x\in(a,b)\cap(c,d)$. Fortunately, two intervals of real numbers may only overlap in one way: this means either $c or $a. Without loss of generality, suppose it is the former case, that $c. Then $(a,b)\cap(c,d)=(c,b)$, and it is so that $x\in(c,b)$, an open interval contained in $A\cap B$ (precisely as desired), and it follows $A\cap B\in\mathscr{T}$.

To show $(2)$ is much easier. Let $\{T_\alpha\}_{\alpha\in\mathscr{A}}$ be a collection4 of sets belonging to $\mathscr{T}$, and suppose $x\in\bigcup_{\alpha\in\mathscr{A}}T_\alpha$. Then there exists an index, say $\alpha_0\in\mathscr{A}$, such that $x\in T_{\alpha_0}$. Since $T_{\alpha_0}\in\mathscr{T}$, there exist real numbers $a$ and $b$ such that $x\in(a,b)\subset T_{\alpha_0}$. But this means $x\in(a,b)\subset\bigcup_{\alpha\in\mathscr{A}}T_\alpha$. Since $x$ was chosen arbitrarily, it follows $\bigcup_{\alpha\in\mathscr{A}}T_\alpha\in\mathscr{T}$.

The proof above shows $(\mathbb{R},\mathscr{T})$ is a topological space; the collection $\mathscr{T}$ is referred to as the standard topology on $\mathbb{R}$. The open sets in this space are all the subsets of real numbers contained in open intervals. Fittingly, then, open intervals are open sets in the standard topology.

## Conclusion

This first post is meant to express the simple starting points of topology as a subject of study. It only takes the two criteria mentioned here to define a topology of sets, and yet an entire realm of theory rests upon them. This is a recurring theme in topology, algebra, and mathematics in general. Building the fully-featured universes that hold the answers for more specific inquiry: the complete ordered field of real numbers5 $\mathbb{R}$, the space $\mathcal{C}^{\infty}(\mathbb{C})$ of infinitely differentiable functions $f\mathrel{:}\mathbb{C}\to\mathbb{C}$, the class of all real-valued Lebesgue-integrable functions on $\mathbb{R}$, each of these requires a well-made foundation.

The next post in this series will cover the nature of sequences in topological spaces, particularly those instances where the convenient features afforded by the real numbers are no longer available. With the metric space structure stripped away, how does one define convergence and limit of sequences? What does it mean for elements in topological spaces to be close when distance is otherwise without definition?