Basic Topology of $\mathbb R$

Basics

In calculus, we discuss two types of intervals:

• Open: $(a,b) = \{x\in\mathbb R: a < x < b\}$ and
• Closed: $[a,b] = \{x\in\mathbb R: a \leq x \leq b\}$.

The distinction, of course, is that the closed intervals include their endpoints, while open intervals do not.

Illustration

An open interval:

A closed interval:

Open and closed intevals have some nice properties with respect to limiting operations. For example,

• Closed intervals contain their limit points.
• If $I$ is an open interval and $x\in I$, then points close enough to $x$ are also in $I$.

Thus, we'd like to generalize these ideas to broader classes of sets that also have these properties.

Open sets

The notion of an open set is defined in terms of a neighborhood.

Def 2.2.4: $\varepsilon$-neighborhood about a point

Given a real number $a\in\mathbb R$ and a positive number $\varepsilon > 0$, the set $$V_{\varepsilon}(a) = \{x \in R: |x − a| < \varepsilon\}$$ is called the $\varepsilon$-neighborhood of $a$.

Note that $V_{\varepsilon}(a)$ is an open interval, centered at $a$, with radius $\varepsilon$.

Definition

Def 3.2.1 A set $O\subset \mathbb R$ is called open if for all points $a\in O$ there is an $\varepsilon$-neighborhood $V_{\varepsilon}(a) \subset O$.

Examples

• The empty set is vacuously open.
• The set of all real numbers is open.
• An open interval $I=(a,b)$ is open; given $x\in I$, we can choose

$$\varepsilon = \min(\{x-a, b-x\}).$$

A major theorem

Theorem 3.2.3

1. The union of an arbitrary collection of open sets is open.
2. The intersection of a finite collection of open sets is open.

The proof is fairly simple and is on page 89 of our text.

Note, however, that for part (ii) the assumption that the collection be finite is crucial.

For example, $I_n = (-\frac{1}{n},\frac{1}{n})$ is open for every $n\in\mathbb N$. However, $$\bigcap_{n=1}^{\infty} \left(-\frac{1}{n},\frac{1}{n}\right) = \{0\},$$ which is not open.

Closed sets

Closed sets are defined in terms of limit points.

Def 3.2.4 A point $x$ is called a limit point of a set $A$ if every $\varepsilon$-neighborhood $V_{\varepsilon}(x)$ of $x$ intersects the set $A$ at some point other than $x$.

Def 3.2.7 A set $F \subset \mathbb R$ is called closed if it contains all of its limit points.

I guess that it's fairly easy to see that $\varnothing$ is closed, $\mathbb R$ is closed, and that every closed interval is closed.

Closure

Def 3.2.11 Given a set $A \subset \mathbb R$, let $L$ be the set of all limit points of $Q$. The closure of $A$ is defined to be $$\overline{A} = A \cup L.$$

Example: Since $\mathbb Q$ is dense in $\mathbb R$, we have that $$\overline{\mathbb Q} = \mathbb R.$$

Alternative characterization

Thm 3.2.12 For any $A\subset \mathbb R$, the closure $\overline{A}$ is a closed set and is the smallest closed set containing $A$.

The proof is on page 92 of our text.

Complements

Open and closed sets are complements of one another!

Thm 3.2.13 A set $O$ is open if and only if $O^c$ is closed. Likewise, a set $F$ is closed if and only if $F^c$ is open.

The proof is on page 92 of our text.

Unions and intersections of closed sets

Here's a consequence of the complementary nature of open and closed sets:

Theorem 3.2.14

1. The union of a finite collection of closed sets is closed.
2. The intersection of an arbitrary collection of closed sets is closed.

Page 93 of our text indicates that the proof of theorem 3.2.14 is a direct consequence of the complementary statement for open sets (theorem 3.2.3), together with De Morgan's laws:

$$ \left(\bigcup_{\lambda\in\Lambda} E_{\lambda}\right)^c = \bigcap_{\lambda\in\Lambda} E_{\lambda}^c \: \text{ and } \: \left(\bigcap_{\lambda\in\Lambda} E_{\lambda}\right)^c = \bigcup_{\lambda\in\Lambda} E_{\lambda}^c $$

The Cantor Set

The Cantor set is an object of tremendous importance in the history of mathematics with a number of non-intuitive properties. In our context, it illustrates why we might need to think about the basic topological properties somewhat carefully.

Step 0

To construct the Cantor set, we start with a closed interval:

Step 1

We then remove the open middle third of that closed interval; we are left with two closed intervals.

Step 2

We then repeat that process with the remaining intervals:

Step 3

Which we do again:

Step 4

And again:

Step 5

And again:

Ad infinitum

In the limit, we're left with a very sparse, dust-like set:

Properties of the Cantor set

• The Cantor set is closed.
• The Cantor set is infinite.
• The length of the Cantor set is zero.
• The Cantor set is uncountable.
• The Cantor set is a self-similar fractal with dimension $$\frac{\log(2)}{\log(3)} \approx 0.63093.$$

The Cantor set is closed

Let's denote the Cantor set by $C$ and the $n^{\text{th}}$ approximation to the Cantor set by $C_n$. Note that $$C = \bigcap_{i=0}^{\infty} C_n.$$

Now, each $C_n$ is a finite union of closed intervals and, therefore, closed. Thus, $C$ is the intersection of a collection of closed sets and, therefore, also closed.

Similarly, the complement of $C$ is open, since it's a union of the open intervals removed during the construction.

The Cantor set is infinite

Note that $C_n$ consists of $2^n$ closed intervals and we never remove any of the endpoints of those intervals. Thus, there are infinitely points left over after the construction.

For concreteness sake, note that $1/3^n \in C$ for every $n\in\mathbb N$.

There's still a whole lot more to the Cantor set, however!

The length of the Cantor set is zero

The $n^{\text{th}}$ approximation $C_n$ to $C$ consists of $2^n$ intervals of length $3^{-n}$. Thus, the total length of the intervals in $C_n$ is $$\frac{2^n}{3^n} \to 0 \: \text{ as } \: n \to \infty.$$

An alternative approach is to add the lengths of the intervals removed from the original interval.

The Cantor set is uncountable

We begin by labeling the two intervals in $C_1$:

We then extend the labels in our labeling of $C_1$:

And we continue this process recursively:

Note that each sequence of 0s and 2s determines a specific sequence of intervals that collapse down to a particular point in the Cantor set. In this way, $C$ may be placed into 1-1 correspondence with the set of all sequences of 0s and 2s. Since the set of all of those sequences is uncountable, so is the Cantor set!

The Cantor set is a fractal

That's a whole other class!