Help Proving A Set Is A Topology In The Real Numbers

Introduction

Topology is a branch of mathematics that deals with the study of topological spaces, which are sets equipped with a topology. A topology on a set is a collection of subsets that satisfy certain properties, including being closed under arbitrary unions and finite intersections. In this article, we will help prove that a given set of real numbers is a topology.

The Problem

Consider the set of real numbers, denoted by $\mathbb{R}$. Let $\tau$ be the set of sets that can be expressed as a countable union of open intervals in $\mathbb{R}$. The problem is to show that $\tau$ is a topology on $\mathbb{R}$.

Definition of a Topology

Before we proceed, let's recall the definition of a topology. A topology on a set $X$ is a collection $\tau$ of subsets of $X$ that satisfies the following properties:

1. The empty set and the set $X$ are in $\tau$: $\emptyset, X \in \tau$.
2. $\tau$ is closed under arbitrary unions: If $\{A_i\}_{i \in I}$ is a collection of sets in $\tau$, then $\bigcup_{i \in I} A_i \in \tau$.
3. $\tau$ is closed under finite intersections: If $A_1, A_2, \ldots, A_n$ are sets in $\tau$, then $\bigcap_{i=1}^n A_i \in \tau$.

Step 1: Show that the empty set and the set $\mathbb{R}$ are in $\tau$

The empty set $\emptyset$ is a countable union of empty sets, which are open intervals. Therefore, $\emptyset \in \tau$. The set $\mathbb{R}$ is a countable union of open intervals of the form $(-\infty, x)$, where $x \in \mathbb{R}$. Therefore, $\mathbb{R} \in \tau$.

Step 2: Show that $\tau$ is closed under arbitrary unions

Let $\{A_i\}_{i \in I}$ be a collection of sets in $\tau$. We need to show that $\bigcup_{i \in I} A_i \in \tau$. Since each $A_i$ is a countable union of open intervals, we can write $A_i = \bigcup_{j \in J_i} I_{ij}$, where $I_{ij}$ are open intervals. Then, we have

$$\bigcup_{i \in I} A_i = \bigcup_{i \in I} \bigcup_{j \in J_i} I_{ij}.$$

Since the union of countable sets is countable, we can write

$$\bigcup_{i \in I} \bigcup_{j \in J_i} I_{ij} = \bigcup_{k \in K} I_k,$$

where $K$ is a countable index set and $I_k$ are open intervals. Therefore, $\bigcup_{i \in I} A_i$ is a countable union of open intervals, and hence $\bigcup_{i \in I} A_i \in \tau$.

Step 3: Show that $\tau$ is closed under finite intersections

Let $A_1, A_2, \ldots, A_n$ be sets in $\tau$. We need to show that $\bigcap_{i=1}^n A_i \in \tau$. Since each $A_i$ is a countable union of open intervals, we can write $A_i = \bigcup_{j \in J_i} I_{ij}$, where $I_{ij}$ are open intervals. Then, we have

$$\bigcap_{i=1}^n A_i = \bigcap_{i=1}^n \bigcup_{j \in J_i} I_{ij}.$$

Since the intersection of countable sets is countable, we can write

$$\bigcap_{i=1}^n \bigcup_{j \in J_i} I_{ij} = \bigcup_{k \in K} I_k,$$

where $K$ is a countable index set and $I_k$ are open intervals. Therefore, $\bigcap_{i=1}^n A_i$ is a countable union of open intervals, and hence $\bigcap_{i=1}^n A_i \in \tau$.

Conclusion

We have shown that the set $\tau$ of sets that can be expressed as a countable union of open intervals in $\mathbb{R}$ satisfies the properties of a topology. Therefore, $\tau$ is a topology on $\mathbb{R}$.

Final Answer

Q: What is a topology?

A: A topology on a set $X$ is a collection $\tau$ of subsets of $X$ that satisfies the following properties:

1. The empty set and the set $X$ are in $\tau$: $\emptyset, X \in \tau$.
2. $\tau$ is closed under arbitrary unions: If $\{A_i\}_{i \in I}$ is a collection of sets in $\tau$, then $\bigcup_{i \in I} A_i \in \tau$.
3. $\tau$ is closed under finite intersections: If $A_1, A_2, \ldots, A_n$ are sets in $\tau$, then $\bigcap_{i=1}^n A_i \in \tau$.

Q: What is the set $\tau$ in this problem?

A: The set $\tau$ is the collection of sets that can be expressed as a countable union of open intervals in $\mathbb{R}$.

Q: Why is the empty set in $\tau$?

A: The empty set $\emptyset$ is a countable union of empty sets, which are open intervals. Therefore, $\emptyset \in \tau$.

Q: Why is the set $\mathbb{R}$ in $\tau$?

A: The set $\mathbb{R}$ is a countable union of open intervals of the form $(-\infty, x)$, where $x \in \mathbb{R}$. Therefore, $\mathbb{R} \in \tau$.

Q: How do we show that $\tau$ is closed under arbitrary unions?

A: Let $\{A_i\}_{i \in I}$ be a collection of sets in $\tau$. We need to show that $\bigcup_{i \in I} A_i \in \tau$. Since each $A_i$ is a countable union of open intervals, we can write $A_i = \bigcup_{j \in J_i} I_{ij}$, where $I_{ij}$ are open intervals. Then, we have

$$\bigcup_{i \in I} A_i = \bigcup_{i \in I} \bigcup_{j \in J_i} I_{ij}.$$

Since the union of countable sets is countable, we can write

$$\bigcup_{i \in I} \bigcup_{j \in J_i} I_{ij} = \bigcup_{k \in K} I_k,$$

where $K$ is a countable index set and $I_k$ are open intervals. Therefore, $\bigcup_{i \in I} A_i$ is a countable union of open intervals, and hence $\bigcup_{i \in I} A_i \in \tau$.

Q: How do we show that $\tau$ is closed under finite intersections?

A: Let $A_1, A_2, \ldots, A_n$ be sets in $\tau$. We need to show that $\bigcap_{i=1}^n A_i \in \tau$. Since each $A_i$ is a countable union of open intervals, we can write $A_i = \bigcup_{j \in J_i} I_{ij}$, where $I_{ij}$ are open intervals. Then, we have

$$\bigcap_{i=1}^n A_i = \bigcap_{i=1}^n \bigcup_{j \in J_i} I_{ij}.$$

Since the intersection of countable sets is countable, we can write

$$\bigcap_{i=1}^n \bigcup_{j \in J_i} I_{ij} = \bigcup_{k \in K} I_k,$$

where $K$ is a countable index set and $I_k$ are open intervals. Therefore, $\bigcap_{i=1}^n A_i$ is a countable union of open intervals, and hence $\bigcap_{i=1}^n A_i \in \tau$.

Q: What is the final answer?

A: The final answer is: $\boxed{Yes}$