Sets of real numbers – Steven Golovkine

This post concerns sets of real numbers. It is based on the book Introduction to Topology, exercise 5, page 9, of Menselson [1]. It aims to prove some properties of the intersection and union of sets of real numbers.

Let $S$ be a set. We denote by $\{A_x\}_{x \in S}$ an indexed family of subsets of $S$. The union of the subsets is noted $\bigcup_{x \in S} A_x$. It represents the set of all elements in $S$ that belongs to at least one subset $A_x$. The intersection of the subsets is noted $\bigcap_{x \in S} A_x$. It represents the set of all elements in $S$ that belongs to every subset $A_x$.

Here, we will focus on indexed families of subsets of positive real numbers to manipulate families of sets. The proofs will involve classical techniques: double inclusion and contradiction.

Let $I$ be the set of real numbers that are greater than $0$. Note $\forall x \in I, A_x = (0, x)$ and $B_x = [0, x]$.

The intersection of open sets of positive real numbers is empty,

\[\bigcap_{x \in I} A_x = \emptyset.\]

The proof is by contradiction. Assume that $\bigcap_{x \in I} A_x \neq \emptyset$. It exists $y \in I$ such that $y \in \bigcap_{x \in I} A_x$. By definition of $A_y$, $y \notin A_y$. By definition of the intersection of sets, $y \notin \bigcap_{x \in I} A_x$, which contradicts the assumption. So, $\bigcap_{x \in I} A_x = \emptyset$.

The union of open sets of positive real numbers is the set of positive real numbers,

\[\bigcup_{x \in I} A_x = I.\]

The proof is by double inclusion. Let $y \in \bigcup_{x \in I} A_x$. By definition of the union, it exists $x \in I$ such that $y \in A_x$. As $A_x \subset I$, then $y \in I$. So $\bigcup_{x \in I} A_x \subset I$. For the other way, let $y \in I$. It exists $z \in \mathbb{N}$ such that $z > y$. So, $y \in A_z$. By definition of the union, $y \in \bigcup_{x \in I} A_x$. So $I \subset \bigcup_{x \in I} A_x$. Finally, $\bigcup_{x \in I} A_x = I$.

The intersection of closed sets of positive real numbers is the singleton $\{0\}$,

\[\bigcap_{x \in I} B_x = \{0\}.\]

The proof is by double inclusion. For all $x \in I, 0 \in B_x$. It implies that $0 \in \bigcap_{x \in I} B_x$ and $\{0\} \subset \bigcap_{x \in I} B_x$. For the other way, assume that it exists $y \in I$ such that $y \in \bigcap_{x \in I} B_x$. By definition, $y \notin B_{y / 2}$ and by definition of the intersection, $y \notin \bigcap_{x \in I} B_x$. So, there is no element of $I$ in $\bigcap_{x \in I} B_x$ and $\bigcap_{x \in I} B_x \subset \{0\}$. Finally, $\bigcap_{x \in I} B_x = \{0\}$.

The union of closed sets of positive real numbers is the union of the set of positive real numbers and the singleton $\{0\}$,

\[\bigcup_{x \in I} B_x = I \cup \{0\}.\]

The proof is by double inclusion. Let $y \in \bigcup_{x \in I} B_x$. By definition of the union, it exists $x \in I, y \in B_x \subset I$. By the definition of a subset, $y = 0$ or $y \in I$. And by definition of the union, $\bigcup_{x \in I} B_x \subset I \cup \{0\}$. For the other way, let $y \in I \cup \{0\}$. If $y = 0$, then by definition, $y \in I \cup \{0\}$. It exists $z \in \mathbb{N}$ such that $z > y$. It implies that $y \in B_z$. By definition of the union, $y \in \bigcup_{x \in I} B_x$ and $I \cup \{0\} \subset \bigcup_{x \in I} B_x$. Finally, $\bigcup_{x \in I} B_x = I \cup \{0\}$.

References:

[1] Mendelson, B., 2012. Introduction to Topology: Third Edition. Courier Corporation.

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--- title: Sets of real numbers author: Steven Golovkine date: '2023-11-13' categories: ['Topology'] description: | Intersection and union of sets of real numbers. image: "../../posts/sets_real_numbers/logo.jpg" image-alt: | A square logo of family of sets created with DALL-E3. --- This post concerns sets of real numbers. It is based on the book *Introduction to Topology*, exercise 5, page 9, of Menselson [1]. It aims to prove some properties of the intersection and union of sets of real numbers. Let $S$ be a set. We denote by $\{A_x\}_{x \in S}$ an indexed family of subsets of $S$. The union of the subsets is noted $\bigcup_{x \in S} A_x$. It represents the set of all elements in $S$ that belongs to at least one subset $A_x$. The intersection of the subsets is noted $\bigcap_{x \in S} A_x$. It represents the set of all elements in $S$ that belongs to every subset $A_x$. Here, we will focus on indexed families of subsets of positive real numbers to manipulate families of sets. The proofs will involve classical techniques: double inclusion and contradiction. Let $I$ be the set of real numbers that are greater than $0$. Note $\forall x \in I, A_x = (0, x)$ and $B_x = [0, x]$. 1. The intersection of open sets of positive real numbers is empty, $$\bigcap_{x \in I} A_x = \emptyset.$$ The proof is by contradiction. Assume that $\bigcap_{x \in I} A_x \neq \emptyset$. It exists $y \in I$ such that $y \in \bigcap_{x \in I} A_x$. By definition of $A_y$, $y \notin A_y$. By definition of the intersection of sets, $y \notin \bigcap_{x \in I} A_x$, which contradicts the assumption. So, $\bigcap_{x \in I} A_x = \emptyset$. 2. The union of open sets of positive real numbers is the set of positive real numbers, $$\bigcup_{x \in I} A_x = I.$$ The proof is by double inclusion. Let $y \in \bigcup_{x \in I} A_x$. By definition of the union, it exists $x \in I$ such that $y \in A_x$. As $A_x \subset I$, then $y \in I$. So $\bigcup_{x \in I} A_x \subset I$. For the other way, let $y \in I$. It exists $z \in \mathbb{N}$ such that $z > y$. So, $y \in A_z$. By definition of the union, $y \in \bigcup_{x \in I} A_x$. So $I \subset \bigcup_{x \in I} A_x$. Finally, $\bigcup_{x \in I} A_x = I$. 3. The intersection of closed sets of positive real numbers is the singleton $\{0\}$, $$\bigcap_{x \in I} B_x = \{0\}.$$ The proof is by double inclusion. For all $x \in I, 0 \in B_x$. It implies that $0 \in \bigcap_{x \in I} B_x$ and $\{0\} \subset \bigcap_{x \in I} B_x$. For the other way, assume that it exists $y \in I$ such that $y \in \bigcap_{x \in I} B_x$. By definition, $y \notin B_{y / 2}$ and by definition of the intersection, $y \notin \bigcap_{x \in I} B_x$. So, there is no element of $I$ in $\bigcap_{x \in I} B_x$ and $\bigcap_{x \in I} B_x \subset \{0\}$. Finally, $\bigcap_{x \in I} B_x = \{0\}$. 4. The union of closed sets of positive real numbers is the union of the set of positive real numbers and the singleton $\{0\}$, $$\bigcup_{x \in I} B_x = I \cup \{0\}.$$ The proof is by double inclusion. Let $y \in \bigcup_{x \in I} B_x$. By definition of the union, it exists $x \in I, y \in B_x \subset I$. By the definition of a subset, $y = 0$ or $y \in I$. And by definition of the union, $\bigcup_{x \in I} B_x \subset I \cup \{0\}$. For the other way, let $y \in I \cup \{0\}$. If $y = 0$, then by definition, $y \in I \cup \{0\}$. It exists $z \in \mathbb{N}$ such that $z > y$. It implies that $y \in B_z$. By definition of the union, $y \in \bigcup_{x \in I} B_x$ and $I \cup \{0\} \subset \bigcup_{x \in I} B_x$. Finally, $\bigcup_{x \in I} B_x = I \cup \{0\}$. References: [1] Mendelson, B., 2012. Introduction to Topology: Third Edition. Courier Corporation.