Sets of real numbers

Intersection and union of sets of real numbers.

Author

Steven Golovkine

Published

November 13, 2023

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 {Ax}xS an indexed family of subsets of S. The union of the subsets is noted xSAx. It represents the set of all elements in S that belongs to at least one subset Ax. The intersection of the subsets is noted xSAx. It represents the set of all elements in S that belongs to every subset Ax.

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 xI,Ax=(0,x) and Bx=[0,x].

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

xIAx=.

The proof is by contradiction. Assume that xIAx. It exists yI such that yxIAx. By definition of Ay, yAy. By definition of the intersection of sets, yxIAx, which contradicts the assumption. So, xIAx=.

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

xIAx=I.

The proof is by double inclusion. Let yxIAx. By definition of the union, it exists xI such that yAx. As AxI, then yI. So xIAxI. For the other way, let yI. It exists zN such that z>y. So, yAz. By definition of the union, yxIAx. So IxIAx. Finally, xIAx=I.

  1. The intersection of closed sets of positive real numbers is the singleton {0},

xIBx={0}.

The proof is by double inclusion. For all xI,0Bx. It implies that 0xIBx and {0}xIBx. For the other way, assume that it exists yI such that yxIBx. By definition, yBy/2 and by definition of the intersection, yxIBx. So, there is no element of I in xIBx and xIBx{0}. Finally, xIBx={0}.

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

xIBx=I{0}.

The proof is by double inclusion. Let yxIBx. By definition of the union, it exists xI,yBxI. By the definition of a subset, y=0 or yI. And by definition of the union, xIBxI{0}. For the other way, let yI{0}. If y=0, then by definition, yI{0}. It exists zN such that z>y. It implies that yBz. By definition of the union, yxIBx and I{0}xIBx. Finally, xIBx=I{0}.

References:

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