Math Problem Statement

Prove that the union of intervals [1/n, 1] for n ∈ ℕ equals (0, 1].

Solution

The question in the image asks to prove the following set equality:

[ \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right] = (0, 1] ]

Proof Outline:

We need to show two inclusions:

  1. Inclusion \subseteq: First, we want to prove that: [ \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right] \subseteq (0, 1] ]

    • Take any xnN[1n,1]x \in \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right].
    • By the definition of the union, there exists some nNn \in \mathbb{N} such that x[1n,1]x \in \left[\frac{1}{n}, 1\right]. This means that 1nx1\frac{1}{n} \leq x \leq 1.
    • Since 1n>0\frac{1}{n} > 0 for all nNn \in \mathbb{N}, we have x>0x > 0 and x1x \leq 1.
    • Therefore, x(0,1]x \in (0, 1].

    Thus, nN[1n,1](0,1]\bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right] \subseteq (0, 1].

  2. Inclusion \supseteq: Next, we show that: (0,1]nN[1n,1](0, 1] \subseteq \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right]

    • Take any x(0,1]x \in (0, 1]. We need to show that x[1n,1]x \in \left[\frac{1}{n}, 1\right] for some nNn \in \mathbb{N}.
    • Since x>0x > 0, we can find an nNn \in \mathbb{N} such that 1n<x\frac{1}{n} < x. This is possible because 1n0\frac{1}{n} \to 0 as nn \to \infty.
    • Therefore, x[1n,1]x \in \left[\frac{1}{n}, 1\right] for sufficiently large nn.
    • This shows that any x(0,1]x \in (0, 1] is also in nN[1n,1]\bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right].

Thus, we have shown both inclusions: [ \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right] = (0, 1] ]

Do you want further details or have any questions?

Related Questions:

  1. What is the difference between a closed interval [a,b][a, b] and an open interval (a,b)(a, b)?
  2. Can you explain how the union of sets works in more detail?
  3. How can we rigorously define the natural numbers, N\mathbb{N}, used in this proof?
  4. What would change if the upper bound were open, i.e., nN[1n,1)\bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right)?
  5. How does the concept of limits play a role in this proof?

Tip:

Understanding the behavior of intervals and how unions work in set theory is essential for more advanced topics in real analysis, particularly when dealing with open and closed sets.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Set Theory
Real Analysis
Intervals
Union of Sets

Formulas

Union of sets: \bigcup_{n \in \mathbb{N}} [1/n, 1]
Interval notation (a, b]

Theorems

Properties of real intervals
Limit of 1/n as n approaches infinity

Suitable Grade Level

University Level (Real Analysis)