Math Problem Statement
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:
-
Inclusion : First, we want to prove that: [ \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right] \subseteq (0, 1] ]
- Take any .
- By the definition of the union, there exists some such that . This means that .
- Since for all , we have and .
- Therefore, .
Thus, .
-
Inclusion : Next, we show that:
- Take any . We need to show that for some .
- Since , we can find an such that . This is possible because as .
- Therefore, for sufficiently large .
- This shows that any is also in .
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:
- What is the difference between a closed interval and an open interval ?
- Can you explain how the union of sets works in more detail?
- How can we rigorously define the natural numbers, , used in this proof?
- What would change if the upper bound were open, i.e., ?
- 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.
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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)
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