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

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

2. **Inclusion \( \supseteq \):**
   Next, we show that:
   \[
   (0, 1] \subseteq \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right]
   \]
   
   - Take any \( x \in (0, 1] \). We need to show that \( x \in \left[\frac{1}{n}, 1\right] \) for some \( n \in \mathbb{N} \).
   - Since \( x > 0 \), we can find an \( n \in \mathbb{N} \) such that \( \frac{1}{n} < x \). This is possible because \( \frac{1}{n} \to 0 \) as \( n \to \infty \).
   - Therefore, \( x \in \left[\frac{1}{n}, 1\right] \) for sufficiently large \( n \).
   - This shows that any \( x \in (0, 1] \) is also in \( \bigcup_{n \in \mathbb{N}} \left[\frac{1}{n}, 1\right] \).

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

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

In this problem, we prove that the union of intervals [1/n, 1] for n belonging to natural numbers results in the interval (0, 1]. This solution uses concepts from set theory and real analysis, focusing on the behavior of intervals and the limit as n approaches infinity.

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