Why sequence 1/n fails to converge in set of all positive real numbers with the metric d(x,y)=|x-y|

The sequence \(\left(\frac{1}{n}\right)\), where \(n\) is a positive integer, is a sequence in the set of all positive real numbers. It is important to understand what it means for a sequence to converge in a particular set with a given metric.

### Convergence in a Metric Space

In a metric space \((X, d)\), a sequence \((x_n)\) converges to a limit \(L \in X\) if, for every \(\epsilon > 0\), there exists an integer \(N\) such that for all \(n \geq N\), the distance \(d(x_n, L) < \epsilon\). 

### The Sequence \(\left(\frac{1}{n}\right)\) in \(\mathbb{R}^+\)

Consider the sequence \(\left(\frac{1}{n}\right)\) in the set of all positive real numbers \(\mathbb{R}^+ = (0, \infty)\):

1. **Sequence Behavior**: The sequence \(\left(\frac{1}{n}\right)\) starts at 1 (when \(n = 1\)) and approaches 0 as \(n \to \infty\). 
   
2. **Limit in \(\mathbb{R}^+\)**: The limit of \(\left(\frac{1}{n}\right)\) as \(n \to \infty\) is 0.

3. **Metric Consideration**: The metric given is the standard absolute value metric \(d(x, y) = |x - y|\), which is a common way to measure distance in \(\mathbb{R}\).

### Why the Sequence \(\left(\frac{1}{n}\right)\) Fails to Converge in \(\mathbb{R}^+\)

Now, let’s discuss why the sequence \(\left(\frac{1}{n}\right)\) fails to converge within the set of positive real numbers \(\mathbb{R}^+ = (0, \infty)\):

1. **Limit Outside the Set**: The sequence \(\left(\frac{1}{n}\right)\) converges to 0, but 0 is **not** in the set of positive real numbers \(\mathbb{R}^+\). The set \(\mathbb{R}^+\) consists of all real numbers greater than zero. Since the limit of the sequence, 0, is not a positive number, the sequence does not converge within \(\mathbb{R}^+\).

2. **Convergence Requirement**: For a sequence to converge in a set, the limit of the sequence must belong to that set. Even though \(\left(\frac{1}{n}\right)\) converges to 0 in the larger space \(\mathbb{R}\), this limit is not included in \(\mathbb{R}^+\), so within the context of \(\mathbb{R}^+\) as a subset of \(\mathbb{R}\), the sequence does not converge.

3. **Closed Sets and Convergence**: In metric spaces, convergence of a sequence within a set \(X\) requires the limit to be within \(X\). The set \(\mathbb{R}^+\) is not closed in \(\mathbb{R}\) because it does not contain all its limit points (specifically, it does not contain 0). Therefore, \(\left(\frac{1}{n}\right)\) does not converge in \(\mathbb{R}^+\).

### Summary

The sequence \(\left(\frac{1}{n}\right)\) fails to converge in the set of positive real numbers \(\mathbb{R}^+\) because its limit, 0, is not a positive real number. Convergence in a set requires that the limit point belongs to the set, and since \(\mathbb{R}^+\) does not include 0, the sequence does not converge in \(\mathbb{R}^+\).

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Would you like more details or have any questions?

Here are some related questions to explore:

1. What is the difference between convergence in a set and convergence in a metric space?
2. How do you determine if a set is closed in a metric space?
3. What are some examples of sequences that do converge in \(\mathbb{R}^+\)?
4. Can you explain the concept of a limit point in a set?
5. How does the concept of convergence change if we consider the set \([0, \infty)\) instead of \((0, \infty)\)?

**Tip**: When studying convergence in different sets or spaces, always check if the limit of the sequence lies within the set or space you are considering.

Understand why the sequence 1/n fails to converge in the set of all positive real numbers with the metric d(x,y)=|x-y|. Learn about convergence in metric spaces, the behavior of the sequence 1/n as n approaches infinity, and why the limit of 1/n, which is 0, is not included in the set of positive real numbers.

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