Let S ⊆ R be nonempty. Suppose that u ∈ R satisfies the following properties:
(i) for every n∈N, u+1/n is not a lower bound of S, (ii) for every n∈N, u−1/n is a lower bound of S.
show that u=inf S.


To show that \( u = \inf S \), we need to demonstrate two things:
1. \( u \) is a lower bound of \( S \).
2. \( u \) is the greatest lower bound of \( S \).

### 1. \( u \) is a Lower Bound of \( S \)

Given:
- For every \( n \in \mathbb{N} \), \( u - \frac{1}{n} \) is a lower bound of \( S \).

This means for all \( s \in S \) and for each \( n \in \mathbb{N} \),
\[
u - \frac{1}{n} \leq s.
\]

Since this inequality holds for every \( n \in \mathbb{N} \), taking the limit as \( n \to \infty \):
\[
\lim_{n \to \infty} \left( u - \frac{1}{n} \right) = u.
\]
So, \( u \leq s \) for all \( s \in S \). Therefore, \( u \) is a lower bound of \( S \).

### 2. \( u \) is the Greatest Lower Bound of \( S \)

To show \( u = \inf S \), we need to prove that \( u \) is the greatest lower bound, i.e., if \( l \) is any lower bound of \( S \), then \( l \leq u \).

Given:
- For every \( n \in \mathbb{N} \), \( u + \frac{1}{n} \) is not a lower bound of \( S \).

This implies that for each \( n \in \mathbb{N} \), there exists some \( s_n \in S \) such that:
\[
u + \frac{1}{n} > s_n.
\]

Since \( s_n \in S \), and \( u \) is a lower bound of \( S \), we have \( u \leq s_n \). Therefore, for each \( n \):
\[
u \leq s_n < u + \frac{1}{n}.
\]

Thus,
\[
s_n - u < \frac{1}{n}.
\]

Taking the limit as \( n \to \infty \):
\[
\lim_{n \to \infty} (s_n - u) = 0,
\]
which means
\[
\lim_{n \to \infty} s_n = u.
\]

Since for any \( \epsilon > 0 \), we can choose \( n \) large enough such that \( \frac{1}{n} < \epsilon \) and \( u + \frac{1}{n} \) is not a lower bound, we find \( s_n \) that is within \( \epsilon \) of \( u \). Therefore, \( u \) is the greatest number such that \( u \) is less than or equal to all elements of \( S \).

Since \( u \) is a lower bound and no number greater than \( u \) can be a lower bound (because there exists \( s_n \in S \) arbitrarily close to \( u \) from above), \( u \) is the greatest lower bound of \( S \).

Hence, \( u = \inf S \).

Would you like to see more details on any specific part of this proof, or do you have any other questions? 

Here are some related questions for further exploration:
1. What are the implications of a set having an infimum in the real numbers?
2. How would the proof change if we considered the supremum instead of the infimum?
3. Can \( u \) be an element of \( S \)? If so, under what conditions?
4. How would this proof differ if \( S \) were a subset of \( \mathbb{Q} \) (rationals) instead of \( \mathbb{R} \)?
5. Can we use this method to approximate the infimum of a set \( S \) numerically?

**Tip:** Remember, the infimum of a set \( S \subseteq \mathbb{R} \) might not be an element of \( S \), but it is the greatest lower bound that \( S \) has in the real numbers.

Let S ⊆ R be nonempty. Suppose that u ∈ R satisfies the following properties: (i) for every n ∈ N, u + 1/n is not a lower bound of S, (ii) for every n ∈ N, u − 1/n is a lower bound of S. Show that u = inf S.

Explore a detailed proof demonstrating that u is the infimum of a nonempty subset S of real numbers, using properties of lower bounds and limits. This problem focuses on advanced concepts in real analysis and is suitable for undergraduate students.

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