To prove Theorem 1.4.10, which states the conditions for $l \in \mathbb{R}$ being the limit superior ($\limsup x_n$) of a sequence $\{x_n\}$, we need to break it down step by step using the given conditions and definitions.

Theorem 1.4.10:

Let $l \in \mathbb{R}$. Then $l = \limsup x_n$ if and only if:

1. Given $\varepsilon > 0$, there exists $n \in \mathbb{N}$ such that $x_k < l + \varepsilon$ for all $k \geq n$, and
2. Given $\varepsilon > 0$, and given $n \in \mathbb{N}$, there exists $k \geq n$ such that $x_k > l - \varepsilon$.

This essentially combines the definition of limit superior ($\limsup x_n$) with two conditions involving $\varepsilon$-neighborhoods around $l$.

Restatement:

We can restate $l = \limsup x_n$ as follows:

• For every $\varepsilon > 0$, the interval $(l - \varepsilon, l + \varepsilon)$ contains infinitely many $x_n$'s, and
• The interval $(l + \varepsilon, +\infty)$ contains only finitely many $x_n$'s.

Proof:

Step 1: "If" direction ($\limsup x_n = l \Rightarrow$ the two conditions hold)

Assume $l = \limsup x_n$. By definition:

• $\limsup x_n$ is the largest accumulation point of the sequence $\{x_n\}$.
• Hence, for any $\varepsilon > 0$, eventually $x_n$ must stay below $l + \varepsilon$ (Condition 1).
• Similarly, for any $\varepsilon > 0$, there will be infinitely many $x_n$'s greater than $l - \varepsilon$ (Condition 2).

Thus, we check that the two conditions:

1. There exists some $n \in \mathbb{N}$ such that for all $k \geq n$, $x_k < l + \varepsilon$ (follows from the fact that the sequence eventually stays below $l + \varepsilon$).
2. For every $n \in \mathbb{N}$, there exists some $k \geq n$ such that $x_k > l - \varepsilon$ (follows from the fact that infinitely many $x_n$'s are close to $l$, but not less than $l - \varepsilon$).

Step 2: "Only if" direction (The two conditions hold $\Rightarrow l = \limsup x_n$)

Now, assume the two conditions hold:

1. There exists some $n \in \mathbb{N}$ such that for all $k \geq n$, $x_k < l + \varepsilon$.
2. For every $n \in \mathbb{N}$, there exists some $k \geq n$ such that $x_k > l - \varepsilon$.

We need to show that $l = \limsup x_n$.

• Condition 1 implies that for any $\varepsilon > 0$, eventually, all terms of the sequence $\{x_n\}$ lie below $l + \varepsilon$, so the limit superior cannot exceed $l$.
• Condition 2 implies that for every $\varepsilon > 0$, there are infinitely many terms of the sequence $\{x_n\}$ that exceed $l - \varepsilon$, so the limit superior must be at least $l$.

Together, these conditions ensure that $l = \limsup x_n$.

This completes the proof of Theorem 1.4.10.

Would you like further details on any step, or clarification on how $\limsup$ is rigorously defined in this context?

Related Questions:

1. What is the difference between $\limsup x_n$ and $\liminf x_n$?
2. How can we intuitively understand the concept of $\limsup$ and $\liminf$?
3. What are some examples of sequences where $\limsup \neq \liminf$?
4. Can a sequence have $\limsup x_n = \infty$? Under what conditions?
5. How do the two conditions in the theorem relate to bounded sequences?

Tip: Understanding limit superior is crucial in advanced analysis, especially when working with bounded sequences or oscillatory behaviors in series.