To understand why this inequality holds for a Cauchy sequence $\{x_n\}$, let's go through the definition of a Cauchy sequence and apply it to this proof.

Step-by-Step Proof:

1. Definition of a Cauchy Sequence: By definition, a sequence $\{x_n\}$ is Cauchy if: $\forall \varepsilon > 0, \exists M \in \mathbb{N} \text{ such that for all } n, k \geq M, \, |x_n - x_k| < \varepsilon.$

2. Applying the Definition with $\varepsilon = 1$: Here, we are given that $\{x_n\}$ is a Cauchy sequence, so it must satisfy the Cauchy condition for any positive $\varepsilon$. Set $\varepsilon = 1$.

   Since $\{x_n\}$ is Cauchy, there exists an integer $M \in \mathbb{N}$ such that for all $n, k \geq M$, the following holds: $|x_n - x_k| < 1.$

3. Conclusion: This directly shows that there exists some $M$ such that for all $n, k \geq M$, $|x_n - x_k| < 1$. Therefore, the inequality in the proof statement holds because it’s a direct application of the Cauchy definition with $\varepsilon = 1$.

Summary:

The reason why the inequality $|x_n - x_k| < 1$ holds for sufficiently large $n$ and $k$ is because, for a Cauchy sequence, the terms become arbitrarily close to each other as $n$ and $k$ grow. Setting $\varepsilon = 1$ gives this specific bound.

Would you like further details or have any other questions?

Here are 5 related questions to deepen your understanding:

1. What is the importance of the Cauchy criterion in determining convergence in metric spaces?
2. How does the Cauchy criterion differ from the convergence criterion in real analysis?
3. Can a sequence be bounded but not Cauchy? Why or why not?
4. In what spaces does every Cauchy sequence converge, and why is this significant?
5. What happens if we choose a different $\varepsilon$, say $\varepsilon = 0.5$, in the Cauchy condition?

Tip: In a metric space, if every Cauchy sequence converges, the space is called complete.