Prove that every convergent sequence in a metric space is cauchy sequence 

To prove that every **convergent sequence** in a **metric space** is a **Cauchy sequence**, let's go through it step by step.

### Definitions:
1. **Convergent Sequence**: A sequence \((x_n)\) in a metric space \( (X, d) \) converges to a point \( x \in X \) if for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(n \geq N\), \(d(x_n, x) < \epsilon\).
   
2. **Cauchy Sequence**: A sequence \((x_n)\) in a metric space \( (X, d) \) is called a Cauchy sequence if for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(m, n \geq N\), \(d(x_n, x_m) < \epsilon\).

### Proof:
Let \((x_n)\) be a sequence that **converges** to \(x \in X\) in the metric space \((X, d)\). We need to show that \((x_n)\) is a **Cauchy sequence**.

#### Step 1: Convergence implies small distances to the limit
Since \((x_n)\) converges to \(x\), by definition:
\[
\forall \epsilon > 0, \exists N \in \mathbb{N} \text{ such that } \forall n \geq N, \ d(x_n, x) < \frac{\epsilon}{2}.
\]
That is, for all \(n \geq N\), the distance between \(x_n\) and \(x\) is less than \(\frac{\epsilon}{2}\).

#### Step 2: Use the triangle inequality
Now, consider any \(m, n \geq N\). We want to show that \(d(x_n, x_m) < \epsilon\). Using the **triangle inequality**:
\[
d(x_n, x_m) \leq d(x_n, x) + d(x_m, x).
\]
Since \(n, m \geq N\), we know from Step 1 that both \(d(x_n, x) < \frac{\epsilon}{2}\) and \(d(x_m, x) < \frac{\epsilon}{2}\). Substituting these into the inequality:
\[
d(x_n, x_m) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
\]

#### Step 3: Conclusion
Therefore, for all \(m, n \geq N\), we have \(d(x_n, x_m) < \epsilon\), which is precisely the definition of a **Cauchy sequence**.

Thus, we have shown that every convergent sequence in a metric space is a Cauchy sequence.

### Would you like further details or have any questions?

### Relative questions:
1. Can you give an example of a sequence that is Cauchy but not convergent?
2. What is the difference between Cauchy and convergent sequences in non-complete metric spaces?
3. How does the proof change if the metric space is not complete?
4. Can you explain the role of the triangle inequality in this proof?
5. What are some examples of metric spaces where every Cauchy sequence converges?

### Tip:
In any metric space, a **Cauchy sequence** doesn’t always have to converge unless the space is **complete**. However, every **convergent sequence** is always **Cauchy**.

Prove that every convergent sequence in a metric space is a Cauchy sequence

Explore the detailed proof that every convergent sequence in a metric space is a Cauchy sequence. This solution covers key mathematical concepts like the triangle inequality and the definitions of convergent and Cauchy sequences. Suitable for undergraduate students studying real analysis or metric spaces.

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