Theorem 2. Let ⟨ pn⟩n=1^∞ be a Cauchy sequence in a metric space (E, d). Assume that this sequen

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Theorem 2. Let \( \left\langle p_{n}\right\rangle_{n=1}^{\infty} \) be a Cauchy sequence in a metric space \( (E, d) \). Assume that this sequence has a convergent subsequence \( \left\langle p_{n_{k}}\right\rangle_{k=1}^{\infty} \). Then the original sequence \( \left\langle p_{n}\right\rangle_{n=1}^{\infty} \) converges. Put more succinctly: Cauchy plus convergent subsequence implies convergent. Problem 3. Prove Theorem 2. Hint: The \( \left\langle p_{n_{k}}\right\rangle_{k=1}^{\infty} \) converges, so there is a \( p \in E \) such that \[ \lim _{k \rightarrow \infty} a_{n_{k}}=p . \] The basic idea of the proof is to note that by the triangle inequality that \[ d\left(p, p_{n}\right) \leq d\left(p, p_{n_{k}}\right)+d\left(p_{n_{k}}, p_{n}\right) . \] Because the subsequence converges, the first term on the right side of the inequality can be made small by making \( k \) large. Because the original sequence is Cauchy the second term on the right can be made small by making both \( n_{k} \) and \( n \) large. Here is an outline of making this precise. Let \( \varepsilon>0 \). (a) Explain why there is a \( N \) such that \[ m, n \geq N \quad \text { implies } \quad d\left(a_{m}, a_{n}\right)<\frac{\varepsilon}{2} \text {. } \] (b) Explain there is a \( n_{k} \geq N \) \[ d\left(p, p_{n_{k}}\right)<\frac{\varepsilon}{2} . \] (c) Show \[ n \geq N \quad \text { implies } \quad d\left(p, p_{n}\right)<\varepsilon \] which completes the proof.

          Theorem 2. Let \( \left\langle p_{n}\right\rangle_{n=1}^{\infty} \) be a Cauchy sequence in a metric space \( (E, d) \). Assume that this sequence has a convergent subsequence \( \left\langle p_{n_{k}}\right\rangle_{k=1}^{\infty} \). Then the original sequence \( \left\langle p_{n}\right\rangle_{n=1}^{\infty} \) converges.

Put more succinctly: Cauchy plus convergent subsequence implies convergent.
Problem 3. Prove Theorem 2. Hint: The \( \left\langle p_{n_{k}}\right\rangle_{k=1}^{\infty} \) converges, so there is a \( p \in E \) such that
\[
\lim _{k \rightarrow \infty} a_{n_{k}}=p .
\]

The basic idea of the proof is to note that by the triangle inequality that
\[
d\left(p, p_{n}\right) \leq d\left(p, p_{n_{k}}\right)+d\left(p_{n_{k}}, p_{n}\right) .
\]

Because the subsequence converges, the first term on the right side of the inequality can be made small by making \( k \) large. Because the original sequence is Cauchy the second term on the right can be made small by making both \( n_{k} \) and \( n \) large. Here is an outline of making this precise. Let \( \varepsilon>0 \).
(a) Explain why there is a \( N \) such that
\[
m, n \geq N \quad \text { implies } \quad d\left(a_{m}, a_{n}\right)<\frac{\varepsilon}{2} \text {. }
\]
(b) Explain there is a \( n_{k} \geq N \)
\[
d\left(p, p_{n_{k}}\right)<\frac{\varepsilon}{2} .
\]
(c) Show
\[
n \geq N \quad \text { implies } \quad d\left(p, p_{n}\right)<\varepsilon
\]
which completes the proof.

Theorem 2. Let ⟨ pn⟩n=1^∞ be a Cauchy sequence in a metric space (E, d). Assume that this sequence has a convergent subsequence ⟨ pnk⟩k=1^∞. Then the original sequence ⟨ pn⟩n=1^∞ converges.

Put more succinctly: Cauchy plus convergent subsequence implies convergent.
Problem 3. Prove Theorem 2. Hint: The ⟨ pnk⟩k=1^∞ converges, so there is a p ∈ E such that

limk →∞ ank=p .

The basic idea of the proof is to note that by the triangle inequality that

d(p, pn) ≤ d(p, pnk)+d(pnk, pn) .

Because the subsequence converges, the first term on the right side of the inequality can be made small by making k large. Because the original sequence is Cauchy the second term on the right can be made small by making both nk and n large. Here is an outline of making this precise. Let ε>0.
(a) Explain why there is a N such that

m, n ≥ N implies d(am, an)<(ε)/(2).

(b) Explain there is a nk≥ N

d(p, pnk)<(ε)/(2) .

(c) Show

n ≥ N implies d(p, pn)<ε

which completes the proof.

Added by Jonathan T.

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