Cauchy's Convergence Criterion/Real Numbers/Sufficient Condition/Proof 2

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Theorem

Let $\sequence {x_n}$ be a sequence in $\R$.

Let $\sequence {x_n}$ be a Cauchy sequence.


Then $\sequence {x_n}$ is convergent.


Proof

Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.

By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded.

By the Bolzano-Weierstrass Theorem, $\sequence {a_n}$ has a convergent subsequence $\sequence {a_{n_r} }$.

Let $a_{n_r} \to l$ as $r \to \infty$.

It is to be shown that $a_n \to l$ as $n \to \infty$.


Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.

Then $\dfrac \epsilon 2 > 0$.

Hence:

$(1): \quad \exists R \in \R: \forall r > R: \size {a_{n_r} - l} < \dfrac \epsilon 2$


We have that $\sequence {a_n}$ is a Cauchy sequence.

Hence:

$(2): \quad \exists N \in \R: \forall m > N, n > N: \size {x_m - x_n} \le \dfrac \epsilon 2$

Let $n > N$.

Let $r \in \N$ be sufficiently large that:

$n_r > N$

and:

$r > R$

Then $(1)$ is satisfied, and $(2)$ is satisfied with $m = n_r$.

So:

\(\ds \forall n > N: \, \) \(\ds \size {a_n - l}\) \(=\) \(\ds \size {a_n - a_{n_r} + a_{n_r} - l}\)
\(\ds \) \(\le\) \(\ds \size {a_n - a_{n_r} } + \size{a_{n_r} - l}\) Triangle Inequality for Real Numbers
\(\ds \) \(<\) \(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

So, given $\epsilon > 0$, we have found $n \in \R$ such that:

$\forall n > N: \size {a_n - l} < \epsilon$

Thus:

$x_n \to l$ as $n \to \infty$.

$\blacksquare$


Sources

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