Convergent Sequence is Cauchy Sequence

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Every convergent sequence in $$M$$ is a Cauchy sequence.

When $$M$$ is the real number line, the converse also holds:

Every Cauchy sequence in $$\R$$ is convergent.

Remark
A metric space in which the converse also holds is called complete. An example of a complete metric space is given by the real number line.

Proof
Let $$\left \langle {x_n} \right \rangle$$ be a sequence in $A$ that converges to the limit $$l\in A$$.

Let $$\epsilon > 0$$.

Then also $$\frac \epsilon 2 > 0$$.

Because $$\left \langle {x_n} \right \rangle$$ converges to $$l$$,$$\exists N: \forall n > N: d \left({x_n, l}\right) < \frac \epsilon 2$$.

In the same way, $$\forall m > N: d \left({x_m, l}\right) < \frac \epsilon 2$$

So if $$m > N$$ and $$n > N$$, then

$$ $$ $$

Thus $$\left \langle {x_n} \right \rangle$$ is a Cauchy sequence.


 * Next we show that every Cauchy sequence in $$\R$$ is convergent.

Let $$\left \langle {x_n} \right \rangle$$ be a Cauchy sequence in $$\R$$.

As Every Cauchy Sequence is Bounded‎, $$\left \langle {x_n} \right \rangle$$ is bounded.

Hence, by the Bolzano-Weierstrass Theorem, it has a convergent subsequence, which we will call $$\left \langle {x_{n_r}} \right \rangle$$.

Suppose $$x_{n_r} \to l$$ as $$r \to \infty$$.

We need to show that $$x_n \to l$$ as $$r \to \infty$$.

So, let $$\epsilon > 0 \Longrightarrow \frac \epsilon 2 > 0$$.

Hence $$\exists R: \forall r > R: \left|{x_{n_r} - l}\right| < \frac \epsilon 2$$.

Since $$\left \langle {x_n} \right \rangle$$ is a Cauchy sequence, $$\exists N: m, n > N: \left|{x_m - x_n}\right| < \frac \epsilon 2$$.

Now let $$n > N$$.

Let $$r$$ be chosen such that $$n_r > N$$ and $$r > R$$.

Then $$\left|{x_{n_r} - l}\right| < \frac \epsilon 2$$.

Also, putting $$m = n_r$$, we have $$\left|{x_m - x_n}\right| < \frac \epsilon 2$$.

So, for any $$n > N$$:

$$ $$ $$ $$

So we have shown that, given any $$\epsilon > 0$$, $$\exists N: \forall n > N: \left|{x_n - l}\right| < \epsilon$$.

So $$\left \langle {x_n} \right \rangle$$ is a sequence in $\R$.