Limit of Subsequence equals Limit of Sequence/Metric Space

Theorem
Let $M = \struct {A, d}$ be a metric space.

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

Let $l \in A$ such that:
 * $\ds \lim_{n \mathop \to \infty} x_n = l$

Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.

Then:
 * $\ds \lim_{r \mathop \to \infty} x_{n_r} = l$

That is, the limit of a convergent sequence equals the limit of a subsequence of it.

Proof
Let $\epsilon > 0$.

Since $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that:
 * $\exists N: \forall n > N: \map d {x_n, l} < \epsilon$

Now let $R = N$.

Then from Strictly Increasing Sequence of Natural Numbers‎:
 * $\forall r > R: n_r \ge r$

Thus $n_r > N$ and so:
 * $\map d {x_n, l} < \epsilon$

The result follows.