Limit of Subsequence equals Limit of Sequence/Real Numbers

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

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

Let $\left \langle {x_{n_r}} \right \rangle$ be a subsequence of $\left \langle {x_n} \right \rangle$.

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

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

Proof
Let $\epsilon > 0$.

Since $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows that:
 * $\exists N: \forall n > N: \left|{x_n - l}\right| < \epsilon$

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:
 * $\left|{x_n - l}\right| < \epsilon$

The result follows.