Limit of Subsequence equals Limit of Sequence

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

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

Let $l \in S$ 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 in a topological space equals the limit of any subsequence of it.

Real Numbers
For the real number line under the usual topology, this translates into the following:

Proof
Let $U \in \tau$ be an open set such that $l \in U$.

By definition of convergence, we have:
 * $\exists N \in \N: \forall n > N: x_n \in U$.

When $r > N$, we have $n_r > n_N > N$ by Strictly Increasing Sequence of Natural Numbers.

It follows that:
 * $\exists N \in \N: \forall r > N: x_{n_r} \in U$.

Therefore, as $U$ was arbitrary, we have established $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$, by definition of convergence.