Limit of Subsequence equals Limit of Sequence/Normed Vector Space

Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

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

Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:


 * $\ds \lim_{n \mathop \to \infty} x_n = l$

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

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

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:
 * $\exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - l} < \epsilon$

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:
 * $\norm {x_{n_r} - l} < \epsilon$

The result follows.