Limit of Subsequence equals Limit of Sequence/Metric Space

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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 $\sequence {x_n}$.


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

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


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.



although the question calls for a separate proof for each of $3$ specific metrics on a specific space.