Limit of Subsequence equals Limit of Sequence/Metric Space

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Theorem

Let $M = \left({A, d}\right)$ be a metric space.

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

Let $l \in A$ 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 from the definition of limit that:

$\exists N: \forall n > N: d \left({x_n, l}\right) < \epsilon$


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:

$d \left({x_n, l}\right) < \epsilon$


The result follows.

$\blacksquare$


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