Limit of Subsequence equals Limit of Sequence/Real Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $l \in \R$ such that:

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

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


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 $\epsilon > 0$.

As $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows that:

$\exists N: \forall n > N: \size {x_n - l} < \epsilon$


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:

$\size {x_n - l} < \epsilon$


The result follows.

$\blacksquare$


Sources