Limit of Subsequence equals Limit of Sequence/Real Numbers

Theorem

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

Let $l \in \R$ such that:

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

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

Then:

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

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

Proof

Let $\epsilon > 0$.

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

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: $\S 5.2$