Limit of a Subsequence

Theorem

Let $T = \left({A, \vartheta}\right)$ be a Hausdorff space.

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

Let $l \in A$ such that $\displaystyle \lim_{n \to \infty} x_n = l$.

Let $\left \langle {x_{n_r}} \right \rangle$ be a subsequence of $\left \langle {x_n} \right \rangle$.

Then $\displaystyle \lim_{r \to \infty} x_{n_r} = l$.

That is, the limit of a convergent sequence in a Hausdorff space equals the limit of any subsequence of it.

Proof

Proof for Real Numbers (with the metric topology)

Let $\epsilon > 0$.

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

$\exists N: \forall n > N: \left|{x_n - l}\right| < \epsilon$

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:

$\left|{x_n - l}\right| < \epsilon$

The result follows.

$\blacksquare$

Proof for General / Hausdorff Space

Let $U \in \vartheta$ be an open set such that $l \in U$.

By definition of convergence, we have:

$\exists N \in \N: \forall n > N: x_n \in U$.

When $r > N$, we have $n_r > n_N > N$ by Strictly Increasing Sequence of Natural Numbers.

It follows that:

$\exists N \in \N: \forall r > N: x_{n_r} \in U$.

Therefore, as $U$ was arbitrary, we have established $\displaystyle \lim_{r \to \infty} x_{n_r} = l$, by definition of convergence.

$\blacksquare$

Remark

Even if $T$ is not a Hausdorff space, subsequences of sequences converge to the same points. However, this limit may not be unique. The proof is identical.

Sources

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