Limit of Subsequence equals Limit of Sequence

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

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

Let $l \in S$ 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 in a topological space equals the limit of any subsequence of it.


Metric Space

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:

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


Normed Division Ring

Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.

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

Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:

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

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

Then:

$\sequence {x_{n_r}}$ is convergent and $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$


Real Numbers

For the real number line under the usual (Euclidean) topology, this translates into the following:


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:

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


Proof

Let $U \in \tau$ 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 $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$, by definition of convergence.

$\blacksquare$