# Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit

Jump to navigation Jump to search

## Theorem

Let $\struct {X, \tau}$ be a first-countable topological space.

Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $X$.

Let $x$ be an accumulation point of $\sequence {x_n}$.

Then $x$ is a subsequential limit of $\sequence {x_n}$.

## Proof

By the definition of a first-countable space, there exists a countable local basis $\BB$ at $x$.

By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to \BB$.

For all $n \in \N$, define the set:

$\displaystyle U_n = \bigcap_{k \mathop = 0}^n \map \phi k$

By General Intersection Property of Topological Space, it follows that $U_n$ is an open neighborhood of $x$.

Using the Principle of Recursive Definition, we construct a strictly increasing sequence $\sequence {n_k}_{k \mathop \in \N}$ in $\N$.

By the definition of an accumulation point, we can choose $n_0 \in \N$ such that $x_{n_0} \in U_0$.

For all $k \in \N$, let $n_{k + 1} > n_k$ be the (unique) smallest natural number such that $x_{n_{k + 1} } \in U_{k + 1}$.

Such an $n_{k + 1}$ exists by the definition of an accumulation point, and by the well-ordering principle.

We now show that $x$ is a limit point of $\sequence {x_{n_k} }$.

Let $U$ be an open neighborhood of $x$.

By the definition of a local basis, there exists an $H \in \BB$ such that $H \subseteq U$.

By the definition of a surjection, there exists a natural number $m$ such that $H = \map \phi m$.

By construction, we have:

$\forall k \in \N: k > m \implies x_{n_k} \in U_k \subseteq \map \phi m = H$

Hence the result, by the definition of a limit point.

$\blacksquare$