# 1 is Limit Point of Sequence in Sierpiński Space

## Theorem

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.

The sequence in $T$:

$\sigma = \sequence {0, 1, 0, 1, \ldots}$

has $1$ as a limit point.

## Proof

By definition, $\alpha$ is a limit point of $\sigma$ if and only if:

$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \notin U}$ is finite.

The only open set of $T$ containing $1$ is $\set {0, 1}$.

It contains all but a finite number (that is: $0$) elements of $\sigma$.

Hence, by definition, $1$ is a limit point of $\sigma$.

$\blacksquare$