1 is Limit Point of Sequence in Sierpiński Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({\left\{{0, 1}\right\}, \tau_0}\right)$ be a Sierpiński space.

The sequence in $T$:

$\sigma = \left\langle{0, 1, 0, 1, \ldots}\right\rangle$

has $1$ as a limit point.


Proof

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

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

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

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

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

$\blacksquare$


Sources