1 is Limit Point of Sequence in Sierpiński Space
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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.
Proof
By definition, $\alpha$ is a limit 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 of $\sigma$.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $11$. Sierpinski Space: $18$