Sequence in Indiscrete Space converges to Every Point
Jump to navigation
Jump to search
Theorem
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $\sequence {s_n}$ be a sequence in $T$.
Then $\sequence {s_n}$ converges to every point of $S$.
Proof
Let $\alpha \in S$.
By definition, $\sequence {s_n}$ converges to $\alpha$ if every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {s_n}$.
But as $T$ has only one open set containing any points at all, every point of $\sequence {s_n}$ is contained in every open set in $T$ containing $\alpha$.
Hence the result.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $4$