Sequence in Indiscrete Space converges to Every Point

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Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Let $\left \langle {s_k} \right \rangle$ be a sequence in $T$.

Then $\left \langle {s_k} \right \rangle$ converges to every point of $S$.


Let $\alpha \in S$.

By definition, $\left \langle {s_k} \right \rangle$ converges to $\alpha$ if every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\left \langle {s_n} \right \rangle$.

But as $T$ has only one open set containing any points at all, every point of $\left \langle {s_n} \right \rangle$ is contained in every open set in $T$ containing $\alpha$.

Hence the result.