Arc connectedness of Telophase topology In Counterexamples in Topology Book by Lynn Steen i found that Telophase topology is arc connected. How can we build an arc in this topology ?
And also could you give me an idea of how to prove that this space is sequencialy compact ?
Thank you very much! 

As always, an arc is just a continuous map from $[0,1]$ to $X$. An arc from $a$ to $b$ where neither of $a,b$ equals $1^*$ is obtained by taking $[0,1]\to X$, $t\mapsto a+(b-a)t$, of course. For an arc where $b=1^*$ and $a\notin \{1,1^*\}$, take an arc from $a$ to $1$ and replace $1$ with $1^*$. For an arc from $1$ to $1^*$, connect an arc from $1$ to $0$ with an arc from $0$ to $1^*$, for example.
Sequence compactness: Let $(x_n)_{n\in\Bbb N}$ be any sequence in $X$.
If infinitely many terms of the sequence are $\ne 1^*$, then we obtain a convergent subsequence because $[0,1]$ is sequentially compact. And otherwise, i.e., if all but finitely many terms are $=1^*$, then we of course obtain a subsequence converging to $1^*$.
