An example in a book I am reading on general topology demonstrates the concept of compactness by showing that the set $E = \{s_n : n = 0,1,2,3 \ldots \}$ is compact in some topological space S. The text assigns the symbol G to a family of S-open sets that covers E. Then it claims that some member $G_0$ contains $s_0$. (And then uses the definition of convergence to show that $s_n$ is in this open set for all n greater than some N, so that if $G_i$ is a member of the covering that contains $s_i$, the sets $\{G_0,G_1 \ldots G_N\}$ cover E).
However, I am confused about why the member $G_0$ contains $s_0$, though this would make sense to me if we were considering the closure of the set of elements in the sequence to be E. As it stands, $s_0$ is not necessarily an element of the set E. (For example, if our space was the real numbers under the regular topology and our sequence 1/n for every n in the naturals we could construct an open covering of increasingly smaller balls around each element.) Is there something I am missing, or is there a typo somewhere in this example? (The latter seems more likely to me, particularly since this example followed a theorem that showed that a closed subset of a compact space is compact, but I just want to be sure as I am self-teaching and cannot let minor confusion rest.)
Thank you all in advance for your help.
(The book is Topology and Normed Spaces by G.J.O Jameson, a book which I have found to be extremely lucid and well-thought out so far.)
[Edit: As was pointed out below, E contains $s_0$ by definition. There is no cause for alarm.]