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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.]

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If $E=\{s_n:n=0,1,2,3,\dots\}$, then by definition $s_0\in E$; $E=\{s_0,s_1,s_2,s_3,\dots\}$. –  Brian M. Scott Dec 23 '12 at 22:28
    
The claim is false as stated, as your counterexample shows. However, there seems to be an error in your statement, since as Brian points out $s_0\in E$ by definition. Are you using $s_0$ to refer to the limit of the sequence? –  joriki Dec 23 '12 at 22:33
    
Yes, now I see that 0 and feel really silly. Thank you both. –  AreaMan Dec 23 '12 at 22:34
    
@user54092: That hasn't cleared things up yet. If $s_0$ is just the first sequence element, then the proof is flawed, since $G_0$ need not contain any of the other $s_i$. That argument only works if $s_0$ is being used to denote the limit. (In fact the body of the question doesn't even state that the sequence is convergent; that can only be assumed from the title.) –  joriki Dec 23 '12 at 22:36
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$s_0$ is being used by the book to denote the limit. The proof is correct as written in the book, but my transcription above (which reflected what I thought the author was saying because my brain passed over the 0 in in the definition of E even as I typed it up) is flawed. As always the solutions to my mathematical confusion seem to be connected to an inadequate attention to detail, so this is a lesson that I take to heart. –  AreaMan Dec 23 '12 at 22:42

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