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In Lang's Complex Analysis, page 21, he gives the following definition of compactness:

We define a set of complex numbers to be compact if every sequence of elements of S has an accumulation point in S.

Where an accumulation point is defined (earlier) as:

A point v is an accumulation point of a set S if given any open set containing v it also contains infinitely many elements of S.

Question: in the definition of compactness, is the accumulation in reference to the set of points in the sequence, or the underlying set?

If it's the underlying set, then if we consider the space $\mathbb{C}$ and the sequence $(n : n \in \mathbb{N})$ (just the natural numbers), then the sequence is contained in the set, and we can trivially consider the first element of the sequence (0 or 1) which is an accumulation point of the set, whereas no element of the sequence is an accumulation point with respect to the sequence set itself.

$\mathbb{C}$ is not compact, but it feels odd that Lang wouldn't have been more specific in his definition of compactness otherwise.

Can someone please confirm that with this definition of compactness we're referring to an accumulation point with respect to the sequence and not the set we're defining to be compact?

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No, $1$ is not an accumulation point of $\mathbb N$, since the open set $D(1,1)$, to which $1$ belongs, only contains one element of $\mathbb N$, not infinitely many of them.

And a set $S$ of complex numbers is compact if, for every sequence $(x_n)_{n\in\mathbb N}$ of elements of $S$, there is a $z\in S$ such that, for any open set $A$ to which $z$ belongs, the set $\{n\in\mathbb N\,|\,x_n\in A\}$ is infinite.

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    $\begingroup$ Just to add on since this is hard to notice the first time one sees this: it should hold for every choice of open set $A$ $\endgroup$ Apr 24 '19 at 18:26
  • $\begingroup$ To clarify, in my question when I said that 1 was an accumulation point, I was identifying 1 ($\mathbb{N}$) with 1 + 0i ($\mathbb{C}$), which is an accumulation point of $\mathbb{C}$. Your answer has cleared up the question though, the answer being that the accumulation is with respect to the set of sequence elements and not the underlying set. $\endgroup$ Apr 24 '19 at 18:56

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