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According to my notes

$K\subset\mathbb{C}$ sequentially compact $\iff$ Every sequence in K has a subsequence which converges to a Point in K

$A\subseteq\mathbb{C}$ closed set $\iff ((a_n)_{n\in\mathbb{N}}$ convergent sequence in A $\Rightarrow \lim_{n\rightarrow \infty}a_n\in A$

I am wondering why the definitions, save for $\mathbb{C}$ are not equivalent

If $A\neq\mathbb{C}$ and $A$ closed set, why does $A$ not have to be necessarily compact? Because the opposite is true. If I take a convergent sequence of a sequentially compact subset $K$ then the convergencepoint has to be by Definition in $K$.

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  • $\begingroup$ sequentially compact (in $\mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples. $\endgroup$ Jan 13, 2019 at 15:05

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Take $A=\{z\in\mathbb{C}\,|\,\lvert z\rvert\geqslant1\}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,\ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.

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  • $\begingroup$ You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$. $\endgroup$
    – RM777
    Jan 13, 2019 at 13:02
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    $\begingroup$ Why? After all, $\lvert2\rvert=2\geqslant1$ and $\lvert3\rvert=3\geqslant1$. $\endgroup$ Jan 13, 2019 at 13:03

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