# Difference between sequentially compact spaces and closed sets in $\mathbb{C}$

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 $$K$$.

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

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