Let $A\subseteq \mathbb{R}$. Thus, the following are $\underline{\text{said}}$ to be equivalent:
$\textbf{1.}$ The accumulation point(s) of $A$ belong to $A$.
$\textbf{2.}$ $A$ is closed and bounded.
$\textbf{3.}$ For every open cover $C$ of $A$, $\exists $ a finite subcover $C_1$ of $C$.
$\textbf{Question:}$ Concerning the equivalence of $1$ and $2$–Why is there the restriction that $A$ must be bounded? Isn't $A$ closed iff the accumulation point(s) of $A$ belong to $A$ as it is?
Note I am getting the definition of compactness partly from here.
Best Answer
The stated theorem is incorrect. In particular, (2) and (3) are indeed equivalent (Heine-Borel) but the point (1) is in general weaker for arbitrary subsets of $\mathbb{R}$. As a counterexample, consider the whole of $\mathbb{R}$ which contains all of its accumulation points trivially but for which the open cover $$\mathcal{U} = \{ (-i,i) \mid i \in \mathbb{N}\}$$ has no finite subcover. Further, $\mathbb{R}$ is not bounded.