I have the next question: Let $K \subset $ $R^1$ consist of $0$ and the numbers 1/$n$, for $n=1,2,3,\ldots$ Prove that $K$ is compact directly from the definition (without using Heine-Borel).
I'm trying to understand compact sets so I would be grateful if someone could give me some examples of open covers and subcovers. Thank you!
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$\begingroup$As you're trying to understand this, I'll try not to give too much away so you can work it out for yourself.
Let $\mathcal{U}$ be a cover of the set $K$. As $\mathcal{U}$ is a cover of $K$, there must exist some open $U_0\in\mathcal{U}$ such that $0\in U_0$.
From this, can you construct a finite subcover of $\mathcal{U}$ for $K$? I would suggest putting the set $U_0$ in your finite subcover.
I'll try to help you establish why the set $U_0$ is so important in this proof.
Suppose that $K'=K\setminus\{0\}$. If we tried to do something similar to the above, we would notice that we can not be entirely certain that there exists some open set in $\mathcal{U}$ which contains $0$. This really is crucial. With this information, we can quite easily construct an infinite cover of $K'$ which has no finite subcover.
Let $\mathcal{U}=\{U_n\}$ where $U_n=\left(\frac{1}{n}-\frac{1}{2}(\frac{1}{n(n+1)}),\frac{1}{n}+\frac{1}{2}(\frac{1}{n(n+1)})\right)$.
Now, the $U_n$ are defined so that $\frac{1}{n}\in U_n$ and no two $U_n$ and $U_{n'}$ intersect in a non-empty subset for $n\neq n'$. So there are no proper subcovers of $\mathcal{U}$ for $K'$, let alone a finite subcover. It follows that $K'$ is not compact.
Because $0$ is no longer in our set, the possible open covers of our set can now include covers which include no open sets which cover all but finitely many elements, and this was the key to the proof for the original $K$ which included $0$.
$\endgroup$ 3 $\begingroup$Every open cover has at least one element that covers $0$. Only finitely many members of $K$ are not in that one member of the cover.
$\endgroup$ $\begingroup$Let $\{U_\alpha\}_{\alpha \in A}$ be a cover of $K$. Then there exists some $\alpha_0 \in A$ such that $U_{\alpha_0}$ contains $0$. Since $U_{\alpha_0}$ is open, for some $\epsilon > 0$, we have an open ball $B(0,\epsilon)$ centered at $0$ such that $B(0,\epsilon) \subset U_{\alpha_0}$. Since $\epsilon > 0$, we can find $n \in \mathbb{N}$ such that $\epsilon > 1/n$. Thus, $B(0,\epsilon)$ contains $\{0,1/n, 1/(n+1), \ldots\}$. Now take elements of the open cover containing $1/2, \dots, 1/(n-1)$. Since only finitely many elements are not in $U_{\alpha_0}$, we only need finitely many open sets in our cover.
$\endgroup$ $\begingroup$Let $\{U_\alpha\}$ be a open covering of $K$. Since it is a covering, there exists a $\beta$ such that $0 \in U_\beta$. Since $U_\beta$ is open, there exists an $N$ such that $B_\frac{1}{N}(0) \subseteq U_\beta$. Hence for all $n > N$, $\frac{1}{n} \in B_\frac{1}{N}(0) \subseteq U_\beta$. Since $\{U_\alpha\}$ covers, there exists $\gamma_1, \ldots, \gamma_N$ such that $\frac{1}{i} \in U_{\gamma_i}$. Hence $U_\beta, U_{\gamma_1}, \ldots, U_{\gamma_N}$ is a finite subcover of $K$.
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