Strategy to prove a set is compact. Case: $S=\{0\}\cup\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$

Question

Verify that $S=\{0\}\cup\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ is compact subset of $\mathbb{R}$ while $\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ is not.

If I take the sequences $\{\frac{1}{i}:i\in\mathbb{N}\}$ there is no convergent subsequnce once $\{0\}\notin \bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ . So $\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$ is not compact.

However when I examine $S=\{0\}\cup\bigcup_\limits{i=1}^{\infty}\{\frac{1}{n}\}$. I do not know how can I address all the coverings in order to prove it is compact.

Questions:

Is my proof right insofar? How should I prove $S$ is not compact? What is the strategy for that sort of cases?

Thanks in advance!

Answer 1

You proved correctly that the set $\left\{\frac1n\,\middle|\,n\in\mathbb{N}\right\}$ is not compact. If you add $\{0\}$, then it becomes compat because for any open cover $(A_\lambda)_{\lambda\in\Lambda}$ of that set, $0\in A_{\lambda_0}$, for some $\lambda_0\in\Lambda$. And, since $\lim_{n\to\infty}\frac1n=0$, there is some $p\in\mathbb N$ such that $n\geqslant p\implies\frac1n\in A_{\lambda_0}$. And if $n<p$, take some $\lambda_n\in\Lambda$ such that $\frac1n\in A_{\lambda_n}$. So$$\{0\}\cup\left\{\frac1n\,\middle|\,n\in\mathbb{N}\right\}\subset A_{\lambda_0}\cup A_{\lambda_1}\cup\cdots\cup A_{\lambda_{p-1}}.$$

Answer 2

The first part is right. The sequence you found has no convergent subsequence.

For $S$ take some open covering $U_i$. Then some $U_j$ must contain $0$. Now observe that this $U_j$ already contains all but finitely many $\frac{1}{n}$ and for the finitely many exceptions you can find finitely many other sets.

Answer 3

Intuitively $S\setminus\{0\}$ is not compact because you can cover it by $\bigcup\limits_{i=1}^{\infty}\ B_n$ where $B_n=B(\frac 1n,\frac 1{4n})$ which is infinite and disjoint so you cannot extract a finite covering out of it.

But a covering of $S$ should at least contain a neighbourhood of $0$ or to simplify a ball $B(0,\varepsilon)$. Yet $\varepsilon$ is fixed, so it will contain all balls $B_n$ for $n>n_0$ large enough (i.e. $n_0>\frac 1\varepsilon$). Thus you are left with $n_0+1$ balls which is a finite covering.

If instead of taking balls, you take random open sets, just consider the intersections of all sets not containing $0$ with the open set containing $0$. Similarly the infinity of sets vanishes in the set containing $0$ and it remains only a finite number of them for the rest of the points away for $0$.

Answer 4

You can prove in general that if $(X,d)$ is a metric space and $x_n \to x$ in $X$ for the metric $d$, then $\{x_n \mid n \in \mathbb{N}\}\cup \{x\}$ is compact. Your question is a special case of this.