An open cover for a subset of $R$ where it has no finite subcover.

Question

I don't know what to do here........

Answer 1

Since $S$ is not closed so it is not compact by Heine-Borel Theorem.

Hence there exists an open cover of $S$ which has no finite sub-cover.

Answer 2

Every compact subset of $\mathbb{R}$ is necessarily closed. Thus, if $S$ is not closed, it can't be compact, which is to say there exists an open cover of $S$ with no finite subcover.

Answer 3

That's pretty much it — you're essentially done. Your open cover of $S$ consists of the sets $A_{\varepsilon}=(-\infty,x-\varepsilon)\cup(x+\varepsilon,+\infty)$. (Or we can take them only for values of $\varepsilon=\frac{1}{n}$ if for whatever reason we want only countably many sets.) You just need to carefully demonstrate that this collection $\{A_{\varepsilon}\}_{\varepsilon>0}$ satisfies all the requirements.

First of all, $\{A_{\varepsilon}\}_{\varepsilon>0}$ covers $S$, because $\bigcup\limits_{\varepsilon>0}A_{\varepsilon}=\mathbb{R}\setminus\{x\}$ and $x\notin S$. If a finite subcover exists, then there's the smallest value $\varepsilon_0$ such that $A_{\varepsilon_0}$ is included in the subcover. But then the union of this subcover is equal to $A_{\varepsilon_0}=(-\infty,x-\varepsilon_0)\cup(x+\varepsilon_0,+\infty)$, and $(x-\varepsilon_0,x+\varepsilon_0)\cap S=\varnothing$, contradicting the fact that $x$ is a limit point of $S$.

Answer 4

A correction: the set $U_{\varepsilon}=(-\infty, x - \varepsilon) \cup (x + \varepsilon,+\infty)$ is not in general contained in $S$.

First check that the sets $U_{\varepsilon}$ cover $S$. This is true because $\cup_{\varepsilon > 0}U_{\varepsilon} = \mathbf{R} - \{x\}\supseteq S$, since $x \not\in S$.

Now assume for a contradiction that there is some finite collection of sets $U_{\varepsilon}$, say $U_{\varepsilon_1}, \dots, U_{\varepsilon_n}$, that cover $S$. Then $S \subseteq U_{\varepsilon_1} \cup \dots \cup U_{\varepsilon_n} = U_{\alpha}$, where $\alpha = \min(\varepsilon_1,\dots,\varepsilon_n)$. This shows that $S \cap [x - \alpha,x+\alpha] = \varnothing$, contradicting the fact that $x$ is a limit point of $S$.