Compact Subsets of R .... Sohrab, Proposition 4.1.8 .... ....

[Math Amateur]

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 4: Topology of [FONT=MathJax_AMS]R[/FONT] and Continuity ... ...

I need help in order to fully understand the proof of Proposition 4.1.8 ...Proposition 4.1.8 and its proof read as follows:View attachment 9088In the above proof by Sohrab we read the following:

" ... ... If, to get a contradiction, we assume that \(\displaystyle \xi \notin K\) is a limit point of \(\displaystyle K\), then the open cover \(\displaystyle \{ ( - \infty, \xi - 1/n) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} }\) has no finite subcover ... ... "
My question is as follows:

How would we demonstrate rigorously that the open cover \(\displaystyle \{ ( - \infty, \xi - 1/n) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} }\) has no finite subcover ... ...?
Help will be appreciated ...

Peter
=======================================================================================It may help readers of the above post to have access to Sohrab's definition of a limit point ... so I am providing the relevant text ... as follows ...
View attachment 9089
Hope that helps ...

Peter

 

[Olinguito]

The collection
$$\left\{U_n=\left(-\infty,\,\xi-\dfrac1n\right)\cup\left(\xi+\dfrac1n,\,\infty\right):n\in\mathbb N\right\}$$
is an open cover of $K$ because $\displaystyle\bigcup_{n=1}^\infty U_n=\mathbb R\setminus\{\xi\}$ and $\xi\notin K$. If there were were a finite subcover $\{U_{n_1},\ldots,U_{n_r}\}$ then if $M=\max\{n_1,\ldots,n_r\}$ we would have $K\subseteq U_M$. But then if $I=\left(\xi-\dfrac1{M+1},\,\xi+\dfrac1{M+1}\right)$ then $\xi\in I$ and $I\cap K=\emptyset$, contradicting the assumption of $\xi$ as a limit point of $K$.

 

[Math Amateur]

The collection
$$\left\{U_n=\left(-\infty,\,\xi-\dfrac1n\right)\cup\left(\xi+\dfrac1n,\,\infty\right):n\in\mathbb N\right\}$$
is an open cover of $K$ because $\displaystyle\bigcup_{n=1}^\infty U_n=\mathbb R\setminus\{\xi\}$ and $\xi\notin K$. If there were were a finite subcover $\{U_{n_1},\ldots,U_{n_r}\}$ then if $M=\max\{n_1,\ldots,n_r\}$ we would have $K\subseteq U_M$. But then if $I=\left(\xi-\dfrac1{M+1},\,\xi+\dfrac1{M+1}\right)$ then $\xi\in I$ and $I\cap K=\emptyset$, contradicting the assumption of $\xi$ as a limit point of $K$.

Thanks Olinguito ...

Appreciate your help ...

Peter