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

In summary, the conversation discusses Proposition 4.1.8 in Chapter 4 of Houshang H. Sohrab's book "Basic Real Analysis" (Second Edition) and the proof of the proposition. The conversation also includes a question regarding the demonstration of the open cover \{ ( - \infty, \xi - 1/n) \cup ( \xi + 1/n, \infty ) \}_{ n \in \mathbb{N} } having no finite subcover. The conversation concludes with the addition of Sohrab's definition of a limit point and a thank you message for the help provided.
  • #1
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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
 

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  • #2
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 $K$.
 
  • #3
Olinguito said:
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
 

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

1. What is a compact subset of R?

A compact subset of R is a subset of the real numbers that is both closed and bounded. This means that it contains all of its limit points and can be contained within a finite interval.

2. How is compactness defined in mathematics?

Compactness is a topological property that describes the behavior of a subset of a topological space. It is defined as a set that can be covered by a finite number of open sets from the space's topology.

3. What is Proposition 4.1.8 in Sohrab's work?

Proposition 4.1.8 in Sohrab's work states that a subset of a compact set is also compact. In other words, if a set is contained within a compact set, it is also compact.

4. How is compactness related to continuity?

In general, a continuous function maps compact sets to compact sets. This means that if a function is continuous and its input is a compact set, its output will also be a compact set.

5. What are some examples of compact subsets of R?

Some examples of compact subsets of R include closed intervals, finite sets, and the Cantor set. Any subset that is both closed and bounded can be considered compact in R.

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