MHB Open Sets in R .... .... Willard, Example 2.7 (a) .... ....

[Math Amateur]

I am reading Stephen Willard: General Topology ... ... and am currently focused on Chapter 1: Set Theory and Metric Spaces and am currently focused on Section 2: Metric Spaces ... ...

I need help in order to fully understand Example 2.7(a) ... .. The relevant text reads as follows:View attachment 9688My questions are as follows:Question 1

In the above example from Willard we read the following;

" ... ... If $$A$$ is an open set in $$\mathbb{R}$$, the relation $$x \sim y$$ iff there is some open interval $$(a, b)$$ with $$\{ x, y \} \subset (a, b) \subset A$$ is an equivalence relation on $$A$$ and the resulting equivalence classes are disjoint open intervals whose union is $$A$$ ... ... "Can someone please demonstrate formally and rigorously that the resulting equivalence classes are disjoint open intervals whose union is $$A$$ ... ... ?Question 2

In the above example from Willard we read the following;

" ... ... The fact that there can be only countably many follows since each must contain a distinct rational ... "

I am somewhat lost in trying to understand this statement ... can someone please explain the meaning of "there can be only countably many follows since each must contain a distinct rational ...?
Help will be much appreciated ... ...

Peter

 

[GJA]

Hi Peter,

This is a useful result, so it's good to go through it in some detail.

Question 1

Fix one of the equivalence classes and let this set be represented by $p$; i.e., consider $[p].$

To see that $[p]$ has the interval property, let $x,y\in [p]$, with $x<y$. Since $x\sim y$, there is an open interval $(a,b)$ such that $\{x,y\}\subset (a,b)\subset A$. Since $(a,b)$ is an interval, it follows that $(x,y)\subset (a,b)$. Hence, for any $t$ such that $x<t<y$, $\{x,t\}\subset (a,b)\subset A.$ Hence $x\sim t$, and, therefore $p\sim t$ via transitivity of $\sim$. This proves that $[p]$ has the interval property. (Note: we could have also argued that $t\sim y$ to get the same result).

To prove that $[p]$ is open, let $x\in [p]$. Since $x\in A$ and $A$ is open, there is an open interval $(a,b)$ such that $x\in (a,b)\subset A$. It follows that $x\sim y$ for all $y\in (a,b)$. By transitivity, we have $p\sim y$ for all $y\in (a,b)$, from which we conclude that $x\in (a,b)\subset [p]$. Hence, $[p]$ is open.

Question 2

Because the rationals are dense in $\mathbb{R}$, each of the disjoint open intervals from above contains a rational number (in fact, each of the intervals contains infinitely many rationals but that's not important here). From each interval select a single rational number. Since the intervals are disjoint, it follows that the collection of disjoint open intervals is in bijection with a subset of $\mathbb{Q}.$ Since $\mathbb{Q}$ is countable, there can only be countably many such intervals.Hopefully that helps. Feel free to let me know if anything is still unclear.

 

[Math Amateur]

Hi Peter,

This is a useful result, so it's good to go through it in some detail.

Question 1

Fix one of the equivalence classes and let this set be represented by $p$; i.e., consider $[p].$

To see that $[p]$ has the interval property, let $x,y\in [p]$, with $x<y$. Since $x\sim y$, there is an open interval $(a,b)$ such that $\{x,y\}\subset (a,b)\subset A$. Since $(a,b)$ is an interval, it follows that $(x,y)\subset (a,b)$. Hence, for any $t$ such that $x<t<y$, $\{x,t\}\subset (a,b)\subset A.$ Hence $x\sim t$, and, therefore $p\sim t$ via transitivity of $\sim$. This proves that $[p]$ has the interval property. (Note: we could have also argued that $t\sim y$ to get the same result).

To prove that $[p]$ is open, let $x\in [p]$. Since $x\in A$ and $A$ is open, there is an open interval $(a,b)$ such that $x\in (a,b)\subset A$. It follows that $x\sim y$ for all $y\in (a,b)$. By transitivity, we have $p\sim y$ for all $y\in (a,b)$, from which we conclude that $x\in (a,b)\subset [p]$. Hence, $[p]$ is open.

Question 2

Because the rationals are dense in $\mathbb{R}$, each of the disjoint open intervals from above contains a rational number (in fact, each of the intervals contains infinitely many rationals but that's not important here). From each interval select a single rational number. Since the intervals are disjoint, it follows that the collection of disjoint open intervals is in bijection with a subset of $\mathbb{Q}.$ Since $\mathbb{Q}$ is countable, there can only be countably many such intervals.Hopefully that helps. Feel free to let me know if anything is still unclear.

Thanks so much for the help GJA ...

You write:

" ... ... This proves that $[p]$ has the interval property. ... ..."

I have a followup question: How do we show that the equivalence class are disjoint intervals whose union is A?PeterEDIT! Just realized! Equivalence classes are disjoint! and together make up the whole set!

Is that correct?

Peter

 

[GJA]

EDIT! Just realized! Equivalence classes are disjoint! and together make up the whole set!

Is that correct?

Peter

That's correct, nicely done!