Answer to Question #104753 in Real Analysis for nima

Question #104753
Let G denote the family of all open subsets of the real numbers and F the fam-
ily of all closed subsets of the real numbers. Then

Every set G in G is the union of a sequence of disjoint open intervals (called the compo-
nents of G).
1
Expert's answer
2020-03-09T11:54:02-0400

Let "\\mathcal G" be the set of all open sets of "\\mathbb R", and "\\mathfrak G" be the set of all open intervals in "\\mathbb R".

Let "G\\in\\mathcal G". Since "\\mathfrak G" is the base of topology in "\\mathbb R", we have "G=\\bigcup\\limits_{I\\in K}I" for some "K\\subset\\mathfrak G",

Introduce a binary relation "R_0" in "K": "(I_1,I_2)\\in R_0" if "I_1\\cap I_2\\neq\\varnothing". Let "R" be a transitive closure of "R_0". Prove that "R" is an equivalence relation in "K".

Indeed,

1)It is reflexive. For every "I\\in K" we have "(I,I)\\in R_0\\subset R".

2)It is transitive, because it is transitive closure of "R_0".

3)It is symmetric. If "(I_1,I_2)\\in R", then "(I_1,I_2)\\in R_0^n" for some "n". But since "R_0" is symmetric, "R_0^n" is also symmetric. So "(I_2,I_1)\\in R_0^n\\subset R."

Then "R" is an equivalence relation in "K".


So we have "K\/R" - the factorset "K" over "R". Denote the class of "x\\in K" in "K\/R" by "x_R".

We have "G=\\bigcup\\limits_{I\\in K}I=\\bigcup\\limits_{x\\in K}\\bigcup\\limits_{I\\in x_R}I".

For every "x\\in K" denote "\\bigcup\\limits_{I\\in x_R}I" by "[x]" , then "G=\\bigcup\\limits_{x\\in K}[x]" .

Note that "[x]" is an open set for every "x\\in K". We need to prove

1) "[x]=[y]" or "[x]\\cap [y]=\\varnothing" for every "x,y\\in K".

2) "[x]" is an open interval in "\\mathbb R" for every "x\\in K".


Indeed,

1) "R" is an equivalence relation, we have that "x_R=y_R" or "x_R\\cap y_R=\\varnothing" for every "x,y\\in K".

If "x_R=y_R", then "[x]=\\bigcup\\limits_{I\\in x_R}I=\\bigcup\\limits_{I\\in y_R}I=[y]".

Let "x_R\\cap y_R=\\varnothing". Then for every "I\\in x_R" and "J\\in y_R" we have "(I,J)\\not\\in R" (otherwise "x_R=I_R=J_R=y_R"). Since "R_0\\subset R", we have "(I,J)\\not\\in R_0", that is "I\\cap J=\\varnothing". So we obtain "[x]\\cap [y]=\\left(\\bigcup\\limits_{I\\in x_R}I\\right)\\cap\\left(\\bigcup\\limits_{J\\in y_R}J\\right)=\\bigcup\\limits_{I\\in x_R}\\bigcup\\limits_{J\\in y_R}(I\\cap J)=\\varnothing".

And we proved the statement 1.


2)Take arbiarary "x\\in K". Let "u,v\\in [x]" , then there are "I,J\\in x_R" such that "u\\in I" and "v\\in J".

Since "I,J\\in x_R", we have "(I,J)\\in R", that is "(I,J)\\in R_0^n" for some "n".

So there is "I_0,\\ldots,I_n\\in K" such that "I=I_0", "J=I_n" and "(I_{k-1},I_k)\\in R_0" for every "k=1,\\ldots,n".


That is "I_{k-1}\\cap I_k\\neq\\varnothing" for every "k=1,\\ldots,n". So for every "k=1,\\ldots,n" take "u_k\\in I_{k-1}\\cap I_k". Let "u_0=u" and "u_{n+1}=v".

We have "u_{k-1}, u_k\\in I_k". Since "I_k" is an interval, we have "[u_{k-1},u_k]\\subset I_k", where "[u_{k-1},u_k]" is the line segment, which connects "u_{k-1}" and "u_k", so "[u,v]=[u_0,u_{n+1}]\\subset\\bigcup\\limits_{k=1}^n [u_{k-1},u_k]\\subset\\bigcup\\limits_{k=1}^n I_k".

But "(I,I_k)\\in R_0^k\\subset R" and "(x,I)\\in R", so "I_k\\in x_R", so "[u,v]\\subset\\bigcup\\limits_{k=1}^n I_k\\subset\\bigcup\\limits_{I\\in x_R} I=[x]" .

So we proved that "[x]" is an interval in "\\mathbb R". Since "[x]" is an open set, we have that "[x]" is an open interval.

And we proved the statement 2.


So we obtain that "G=\\bigcup\\limits_{x\\in K}[x]" is an union of disjoint open intervals.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS