Proof for $mathbbR_l$ is Lindelof

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In Munkers there is a proof that $mathbbR_l$ is Lindelof.
$mathcalA$ be the collection of basis elements of the form $ [a_alpha,b_alpha) : alpha in J $ be a covering of $mathbbR$. Let $C= bigcup_alpha in J (a_alpha,b_alpha)$ which is a subset of $mathbbR$.

Then it is shown that $mathbbR - C$ is countable. Then construct a sub collection of basis elements in $mathcalA$ using this $mathbbR -C$ which is countable and cover $mathbbR- C$. In next step we will topologize $C$ as a subspace of $mathbbR$. Then using second countability of $C$ we will construct another countable subcollection of $mathcalA$ which cover $C$. Both the collections together forms a countable subcollection of $mathcalA$ which cover $mathbbR_l$.

My doubt is in the construction of $C$. Is there a chance that $C$ differ from $mathbbR$ ? Even it is equal to $mathbbR$ we are going to get first countable subcollection is going to be non empty.
Can we find a particular covering for $mathbbR_l$ by basis elements of the form $ [a_alpha,b_alpha) $ such that $mathbbR -C$ is non empty ? Provide an example.

general-topology lindelof-spaces

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edited Aug 3 at 7:03

Henno Brandsma

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asked Aug 3 at 2:58

Madhu

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up vote
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In Munkers there is a proof that $mathbbR_l$ is Lindelof.
$mathcalA$ be the collection of basis elements of the form $ [a_alpha,b_alpha) : alpha in J $ be a covering of $mathbbR$. Let $C= bigcup_alpha in J (a_alpha,b_alpha)$ which is a subset of $mathbbR$.

Then it is shown that $mathbbR - C$ is countable. Then construct a sub collection of basis elements in $mathcalA$ using this $mathbbR -C$ which is countable and cover $mathbbR- C$. In next step we will topologize $C$ as a subspace of $mathbbR$. Then using second countability of $C$ we will construct another countable subcollection of $mathcalA$ which cover $C$. Both the collections together forms a countable subcollection of $mathcalA$ which cover $mathbbR_l$.

My doubt is in the construction of $C$. Is there a chance that $C$ differ from $mathbbR$ ? Even it is equal to $mathbbR$ we are going to get first countable subcollection is going to be non empty.
Can we find a particular covering for $mathbbR_l$ by basis elements of the form $ [a_alpha,b_alpha) $ such that $mathbbR -C$ is non empty ? Provide an example.

general-topology lindelof-spaces

share|cite|improve this question

edited Aug 3 at 7:03

Henno Brandsma

91.2k34199

asked Aug 3 at 2:58

Madhu

689922

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

In Munkers there is a proof that $mathbbR_l$ is Lindelof.
$mathcalA$ be the collection of basis elements of the form $ [a_alpha,b_alpha) : alpha in J $ be a covering of $mathbbR$. Let $C= bigcup_alpha in J (a_alpha,b_alpha)$ which is a subset of $mathbbR$.

Then it is shown that $mathbbR - C$ is countable. Then construct a sub collection of basis elements in $mathcalA$ using this $mathbbR -C$ which is countable and cover $mathbbR- C$. In next step we will topologize $C$ as a subspace of $mathbbR$. Then using second countability of $C$ we will construct another countable subcollection of $mathcalA$ which cover $C$. Both the collections together forms a countable subcollection of $mathcalA$ which cover $mathbbR_l$.

My doubt is in the construction of $C$. Is there a chance that $C$ differ from $mathbbR$ ? Even it is equal to $mathbbR$ we are going to get first countable subcollection is going to be non empty.
Can we find a particular covering for $mathbbR_l$ by basis elements of the form $ [a_alpha,b_alpha) $ such that $mathbbR -C$ is non empty ? Provide an example.

general-topology lindelof-spaces

share|cite|improve this question

edited Aug 3 at 7:03

Henno Brandsma

91.2k34199

asked Aug 3 at 2:58

Madhu

689922

In Munkers there is a proof that $mathbbR_l$ is Lindelof.
$mathcalA$ be the collection of basis elements of the form $ [a_alpha,b_alpha) : alpha in J $ be a covering of $mathbbR$. Let $C= bigcup_alpha in J (a_alpha,b_alpha)$ which is a subset of $mathbbR$.

Then it is shown that $mathbbR - C$ is countable. Then construct a sub collection of basis elements in $mathcalA$ using this $mathbbR -C$ which is countable and cover $mathbbR- C$. In next step we will topologize $C$ as a subspace of $mathbbR$. Then using second countability of $C$ we will construct another countable subcollection of $mathcalA$ which cover $C$. Both the collections together forms a countable subcollection of $mathcalA$ which cover $mathbbR_l$.

My doubt is in the construction of $C$. Is there a chance that $C$ differ from $mathbbR$ ? Even it is equal to $mathbbR$ we are going to get first countable subcollection is going to be non empty.
Can we find a particular covering for $mathbbR_l$ by basis elements of the form $ [a_alpha,b_alpha) $ such that $mathbbR -C$ is non empty ? Provide an example.

general-topology lindelof-spaces

share|cite|improve this question

edited Aug 3 at 7:03

Henno Brandsma

91.2k34199

asked Aug 3 at 2:58

Madhu

689922

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 7:03

Henno Brandsma

91.2k34199

edited Aug 3 at 7:03

Henno Brandsma

91.2k34199

edited Aug 3 at 7:03

Henno Brandsma

91.2k34199

91.2k34199

asked Aug 3 at 2:58

Madhu

689922

asked Aug 3 at 2:58

Madhu

689922

asked Aug 3 at 2:58

Madhu

689922

689922

add a comment | 

add a comment | 

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Will this particular case is sufficient ? $mathcalA= [n,n+1) $ where $n$ is an integer. Then even though $mathcalA$ covers $mathbbR_l$, $C$ does not contain any of the integers.

share|cite|improve this answer

edited Aug 3 at 7:05

answered Aug 3 at 3:10

Madhu

689922

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Will this particular case is sufficient ? $mathcalA= [n,n+1) $ where $n$ is an integer. Then even though $mathcalA$ covers $mathbbR_l$, $C$ does not contain any of the integers.

share|cite|improve this answer

edited Aug 3 at 7:05

answered Aug 3 at 3:10

Madhu

689922

add a comment | 

up vote
1
down vote

Will this particular case is sufficient ? $mathcalA= [n,n+1) $ where $n$ is an integer. Then even though $mathcalA$ covers $mathbbR_l$, $C$ does not contain any of the integers.

share|cite|improve this answer

edited Aug 3 at 7:05

answered Aug 3 at 3:10

Madhu

689922

add a comment | 

up vote
1
down vote

up vote
1
down vote

Will this particular case is sufficient ? $mathcalA= [n,n+1) $ where $n$ is an integer. Then even though $mathcalA$ covers $mathbbR_l$, $C$ does not contain any of the integers.

share|cite|improve this answer

edited Aug 3 at 7:05

answered Aug 3 at 3:10

Madhu

689922

Will this particular case is sufficient ? $mathcalA= [n,n+1) $ where $n$ is an integer. Then even though $mathcalA$ covers $mathbbR_l$, $C$ does not contain any of the integers.

share|cite|improve this answer

edited Aug 3 at 7:05

answered Aug 3 at 3:10

Madhu

689922

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 7:05

edited Aug 3 at 7:05

edited Aug 3 at 7:05

answered Aug 3 at 3:10

Madhu

689922

answered Aug 3 at 3:10

Madhu

689922

answered Aug 3 at 3:10

Madhu

689922

689922

add a comment | 

add a comment | 

 

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