Any subcollection $mathcal A_1$ of a locally finite collection $mathcal A$ of subsets of $X$ is locally...
$begingroup$
Any subcollection $mathcal A_1$ of a locally finite collection $mathcal A$ of subsets of $X$ is locally finite.
Proof
Let $mathcal A_1$ be any subcollection of $mathcal A$ i.e., $mathcal A_1subset mathcal A.$
Let $xin X$ and $mathcal A$ is locally finite, so there exists a nbhd $U$ of $x$ that intersects only finitely many elements $A_1, A_2, A_3,..., A_n$ of $mathcal A$.
More precisely,$A_icap Uneq phi$,where $1le i le n $.
Since,$mathcal A_1subset mathcal A.$,so if any element of {$A_i:1le i le n$} is in $mathcal A_1,$ so $U$ intersects only finitely many elements of $mathcal A_1$,making it locally finite collection of subsets of $X$.
Please check the proof critically up to here, especially the use of quantifiers.. If there is some scope of improvement in the proof or some elegant method other than this please let me know...
My question is "What if no element of {$A_i:1le i le n$} is in $mathcal A_1$?"
real-analysis general-topology proof-verification proof-writing quantifiers
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
$endgroup$
add a comment |
$begingroup$
Any subcollection $mathcal A_1$ of a locally finite collection $mathcal A$ of subsets of $X$ is locally finite.
Proof
Let $mathcal A_1$ be any subcollection of $mathcal A$ i.e., $mathcal A_1subset mathcal A.$
Let $xin X$ and $mathcal A$ is locally finite, so there exists a nbhd $U$ of $x$ that intersects only finitely many elements $A_1, A_2, A_3,..., A_n$ of $mathcal A$.
More precisely,$A_icap Uneq phi$,where $1le i le n $.
Since,$mathcal A_1subset mathcal A.$,so if any element of {$A_i:1le i le n$} is in $mathcal A_1,$ so $U$ intersects only finitely many elements of $mathcal A_1$,making it locally finite collection of subsets of $X$.
Please check the proof critically up to here, especially the use of quantifiers.. If there is some scope of improvement in the proof or some elegant method other than this please let me know...
My question is "What if no element of {$A_i:1le i le n$} is in $mathcal A_1$?"
real-analysis general-topology proof-verification proof-writing quantifiers
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
$endgroup$
$begingroup$
Locally Finite: A collection of subsets of a topological space X is said to be locally finite if each point in the space has a neighborhood that intersects only finitely many of the sets in the collection.
$endgroup$
– P.Styles
Dec 24 '18 at 17:57
add a comment |
$begingroup$
Any subcollection $mathcal A_1$ of a locally finite collection $mathcal A$ of subsets of $X$ is locally finite.
Proof
Let $mathcal A_1$ be any subcollection of $mathcal A$ i.e., $mathcal A_1subset mathcal A.$
Let $xin X$ and $mathcal A$ is locally finite, so there exists a nbhd $U$ of $x$ that intersects only finitely many elements $A_1, A_2, A_3,..., A_n$ of $mathcal A$.
More precisely,$A_icap Uneq phi$,where $1le i le n $.
Since,$mathcal A_1subset mathcal A.$,so if any element of {$A_i:1le i le n$} is in $mathcal A_1,$ so $U$ intersects only finitely many elements of $mathcal A_1$,making it locally finite collection of subsets of $X$.
Please check the proof critically up to here, especially the use of quantifiers.. If there is some scope of improvement in the proof or some elegant method other than this please let me know...
My question is "What if no element of {$A_i:1le i le n$} is in $mathcal A_1$?"
real-analysis general-topology proof-verification proof-writing quantifiers
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
$endgroup$
Any subcollection $mathcal A_1$ of a locally finite collection $mathcal A$ of subsets of $X$ is locally finite.
Proof
Let $mathcal A_1$ be any subcollection of $mathcal A$ i.e., $mathcal A_1subset mathcal A.$
Let $xin X$ and $mathcal A$ is locally finite, so there exists a nbhd $U$ of $x$ that intersects only finitely many elements $A_1, A_2, A_3,..., A_n$ of $mathcal A$.
More precisely,$A_icap Uneq phi$,where $1le i le n $.
Since,$mathcal A_1subset mathcal A.$,so if any element of {$A_i:1le i le n$} is in $mathcal A_1,$ so $U$ intersects only finitely many elements of $mathcal A_1$,making it locally finite collection of subsets of $X$.
Please check the proof critically up to here, especially the use of quantifiers.. If there is some scope of improvement in the proof or some elegant method other than this please let me know...
My question is "What if no element of {$A_i:1le i le n$} is in $mathcal A_1$?"
real-analysis general-topology proof-verification proof-writing quantifiers
real-analysis general-topology proof-verification proof-writing quantifiers
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
asked Dec 24 '18 at 17:33
P.StylesP.Styles
1,472927
1,472927
$begingroup$
Locally Finite: A collection of subsets of a topological space X is said to be locally finite if each point in the space has a neighborhood that intersects only finitely many of the sets in the collection.
$endgroup$
– P.Styles
Dec 24 '18 at 17:57
add a comment |
$begingroup$
Locally Finite: A collection of subsets of a topological space X is said to be locally finite if each point in the space has a neighborhood that intersects only finitely many of the sets in the collection.
$endgroup$
– P.Styles
Dec 24 '18 at 17:57
$begingroup$
Locally Finite: A collection of subsets of a topological space X is said to be locally finite if each point in the space has a neighborhood that intersects only finitely many of the sets in the collection.
$endgroup$
– P.Styles
Dec 24 '18 at 17:57
$begingroup$
Locally Finite: A collection of subsets of a topological space X is said to be locally finite if each point in the space has a neighborhood that intersects only finitely many of the sets in the collection.
$endgroup$
– P.Styles
Dec 24 '18 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof is okay.
You could also formulate it like this:
Collection $mathcal A$ is locally finite if and only if for every $xin X$ a neighborhood $U$ of $x$ exists that satisfies:$${Ainmathcal Amid Acap Uneqvarnothing}text{ is a finite set}tag1$$
From $mathcal A_1subseteqmathcal A$ it follows directly that: $${Ainmathcal A_1mid Acap Uneqvarnothing}subseteq{Ainmathcal Amid Acap Uneqvarnothing}tag2$$so if $mathcal A$ is indeed locally finite then from $(1)$ and $(2)$ it follows directly that also:$${Ainmathcal A_1mid Acap Uneqvarnothing}text{ is a finite set}tag3$$
We conclude that $mathcal A_1$ is also locally finite.
Concerning your question:
If ${Ainmathcal A_1mid Acap Uneqvarnothing}$ is empty (in your words no element of ${A_imid 1leq ileq n}$ is in $mathcal A_1$) then the set mentioned under $(3)$ is empty hence is finite.
So nothing is wrong then with the reasoning.
The less elements a collection $mathcal A$ contains the more likely it is that the collection is locally finite.
In specific the empty collection is locally finite.
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
$endgroup$
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Your proof is okay.
You could also formulate it like this:
Collection $mathcal A$ is locally finite if and only if for every $xin X$ a neighborhood $U$ of $x$ exists that satisfies:$${Ainmathcal Amid Acap Uneqvarnothing}text{ is a finite set}tag1$$
From $mathcal A_1subseteqmathcal A$ it follows directly that: $${Ainmathcal A_1mid Acap Uneqvarnothing}subseteq{Ainmathcal Amid Acap Uneqvarnothing}tag2$$so if $mathcal A$ is indeed locally finite then from $(1)$ and $(2)$ it follows directly that also:$${Ainmathcal A_1mid Acap Uneqvarnothing}text{ is a finite set}tag3$$
We conclude that $mathcal A_1$ is also locally finite.
Concerning your question:
If ${Ainmathcal A_1mid Acap Uneqvarnothing}$ is empty (in your words no element of ${A_imid 1leq ileq n}$ is in $mathcal A_1$) then the set mentioned under $(3)$ is empty hence is finite.
So nothing is wrong then with the reasoning.
The less elements a collection $mathcal A$ contains the more likely it is that the collection is locally finite.
In specific the empty collection is locally finite.
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
$endgroup$
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
add a comment |
$begingroup$
Your proof is okay.
You could also formulate it like this:
Collection $mathcal A$ is locally finite if and only if for every $xin X$ a neighborhood $U$ of $x$ exists that satisfies:$${Ainmathcal Amid Acap Uneqvarnothing}text{ is a finite set}tag1$$
From $mathcal A_1subseteqmathcal A$ it follows directly that: $${Ainmathcal A_1mid Acap Uneqvarnothing}subseteq{Ainmathcal Amid Acap Uneqvarnothing}tag2$$so if $mathcal A$ is indeed locally finite then from $(1)$ and $(2)$ it follows directly that also:$${Ainmathcal A_1mid Acap Uneqvarnothing}text{ is a finite set}tag3$$
We conclude that $mathcal A_1$ is also locally finite.
Concerning your question:
If ${Ainmathcal A_1mid Acap Uneqvarnothing}$ is empty (in your words no element of ${A_imid 1leq ileq n}$ is in $mathcal A_1$) then the set mentioned under $(3)$ is empty hence is finite.
So nothing is wrong then with the reasoning.
The less elements a collection $mathcal A$ contains the more likely it is that the collection is locally finite.
In specific the empty collection is locally finite.
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
$endgroup$
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
add a comment |
$begingroup$
Your proof is okay.
You could also formulate it like this:
Collection $mathcal A$ is locally finite if and only if for every $xin X$ a neighborhood $U$ of $x$ exists that satisfies:$${Ainmathcal Amid Acap Uneqvarnothing}text{ is a finite set}tag1$$
From $mathcal A_1subseteqmathcal A$ it follows directly that: $${Ainmathcal A_1mid Acap Uneqvarnothing}subseteq{Ainmathcal Amid Acap Uneqvarnothing}tag2$$so if $mathcal A$ is indeed locally finite then from $(1)$ and $(2)$ it follows directly that also:$${Ainmathcal A_1mid Acap Uneqvarnothing}text{ is a finite set}tag3$$
We conclude that $mathcal A_1$ is also locally finite.
Concerning your question:
If ${Ainmathcal A_1mid Acap Uneqvarnothing}$ is empty (in your words no element of ${A_imid 1leq ileq n}$ is in $mathcal A_1$) then the set mentioned under $(3)$ is empty hence is finite.
So nothing is wrong then with the reasoning.
The less elements a collection $mathcal A$ contains the more likely it is that the collection is locally finite.
In specific the empty collection is locally finite.
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
$endgroup$
Your proof is okay.
You could also formulate it like this:
Collection $mathcal A$ is locally finite if and only if for every $xin X$ a neighborhood $U$ of $x$ exists that satisfies:$${Ainmathcal Amid Acap Uneqvarnothing}text{ is a finite set}tag1$$
From $mathcal A_1subseteqmathcal A$ it follows directly that: $${Ainmathcal A_1mid Acap Uneqvarnothing}subseteq{Ainmathcal Amid Acap Uneqvarnothing}tag2$$so if $mathcal A$ is indeed locally finite then from $(1)$ and $(2)$ it follows directly that also:$${Ainmathcal A_1mid Acap Uneqvarnothing}text{ is a finite set}tag3$$
We conclude that $mathcal A_1$ is also locally finite.
Concerning your question:
If ${Ainmathcal A_1mid Acap Uneqvarnothing}$ is empty (in your words no element of ${A_imid 1leq ileq n}$ is in $mathcal A_1$) then the set mentioned under $(3)$ is empty hence is finite.
So nothing is wrong then with the reasoning.
The less elements a collection $mathcal A$ contains the more likely it is that the collection is locally finite.
In specific the empty collection is locally finite.
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
answered Dec 24 '18 at 18:31
drhabdrhab
103k545136
103k545136
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
add a comment |
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
:your last statement made my day...thank you...Merry Christmas!!
$endgroup$
– P.Styles
Dec 24 '18 at 18:49
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
$begingroup$
Merry Christmas to you as well, of course.
$endgroup$
– drhab
Dec 24 '18 at 18:52
add a comment |
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$begingroup$
Locally Finite: A collection of subsets of a topological space X is said to be locally finite if each point in the space has a neighborhood that intersects only finitely many of the sets in the collection.
$endgroup$
– P.Styles
Dec 24 '18 at 17:57