Collection $mathcal B=${$bar A:Ain mathcal A$} is locally finite,given that $mathcal A$ is locally finite.
$begingroup$
Let $mathcal A$ be a locally finite collection of subsets of X.Then the collection $mathcal B=${$bar A:Ain mathcal A$} of the closures of the element of $mathcal A$ is locally finite.
Proof:
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 $----------------$(1)$
Since,$A_isubset bar A_i$,so from $(1)$,we have $bar A_icap Uneq phi$,where $1le i le n $.
Hence,$U$ can intersect an atmost the same number of elements of $mathcal B$ as that of $mathcal A$.
Is the proof correct?
real-analysis general-topology proof-verification
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
$endgroup$
add a comment |
$begingroup$
Let $mathcal A$ be a locally finite collection of subsets of X.Then the collection $mathcal B=${$bar A:Ain mathcal A$} of the closures of the element of $mathcal A$ is locally finite.
Proof:
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 $----------------$(1)$
Since,$A_isubset bar A_i$,so from $(1)$,we have $bar A_icap Uneq phi$,where $1le i le n $.
Hence,$U$ can intersect an atmost the same number of elements of $mathcal B$ as that of $mathcal A$.
Is the proof correct?
real-analysis general-topology proof-verification
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
$endgroup$
1
$begingroup$
No, the proof is incorrect. There might be other elements of $mathcal B$ intersecting $U$.
$endgroup$
– Crostul
Dec 24 '18 at 18:30
$begingroup$
@Crostul:I also got some intuition that there might be some other elements that can intersect with $U,$ but i was not able to recognize them.If I'm getting you correctly,you meant to say that there exists some $bar A_jin mathcal B$ other than {$A_i:1le i le n$} such that $bar A_j cap U neq phi $?
$endgroup$
– P.Styles
Dec 24 '18 at 18:42
$begingroup$
@Crostul: then how we give the guarantee that there are only finitely many members of $mathcal B$ that intersects with $U$ nontrivially?
$endgroup$
– P.Styles
Dec 24 '18 at 18:46
add a comment |
$begingroup$
Let $mathcal A$ be a locally finite collection of subsets of X.Then the collection $mathcal B=${$bar A:Ain mathcal A$} of the closures of the element of $mathcal A$ is locally finite.
Proof:
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 $----------------$(1)$
Since,$A_isubset bar A_i$,so from $(1)$,we have $bar A_icap Uneq phi$,where $1le i le n $.
Hence,$U$ can intersect an atmost the same number of elements of $mathcal B$ as that of $mathcal A$.
Is the proof correct?
real-analysis general-topology proof-verification
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
$endgroup$
Let $mathcal A$ be a locally finite collection of subsets of X.Then the collection $mathcal B=${$bar A:Ain mathcal A$} of the closures of the element of $mathcal A$ is locally finite.
Proof:
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 $----------------$(1)$
Since,$A_isubset bar A_i$,so from $(1)$,we have $bar A_icap Uneq phi$,where $1le i le n $.
Hence,$U$ can intersect an atmost the same number of elements of $mathcal B$ as that of $mathcal A$.
Is the proof correct?
real-analysis general-topology proof-verification
real-analysis general-topology proof-verification
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
asked Dec 24 '18 at 18:14
P.StylesP.Styles
1,447727
1,447727
1
$begingroup$
No, the proof is incorrect. There might be other elements of $mathcal B$ intersecting $U$.
$endgroup$
– Crostul
Dec 24 '18 at 18:30
$begingroup$
@Crostul:I also got some intuition that there might be some other elements that can intersect with $U,$ but i was not able to recognize them.If I'm getting you correctly,you meant to say that there exists some $bar A_jin mathcal B$ other than {$A_i:1le i le n$} such that $bar A_j cap U neq phi $?
$endgroup$
– P.Styles
Dec 24 '18 at 18:42
$begingroup$
@Crostul: then how we give the guarantee that there are only finitely many members of $mathcal B$ that intersects with $U$ nontrivially?
$endgroup$
– P.Styles
Dec 24 '18 at 18:46
add a comment |
1
$begingroup$
No, the proof is incorrect. There might be other elements of $mathcal B$ intersecting $U$.
$endgroup$
– Crostul
Dec 24 '18 at 18:30
$begingroup$
@Crostul:I also got some intuition that there might be some other elements that can intersect with $U,$ but i was not able to recognize them.If I'm getting you correctly,you meant to say that there exists some $bar A_jin mathcal B$ other than {$A_i:1le i le n$} such that $bar A_j cap U neq phi $?
$endgroup$
– P.Styles
Dec 24 '18 at 18:42
$begingroup$
@Crostul: then how we give the guarantee that there are only finitely many members of $mathcal B$ that intersects with $U$ nontrivially?
$endgroup$
– P.Styles
Dec 24 '18 at 18:46
1
1
$begingroup$
No, the proof is incorrect. There might be other elements of $mathcal B$ intersecting $U$.
$endgroup$
– Crostul
Dec 24 '18 at 18:30
$begingroup$
No, the proof is incorrect. There might be other elements of $mathcal B$ intersecting $U$.
$endgroup$
– Crostul
Dec 24 '18 at 18:30
$begingroup$
@Crostul:I also got some intuition that there might be some other elements that can intersect with $U,$ but i was not able to recognize them.If I'm getting you correctly,you meant to say that there exists some $bar A_jin mathcal B$ other than {$A_i:1le i le n$} such that $bar A_j cap U neq phi $?
$endgroup$
– P.Styles
Dec 24 '18 at 18:42
$begingroup$
@Crostul:I also got some intuition that there might be some other elements that can intersect with $U,$ but i was not able to recognize them.If I'm getting you correctly,you meant to say that there exists some $bar A_jin mathcal B$ other than {$A_i:1le i le n$} such that $bar A_j cap U neq phi $?
$endgroup$
– P.Styles
Dec 24 '18 at 18:42
$begingroup$
@Crostul: then how we give the guarantee that there are only finitely many members of $mathcal B$ that intersects with $U$ nontrivially?
$endgroup$
– P.Styles
Dec 24 '18 at 18:46
$begingroup$
@Crostul: then how we give the guarantee that there are only finitely many members of $mathcal B$ that intersects with $U$ nontrivially?
$endgroup$
– P.Styles
Dec 24 '18 at 18:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Essential is here that for $U$ you can always choose an open neighborhood.
In that situation we have the implication:$$Ucapoverline Aneqvarnothingimplies Ucap Aneqvarnothingtag1$$
Also the converse if this is true, but that is not relevant here.
$(1)$ assures us that the cardinality of the set ${overline Amid A
inmathcal Atext{ and } Ucap overline Aneqvarnothing}$ will not exceed the set the cardinality of the set ${Ainmathcal Amid Ucap Aneqvarnothing}$.
So if ${Ainmathcal Amid Ucap Aneqvarnothing}$ is a finite set then so is ${overline Amid Ainmathcal Atext{ and } Ucap overline Aneqvarnothing}$ (QED).
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
$endgroup$
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
|
show 3 more comments
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1 Answer
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1 Answer
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$begingroup$
Essential is here that for $U$ you can always choose an open neighborhood.
In that situation we have the implication:$$Ucapoverline Aneqvarnothingimplies Ucap Aneqvarnothingtag1$$
Also the converse if this is true, but that is not relevant here.
$(1)$ assures us that the cardinality of the set ${overline Amid A
inmathcal Atext{ and } Ucap overline Aneqvarnothing}$ will not exceed the set the cardinality of the set ${Ainmathcal Amid Ucap Aneqvarnothing}$.
So if ${Ainmathcal Amid Ucap Aneqvarnothing}$ is a finite set then so is ${overline Amid Ainmathcal Atext{ and } Ucap overline Aneqvarnothing}$ (QED).
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
$endgroup$
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
|
show 3 more comments
$begingroup$
Essential is here that for $U$ you can always choose an open neighborhood.
In that situation we have the implication:$$Ucapoverline Aneqvarnothingimplies Ucap Aneqvarnothingtag1$$
Also the converse if this is true, but that is not relevant here.
$(1)$ assures us that the cardinality of the set ${overline Amid A
inmathcal Atext{ and } Ucap overline Aneqvarnothing}$ will not exceed the set the cardinality of the set ${Ainmathcal Amid Ucap Aneqvarnothing}$.
So if ${Ainmathcal Amid Ucap Aneqvarnothing}$ is a finite set then so is ${overline Amid Ainmathcal Atext{ and } Ucap overline Aneqvarnothing}$ (QED).
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
$endgroup$
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
|
show 3 more comments
$begingroup$
Essential is here that for $U$ you can always choose an open neighborhood.
In that situation we have the implication:$$Ucapoverline Aneqvarnothingimplies Ucap Aneqvarnothingtag1$$
Also the converse if this is true, but that is not relevant here.
$(1)$ assures us that the cardinality of the set ${overline Amid A
inmathcal Atext{ and } Ucap overline Aneqvarnothing}$ will not exceed the set the cardinality of the set ${Ainmathcal Amid Ucap Aneqvarnothing}$.
So if ${Ainmathcal Amid Ucap Aneqvarnothing}$ is a finite set then so is ${overline Amid Ainmathcal Atext{ and } Ucap overline Aneqvarnothing}$ (QED).
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
$endgroup$
Essential is here that for $U$ you can always choose an open neighborhood.
In that situation we have the implication:$$Ucapoverline Aneqvarnothingimplies Ucap Aneqvarnothingtag1$$
Also the converse if this is true, but that is not relevant here.
$(1)$ assures us that the cardinality of the set ${overline Amid A
inmathcal Atext{ and } Ucap overline Aneqvarnothing}$ will not exceed the set the cardinality of the set ${Ainmathcal Amid Ucap Aneqvarnothing}$.
So if ${Ainmathcal Amid Ucap Aneqvarnothing}$ is a finite set then so is ${overline Amid Ainmathcal Atext{ and } Ucap overline Aneqvarnothing}$ (QED).
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
edited Dec 24 '18 at 19:12
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
answered Dec 24 '18 at 18:48
drhabdrhab
101k544130
101k544130
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
|
show 3 more comments
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
In my post, I've proved the converse?
$endgroup$
– P.Styles
Dec 25 '18 at 18:18
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
Yes. You proved the irrelevant side.
$endgroup$
– drhab
Dec 25 '18 at 19:00
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:Let $X=mathbb R^2$ with product topology,choosing $U$={$(x,y):x>0,y>0$}=$(0,infty)×(0,infty)$ and fixing $A$={$(0,y):y∈mathbb R$},then $Ucap bar Aneq phi$ but $Ucap A=phi$..
$endgroup$
– P.Styles
Dec 26 '18 at 5:57
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
:In the above argument (1) does not hold. What is the loophole in my example?
$endgroup$
– P.Styles
Dec 26 '18 at 5:58
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
$begingroup$
$A={(0,y)mid yinmathbb R}$ is a closed subset of $mathbb R^2$ so that $A=overline A$.
$endgroup$
– drhab
Dec 26 '18 at 7:23
|
show 3 more comments
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$begingroup$
No, the proof is incorrect. There might be other elements of $mathcal B$ intersecting $U$.
$endgroup$
– Crostul
Dec 24 '18 at 18:30
$begingroup$
@Crostul:I also got some intuition that there might be some other elements that can intersect with $U,$ but i was not able to recognize them.If I'm getting you correctly,you meant to say that there exists some $bar A_jin mathcal B$ other than {$A_i:1le i le n$} such that $bar A_j cap U neq phi $?
$endgroup$
– P.Styles
Dec 24 '18 at 18:42
$begingroup$
@Crostul: then how we give the guarantee that there are only finitely many members of $mathcal B$ that intersects with $U$ nontrivially?
$endgroup$
– P.Styles
Dec 24 '18 at 18:46