Collection $mathcal B=${$bar A:Ain mathcal A$} is locally finite,given that $mathcal A$ is locally finite.












-1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















-1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














-1












-1








-1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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).






share|cite|improve this answer











$endgroup$



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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 19:12

























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


















  • $begingroup$
    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


















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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