Any countable set has measurable zero












1












$begingroup$


To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    $endgroup$
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    $begingroup$
    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:05










  • $begingroup$
    Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    $begingroup$
    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Thanks, Rubik! :)
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:14
















1












$begingroup$


To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    $endgroup$
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    $begingroup$
    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:05










  • $begingroup$
    Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    $begingroup$
    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Thanks, Rubik! :)
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:14














1












1








1


1



$begingroup$


To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory







share|cite|improve this question









$endgroup$




To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?



Thank you :)






real-analysis general-topology elementary-set-theory




real-analysis general-topology elementary-set-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 1:49









Rafael VergnaudRafael Vergnaud

348217




348217








  • 1




    $begingroup$
    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    $endgroup$
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    $begingroup$
    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:05










  • $begingroup$
    Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    $begingroup$
    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Thanks, Rubik! :)
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:14














  • 1




    $begingroup$
    Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
    $endgroup$
    – carmichael561
    Dec 10 '18 at 2:04






  • 1




    $begingroup$
    No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:05










  • $begingroup$
    Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:06








  • 1




    $begingroup$
    Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
    $endgroup$
    – rubikscube09
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Thanks, Rubik! :)
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:14








1




1




$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04




$begingroup$
Are you referring specifically to Lebesgue measure? Because this isn't true for an arbitrary measure.
$endgroup$
– carmichael561
Dec 10 '18 at 2:04




1




1




$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05




$begingroup$
No, it is not sufficient. But, if you have a countable set, you can write it as something similar to the natural numbers. Specifically, you can sequence the set, so that each element has a unique corresponding natural number, i.e $S = {a_1, a_2 ,a_3, cdots}$
$endgroup$
– rubikscube09
Dec 10 '18 at 2:05












$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06






$begingroup$
Hey Carmichael. I'm not sure. My book has only mentioned the term "measure zero," without additional qualifying adjectives. To quote, "A set $A subseteq mathbb{R}$ has measure zero if, for all $epsilon > 0,$ there exists a countable collection of open intervals $O_n$ with the property that $A$ is contained in the union of all of the intervals $O_n$ and the sum of the lengths of all of the intervals is less than or equal to $epsilon.$"
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:06






1




1




$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14






$begingroup$
Yes it is identical. To answer the other question, your book is referring to the Lebesgue measure on $mathbb{R}$. There are other measures of course.
$endgroup$
– rubikscube09
Dec 10 '18 at 2:14






1




1




$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14




$begingroup$
Thanks, Rubik! :)
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:14










1 Answer
1






active

oldest

votes


















2












$begingroup$

(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • $begingroup$
    @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    $endgroup$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Right. Got it! Thanks a lot Holo!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    $begingroup$
    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    $endgroup$
    – Holo
    Dec 10 '18 at 2:29













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1 Answer
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1 Answer
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active

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active

oldest

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2












$begingroup$

(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • $begingroup$
    @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    $endgroup$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Right. Got it! Thanks a lot Holo!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    $begingroup$
    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    $endgroup$
    – Holo
    Dec 10 '18 at 2:29


















2












$begingroup$

(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • $begingroup$
    @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    $endgroup$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Right. Got it! Thanks a lot Holo!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    $begingroup$
    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    $endgroup$
    – Holo
    Dec 10 '18 at 2:29
















2












2








2





$begingroup$

(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.






share|cite|improve this answer











$endgroup$



(Asuuming we are talking about Lebesgue measur)



Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}varepsilon,a_i+2^{-i-1}varepsilon)$.



Now $Asubseteq bigcup A_i$, and $muleft(bigcup A_iright)lesum2^{-i}varepsilon=varepsilon$




Showing that $Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 2:29

























answered Dec 10 '18 at 2:10









HoloHolo

5,75421131




5,75421131












  • $begingroup$
    Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • $begingroup$
    @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    $endgroup$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Right. Got it! Thanks a lot Holo!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    $begingroup$
    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    $endgroup$
    – Holo
    Dec 10 '18 at 2:29




















  • $begingroup$
    Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:11












  • $begingroup$
    @RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
    $endgroup$
    – Holo
    Dec 10 '18 at 2:14








  • 1




    $begingroup$
    Right. Got it! Thanks a lot Holo!
    $endgroup$
    – Rafael Vergnaud
    Dec 10 '18 at 2:15






  • 1




    $begingroup$
    @RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
    $endgroup$
    – Holo
    Dec 10 '18 at 2:29


















$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11






$begingroup$
Hey, Holo. Thanks for the response. My proof was the following: take the union of the intervals $(a_i - frac{epsilon}{2^{i+1}}, a_i + frac{epsilon}{2^{i+1}}).$ The sum would be $sum_{i = 1}^{infty} frac{epsilon}{2} (frac{1}{2})^{i} = epsilon.$ Is this correct? My summation looks different than your's!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:11














$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14






$begingroup$
@RafaelVergnaud you did the exact same things as I did, but you have a little(algebra) mistake, the summation of yours should be with $[...])^{i-1}$, then the $epsilon/2$ will make it $[...])^i$
$endgroup$
– Holo
Dec 10 '18 at 2:14






1




1




$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15




$begingroup$
Right. Got it! Thanks a lot Holo!
$endgroup$
– Rafael Vergnaud
Dec 10 '18 at 2:15




1




1




$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29






$begingroup$
@RafaelVergnaud I added an example where 2 sets has the same cardinality but not the same measure to explain why to show that $mu(Bbb N)=0$ is not enough
$endgroup$
– Holo
Dec 10 '18 at 2:29




















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Verónica Boquete

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Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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