Every finite Borel Measure on a complete separable Metric Space is tight












0












$begingroup$


I'm trying to understand the proof of the Theorem in the title which I found here
https://www.math.leidenuniv.nl/~vangaans/jancol1.pdf. (Theorem 2.6)
The proof is very good in my opinion however I don't understand why he can choose
$bigcup _{ }^{ infty }{ B({ a }_{ n, }delta )=X } $ in the 4th line.
If I'm not fully mistaken, it doesn't hold in general that every complete metric space can be represented as a union of open balls.
Could someone explain to me why he is allowed to make such a choice?










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$endgroup$












  • $begingroup$
    Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
    $endgroup$
    – alexp9
    Jan 20 at 20:18
















0












$begingroup$


I'm trying to understand the proof of the Theorem in the title which I found here
https://www.math.leidenuniv.nl/~vangaans/jancol1.pdf. (Theorem 2.6)
The proof is very good in my opinion however I don't understand why he can choose
$bigcup _{ }^{ infty }{ B({ a }_{ n, }delta )=X } $ in the 4th line.
If I'm not fully mistaken, it doesn't hold in general that every complete metric space can be represented as a union of open balls.
Could someone explain to me why he is allowed to make such a choice?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
    $endgroup$
    – alexp9
    Jan 20 at 20:18














0












0








0





$begingroup$


I'm trying to understand the proof of the Theorem in the title which I found here
https://www.math.leidenuniv.nl/~vangaans/jancol1.pdf. (Theorem 2.6)
The proof is very good in my opinion however I don't understand why he can choose
$bigcup _{ }^{ infty }{ B({ a }_{ n, }delta )=X } $ in the 4th line.
If I'm not fully mistaken, it doesn't hold in general that every complete metric space can be represented as a union of open balls.
Could someone explain to me why he is allowed to make such a choice?










share|cite|improve this question











$endgroup$




I'm trying to understand the proof of the Theorem in the title which I found here
https://www.math.leidenuniv.nl/~vangaans/jancol1.pdf. (Theorem 2.6)
The proof is very good in my opinion however I don't understand why he can choose
$bigcup _{ }^{ infty }{ B({ a }_{ n, }delta )=X } $ in the 4th line.
If I'm not fully mistaken, it doesn't hold in general that every complete metric space can be represented as a union of open balls.
Could someone explain to me why he is allowed to make such a choice?







real-analysis






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edited Jan 20 at 20:05









Bernard

122k740116




122k740116










asked Jan 20 at 19:54









MasterPIMasterPI

21828




21828












  • $begingroup$
    Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
    $endgroup$
    – alexp9
    Jan 20 at 20:18


















  • $begingroup$
    Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
    $endgroup$
    – alexp9
    Jan 20 at 20:18
















$begingroup$
Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
$endgroup$
– alexp9
Jan 20 at 20:18




$begingroup$
Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
$endgroup$
– alexp9
Jan 20 at 20:18










1 Answer
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$begingroup$

Since it's a metric space, the open balls are well defined. In particular, they exist and $B(a_{n},delta)subset X$ holds. This implies that $bigcup_{i}B(a_{i},delta)subset X.$



Now take $xin X$. Since the sequence of points ${a_{n}}_{ninmathbb{N}}$ is dense in X, $exists iinmathbb{N}$ such that $d(a_{i},x)<frac{delta}{2}$. In particular, $xin B(a_{i},delta)$. Since $xin X$ was arbitrary, we conclude that $Xsubset bigcup_{i} B(a_{i},delta)$.






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$endgroup$













  • $begingroup$
    Thanks I get it! actually pretty easy
    $endgroup$
    – MasterPI
    Jan 20 at 20:26











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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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active

oldest

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1












$begingroup$

Since it's a metric space, the open balls are well defined. In particular, they exist and $B(a_{n},delta)subset X$ holds. This implies that $bigcup_{i}B(a_{i},delta)subset X.$



Now take $xin X$. Since the sequence of points ${a_{n}}_{ninmathbb{N}}$ is dense in X, $exists iinmathbb{N}$ such that $d(a_{i},x)<frac{delta}{2}$. In particular, $xin B(a_{i},delta)$. Since $xin X$ was arbitrary, we conclude that $Xsubset bigcup_{i} B(a_{i},delta)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks I get it! actually pretty easy
    $endgroup$
    – MasterPI
    Jan 20 at 20:26
















1












$begingroup$

Since it's a metric space, the open balls are well defined. In particular, they exist and $B(a_{n},delta)subset X$ holds. This implies that $bigcup_{i}B(a_{i},delta)subset X.$



Now take $xin X$. Since the sequence of points ${a_{n}}_{ninmathbb{N}}$ is dense in X, $exists iinmathbb{N}$ such that $d(a_{i},x)<frac{delta}{2}$. In particular, $xin B(a_{i},delta)$. Since $xin X$ was arbitrary, we conclude that $Xsubset bigcup_{i} B(a_{i},delta)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks I get it! actually pretty easy
    $endgroup$
    – MasterPI
    Jan 20 at 20:26














1












1








1





$begingroup$

Since it's a metric space, the open balls are well defined. In particular, they exist and $B(a_{n},delta)subset X$ holds. This implies that $bigcup_{i}B(a_{i},delta)subset X.$



Now take $xin X$. Since the sequence of points ${a_{n}}_{ninmathbb{N}}$ is dense in X, $exists iinmathbb{N}$ such that $d(a_{i},x)<frac{delta}{2}$. In particular, $xin B(a_{i},delta)$. Since $xin X$ was arbitrary, we conclude that $Xsubset bigcup_{i} B(a_{i},delta)$.






share|cite|improve this answer









$endgroup$



Since it's a metric space, the open balls are well defined. In particular, they exist and $B(a_{n},delta)subset X$ holds. This implies that $bigcup_{i}B(a_{i},delta)subset X.$



Now take $xin X$. Since the sequence of points ${a_{n}}_{ninmathbb{N}}$ is dense in X, $exists iinmathbb{N}$ such that $d(a_{i},x)<frac{delta}{2}$. In particular, $xin B(a_{i},delta)$. Since $xin X$ was arbitrary, we conclude that $Xsubset bigcup_{i} B(a_{i},delta)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 20 at 20:17









alexp9alexp9

454314




454314












  • $begingroup$
    Thanks I get it! actually pretty easy
    $endgroup$
    – MasterPI
    Jan 20 at 20:26


















  • $begingroup$
    Thanks I get it! actually pretty easy
    $endgroup$
    – MasterPI
    Jan 20 at 20:26
















$begingroup$
Thanks I get it! actually pretty easy
$endgroup$
– MasterPI
Jan 20 at 20:26




$begingroup$
Thanks I get it! actually pretty easy
$endgroup$
– MasterPI
Jan 20 at 20:26


















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