Every finite Borel Measure on a complete separable Metric Space is tight
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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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add a comment |
$begingroup$
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?
real-analysis
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$begingroup$
Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
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– alexp9
Jan 20 at 20:18
add a comment |
$begingroup$
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?
real-analysis
$endgroup$
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
real-analysis
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
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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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Thanks I get it! actually pretty easy
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– MasterPI
Jan 20 at 20:26
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
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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 $Xsubset bigcup_{i} B(a_{i},delta)$.
$endgroup$
$begingroup$
Thanks I get it! actually pretty easy
$endgroup$
– MasterPI
Jan 20 at 20:26
add a comment |
$begingroup$
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 $Xsubset bigcup_{i} B(a_{i},delta)$.
$endgroup$
$begingroup$
Thanks I get it! actually pretty easy
$endgroup$
– MasterPI
Jan 20 at 20:26
add a comment |
$begingroup$
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 $Xsubset bigcup_{i} B(a_{i},delta)$.
$endgroup$
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 $Xsubset bigcup_{i} B(a_{i},delta)$.
answered Jan 20 at 20:17
alexp9alexp9
454314
454314
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Thanks I get it! actually pretty easy
$endgroup$
– MasterPI
Jan 20 at 20:26
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Just as a comment, the title of your question is misleading. Your problem is entirely about topology.
$endgroup$
– alexp9
Jan 20 at 20:18