Can we write a locally compact metric space as a union of countable compact sets?












-3












$begingroup$


Let $X$ be a locally compact metric space. Can we choose a sequence $H_i$ of compact sets such that $H_i subset operatorname{int}left(H_{i+1}right)$ for all $i geq 1$ and $X =cup_{i=1}^{infty} H_i$.



Ignore the counterexample : uncountable set with discrete metric.
Please, provide me some other examples if known.






general-topology metric-spaces compactness







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Uncountable set with discrete metrics is not necessary. Take any uncountable collection of locally compact metric spaces and then take their disjoint union. The space is metrizable, it is locally compact but doesn't satisfy your condition.
    $endgroup$
    – freakish
    Dec 19 '18 at 17:59










  • $begingroup$
    Compact sets are finite in Discrete metric . So, the countable union of strictly increasing chain of finite sets can be atmost countable.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:05






  • 4




    $begingroup$
    What I've meant is that "ignore the uncountable discrete counterexample" is a very weird assumption. It doesn't make anything easier, cause you can replace each point in an uncountable discrete space with a nontrivial locally compact metric space and you still have a counterexample.
    $endgroup$
    – freakish
    Dec 19 '18 at 18:06












  • $begingroup$
    As freakish has pointed out, obvious counterexamples are uncountable disjoint unions of compact spaces (or more generally locally compact spaces). It seems you want counterexamples which do not have such "trivial decompositions".
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 23:53










  • $begingroup$
    Yeah.. I need some non trivial decompositions.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:41
















-3












$begingroup$


Let $X$ be a locally compact metric space. Can we choose a sequence $H_i$ of compact sets such that $H_i subset operatorname{int}left(H_{i+1}right)$ for all $i geq 1$ and $X =cup_{i=1}^{infty} H_i$.



Ignore the counterexample : uncountable set with discrete metric.
Please, provide me some other examples if known.






general-topology metric-spaces compactness







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Uncountable set with discrete metrics is not necessary. Take any uncountable collection of locally compact metric spaces and then take their disjoint union. The space is metrizable, it is locally compact but doesn't satisfy your condition.
    $endgroup$
    – freakish
    Dec 19 '18 at 17:59










  • $begingroup$
    Compact sets are finite in Discrete metric . So, the countable union of strictly increasing chain of finite sets can be atmost countable.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:05






  • 4




    $begingroup$
    What I've meant is that "ignore the uncountable discrete counterexample" is a very weird assumption. It doesn't make anything easier, cause you can replace each point in an uncountable discrete space with a nontrivial locally compact metric space and you still have a counterexample.
    $endgroup$
    – freakish
    Dec 19 '18 at 18:06












  • $begingroup$
    As freakish has pointed out, obvious counterexamples are uncountable disjoint unions of compact spaces (or more generally locally compact spaces). It seems you want counterexamples which do not have such "trivial decompositions".
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 23:53










  • $begingroup$
    Yeah.. I need some non trivial decompositions.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:41














-3












-3








-3





$begingroup$


Let $X$ be a locally compact metric space. Can we choose a sequence $H_i$ of compact sets such that $H_i subset operatorname{int}left(H_{i+1}right)$ for all $i geq 1$ and $X =cup_{i=1}^{infty} H_i$.



Ignore the counterexample : uncountable set with discrete metric.
Please, provide me some other examples if known.






general-topology metric-spaces compactness







share|cite|improve this question











$endgroup$




Let $X$ be a locally compact metric space. Can we choose a sequence $H_i$ of compact sets such that $H_i subset operatorname{int}left(H_{i+1}right)$ for all $i geq 1$ and $X =cup_{i=1}^{infty} H_i$.



Ignore the counterexample : uncountable set with discrete metric.
Please, provide me some other examples if known.






general-topology metric-spaces compactness




general-topology metric-spaces compactness






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 18:23









Davide Giraudo

126k16151263




126k16151263










asked Dec 19 '18 at 17:46









Abdul Gaffar KhanAbdul Gaffar Khan

386




386








  • 2




    $begingroup$
    Uncountable set with discrete metrics is not necessary. Take any uncountable collection of locally compact metric spaces and then take their disjoint union. The space is metrizable, it is locally compact but doesn't satisfy your condition.
    $endgroup$
    – freakish
    Dec 19 '18 at 17:59










  • $begingroup$
    Compact sets are finite in Discrete metric . So, the countable union of strictly increasing chain of finite sets can be atmost countable.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:05






  • 4




    $begingroup$
    What I've meant is that "ignore the uncountable discrete counterexample" is a very weird assumption. It doesn't make anything easier, cause you can replace each point in an uncountable discrete space with a nontrivial locally compact metric space and you still have a counterexample.
    $endgroup$
    – freakish
    Dec 19 '18 at 18:06












  • $begingroup$
    As freakish has pointed out, obvious counterexamples are uncountable disjoint unions of compact spaces (or more generally locally compact spaces). It seems you want counterexamples which do not have such "trivial decompositions".
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 23:53










  • $begingroup$
    Yeah.. I need some non trivial decompositions.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:41














  • 2




    $begingroup$
    Uncountable set with discrete metrics is not necessary. Take any uncountable collection of locally compact metric spaces and then take their disjoint union. The space is metrizable, it is locally compact but doesn't satisfy your condition.
    $endgroup$
    – freakish
    Dec 19 '18 at 17:59










  • $begingroup$
    Compact sets are finite in Discrete metric . So, the countable union of strictly increasing chain of finite sets can be atmost countable.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:05






  • 4




    $begingroup$
    What I've meant is that "ignore the uncountable discrete counterexample" is a very weird assumption. It doesn't make anything easier, cause you can replace each point in an uncountable discrete space with a nontrivial locally compact metric space and you still have a counterexample.
    $endgroup$
    – freakish
    Dec 19 '18 at 18:06












  • $begingroup$
    As freakish has pointed out, obvious counterexamples are uncountable disjoint unions of compact spaces (or more generally locally compact spaces). It seems you want counterexamples which do not have such "trivial decompositions".
    $endgroup$
    – Paul Frost
    Dec 19 '18 at 23:53










  • $begingroup$
    Yeah.. I need some non trivial decompositions.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:41








2




2




$begingroup$
Uncountable set with discrete metrics is not necessary. Take any uncountable collection of locally compact metric spaces and then take their disjoint union. The space is metrizable, it is locally compact but doesn't satisfy your condition.
$endgroup$
– freakish
Dec 19 '18 at 17:59




$begingroup$
Uncountable set with discrete metrics is not necessary. Take any uncountable collection of locally compact metric spaces and then take their disjoint union. The space is metrizable, it is locally compact but doesn't satisfy your condition.
$endgroup$
– freakish
Dec 19 '18 at 17:59












$begingroup$
Compact sets are finite in Discrete metric . So, the countable union of strictly increasing chain of finite sets can be atmost countable.
$endgroup$
– Abdul Gaffar Khan
Dec 19 '18 at 18:05




$begingroup$
Compact sets are finite in Discrete metric . So, the countable union of strictly increasing chain of finite sets can be atmost countable.
$endgroup$
– Abdul Gaffar Khan
Dec 19 '18 at 18:05




4




4




$begingroup$
What I've meant is that "ignore the uncountable discrete counterexample" is a very weird assumption. It doesn't make anything easier, cause you can replace each point in an uncountable discrete space with a nontrivial locally compact metric space and you still have a counterexample.
$endgroup$
– freakish
Dec 19 '18 at 18:06






$begingroup$
What I've meant is that "ignore the uncountable discrete counterexample" is a very weird assumption. It doesn't make anything easier, cause you can replace each point in an uncountable discrete space with a nontrivial locally compact metric space and you still have a counterexample.
$endgroup$
– freakish
Dec 19 '18 at 18:06














$begingroup$
As freakish has pointed out, obvious counterexamples are uncountable disjoint unions of compact spaces (or more generally locally compact spaces). It seems you want counterexamples which do not have such "trivial decompositions".
$endgroup$
– Paul Frost
Dec 19 '18 at 23:53




$begingroup$
As freakish has pointed out, obvious counterexamples are uncountable disjoint unions of compact spaces (or more generally locally compact spaces). It seems you want counterexamples which do not have such "trivial decompositions".
$endgroup$
– Paul Frost
Dec 19 '18 at 23:53












$begingroup$
Yeah.. I need some non trivial decompositions.
$endgroup$
– Abdul Gaffar Khan
Dec 20 '18 at 19:41




$begingroup$
Yeah.. I need some non trivial decompositions.
$endgroup$
– Abdul Gaffar Khan
Dec 20 '18 at 19:41










2 Answers
2






active

oldest

votes


















1












$begingroup$

This holds (for metric spaces) iff $X$ is separable as well:



If $X$ is locally compact metric and we have the $H_i$ as promised, $X$ is $sigma$-compact, hence Lindelöf and separable and second countable (as these are equivalent in all metric spaces).



If $X$ is locally compact metric and separable, we can reduce the base of open sets with compact closure to a countable base ${B_n: n in omega}$ with all $overline{B_n}$ compact. Then set $H_0 = B_0$ and cover $overline{B_0}$ by finitely many new $B_n$, whose union we then define to be $H_1$ etc. continuing by recursion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How? Any reference please!
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:02






  • 2




    $begingroup$
    @AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 18:04










  • $begingroup$
    Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:43












  • $begingroup$
    @AbdulGaffarKhan Quite true.
    $endgroup$
    – Henno Brandsma
    Dec 20 '18 at 19:44



















0












$begingroup$

In general it is impossible. A space $X$ which is the countable union of compact subsets is called $sigma$-compact. It is well-known that if $X$ is locally compact, then $X$ is $sigma$-compact if and and only it is representable as in your question.



So your question is: Is every locally compact space which is not discrete a $sigma$-compact space?



Consider an uncountable product $P = prod_{alpha in A} I$ of unit intervals $I = [0,1]$ and $x in P$. Then $X = P setminus { x }$ is locally compact. If it were the countable union of compact subsets, then ${ x }$ would be the intersection of countably many open neighborhoods. A basis of open neighborhoods is given by the family of sets $bigcap_{i=1}^n p_{alpha_i} ^{-1}(U_i)$, where $p_alpha : P to I$ is the projection onto the coordinate $alpha$, $U_i subset I$ an open neighborhood of $x_i = p_{alpha_i}(x)$ and $n in mathbb{N}$ is arbitrary. But it is easy to see that any countable intersection of such sets is still an uncountable set since only countably many coordinates are subject to restrictions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for help
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:40











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This holds (for metric spaces) iff $X$ is separable as well:



If $X$ is locally compact metric and we have the $H_i$ as promised, $X$ is $sigma$-compact, hence Lindelöf and separable and second countable (as these are equivalent in all metric spaces).



If $X$ is locally compact metric and separable, we can reduce the base of open sets with compact closure to a countable base ${B_n: n in omega}$ with all $overline{B_n}$ compact. Then set $H_0 = B_0$ and cover $overline{B_0}$ by finitely many new $B_n$, whose union we then define to be $H_1$ etc. continuing by recursion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How? Any reference please!
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:02






  • 2




    $begingroup$
    @AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 18:04










  • $begingroup$
    Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:43












  • $begingroup$
    @AbdulGaffarKhan Quite true.
    $endgroup$
    – Henno Brandsma
    Dec 20 '18 at 19:44
















1












$begingroup$

This holds (for metric spaces) iff $X$ is separable as well:



If $X$ is locally compact metric and we have the $H_i$ as promised, $X$ is $sigma$-compact, hence Lindelöf and separable and second countable (as these are equivalent in all metric spaces).



If $X$ is locally compact metric and separable, we can reduce the base of open sets with compact closure to a countable base ${B_n: n in omega}$ with all $overline{B_n}$ compact. Then set $H_0 = B_0$ and cover $overline{B_0}$ by finitely many new $B_n$, whose union we then define to be $H_1$ etc. continuing by recursion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How? Any reference please!
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:02






  • 2




    $begingroup$
    @AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 18:04










  • $begingroup$
    Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:43












  • $begingroup$
    @AbdulGaffarKhan Quite true.
    $endgroup$
    – Henno Brandsma
    Dec 20 '18 at 19:44














1












1








1





$begingroup$

This holds (for metric spaces) iff $X$ is separable as well:



If $X$ is locally compact metric and we have the $H_i$ as promised, $X$ is $sigma$-compact, hence Lindelöf and separable and second countable (as these are equivalent in all metric spaces).



If $X$ is locally compact metric and separable, we can reduce the base of open sets with compact closure to a countable base ${B_n: n in omega}$ with all $overline{B_n}$ compact. Then set $H_0 = B_0$ and cover $overline{B_0}$ by finitely many new $B_n$, whose union we then define to be $H_1$ etc. continuing by recursion.






share|cite|improve this answer











$endgroup$



This holds (for metric spaces) iff $X$ is separable as well:



If $X$ is locally compact metric and we have the $H_i$ as promised, $X$ is $sigma$-compact, hence Lindelöf and separable and second countable (as these are equivalent in all metric spaces).



If $X$ is locally compact metric and separable, we can reduce the base of open sets with compact closure to a countable base ${B_n: n in omega}$ with all $overline{B_n}$ compact. Then set $H_0 = B_0$ and cover $overline{B_0}$ by finitely many new $B_n$, whose union we then define to be $H_1$ etc. continuing by recursion.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 21:56

























answered Dec 19 '18 at 18:01









Henno BrandsmaHenno Brandsma

110k347116




110k347116












  • $begingroup$
    How? Any reference please!
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:02






  • 2




    $begingroup$
    @AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 18:04










  • $begingroup$
    Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:43












  • $begingroup$
    @AbdulGaffarKhan Quite true.
    $endgroup$
    – Henno Brandsma
    Dec 20 '18 at 19:44


















  • $begingroup$
    How? Any reference please!
    $endgroup$
    – Abdul Gaffar Khan
    Dec 19 '18 at 18:02






  • 2




    $begingroup$
    @AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
    $endgroup$
    – Henno Brandsma
    Dec 19 '18 at 18:04










  • $begingroup$
    Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:43












  • $begingroup$
    @AbdulGaffarKhan Quite true.
    $endgroup$
    – Henno Brandsma
    Dec 20 '18 at 19:44
















$begingroup$
How? Any reference please!
$endgroup$
– Abdul Gaffar Khan
Dec 19 '18 at 18:02




$begingroup$
How? Any reference please!
$endgroup$
– Abdul Gaffar Khan
Dec 19 '18 at 18:02




2




2




$begingroup$
@AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 18:04




$begingroup$
@AbdulGaffarKhan It's clearly necessary as $X$ is then $sigma$-compact hence Lindelöf hence separable (and second countable). If second countable (or separable which is the same for metric spaces) we can use a countable base of sets with compact closure to make the $H_i$.
$endgroup$
– Henno Brandsma
Dec 19 '18 at 18:04












$begingroup$
Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
$endgroup$
– Abdul Gaffar Khan
Dec 20 '18 at 19:43






$begingroup$
Let me call the assumption in question be ( * ). By above remarks and the ( * ), we can conclude the following : Assumption (*) holds in a metric $X$ if and only if $X$ is locally compact and separable if and only if $X$ is locally compact and $sigma$-compact.
$endgroup$
– Abdul Gaffar Khan
Dec 20 '18 at 19:43














$begingroup$
@AbdulGaffarKhan Quite true.
$endgroup$
– Henno Brandsma
Dec 20 '18 at 19:44




$begingroup$
@AbdulGaffarKhan Quite true.
$endgroup$
– Henno Brandsma
Dec 20 '18 at 19:44











0












$begingroup$

In general it is impossible. A space $X$ which is the countable union of compact subsets is called $sigma$-compact. It is well-known that if $X$ is locally compact, then $X$ is $sigma$-compact if and and only it is representable as in your question.



So your question is: Is every locally compact space which is not discrete a $sigma$-compact space?



Consider an uncountable product $P = prod_{alpha in A} I$ of unit intervals $I = [0,1]$ and $x in P$. Then $X = P setminus { x }$ is locally compact. If it were the countable union of compact subsets, then ${ x }$ would be the intersection of countably many open neighborhoods. A basis of open neighborhoods is given by the family of sets $bigcap_{i=1}^n p_{alpha_i} ^{-1}(U_i)$, where $p_alpha : P to I$ is the projection onto the coordinate $alpha$, $U_i subset I$ an open neighborhood of $x_i = p_{alpha_i}(x)$ and $n in mathbb{N}$ is arbitrary. But it is easy to see that any countable intersection of such sets is still an uncountable set since only countably many coordinates are subject to restrictions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for help
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:40
















0












$begingroup$

In general it is impossible. A space $X$ which is the countable union of compact subsets is called $sigma$-compact. It is well-known that if $X$ is locally compact, then $X$ is $sigma$-compact if and and only it is representable as in your question.



So your question is: Is every locally compact space which is not discrete a $sigma$-compact space?



Consider an uncountable product $P = prod_{alpha in A} I$ of unit intervals $I = [0,1]$ and $x in P$. Then $X = P setminus { x }$ is locally compact. If it were the countable union of compact subsets, then ${ x }$ would be the intersection of countably many open neighborhoods. A basis of open neighborhoods is given by the family of sets $bigcap_{i=1}^n p_{alpha_i} ^{-1}(U_i)$, where $p_alpha : P to I$ is the projection onto the coordinate $alpha$, $U_i subset I$ an open neighborhood of $x_i = p_{alpha_i}(x)$ and $n in mathbb{N}$ is arbitrary. But it is easy to see that any countable intersection of such sets is still an uncountable set since only countably many coordinates are subject to restrictions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for help
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:40














0












0








0





$begingroup$

In general it is impossible. A space $X$ which is the countable union of compact subsets is called $sigma$-compact. It is well-known that if $X$ is locally compact, then $X$ is $sigma$-compact if and and only it is representable as in your question.



So your question is: Is every locally compact space which is not discrete a $sigma$-compact space?



Consider an uncountable product $P = prod_{alpha in A} I$ of unit intervals $I = [0,1]$ and $x in P$. Then $X = P setminus { x }$ is locally compact. If it were the countable union of compact subsets, then ${ x }$ would be the intersection of countably many open neighborhoods. A basis of open neighborhoods is given by the family of sets $bigcap_{i=1}^n p_{alpha_i} ^{-1}(U_i)$, where $p_alpha : P to I$ is the projection onto the coordinate $alpha$, $U_i subset I$ an open neighborhood of $x_i = p_{alpha_i}(x)$ and $n in mathbb{N}$ is arbitrary. But it is easy to see that any countable intersection of such sets is still an uncountable set since only countably many coordinates are subject to restrictions.






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$endgroup$



In general it is impossible. A space $X$ which is the countable union of compact subsets is called $sigma$-compact. It is well-known that if $X$ is locally compact, then $X$ is $sigma$-compact if and and only it is representable as in your question.



So your question is: Is every locally compact space which is not discrete a $sigma$-compact space?



Consider an uncountable product $P = prod_{alpha in A} I$ of unit intervals $I = [0,1]$ and $x in P$. Then $X = P setminus { x }$ is locally compact. If it were the countable union of compact subsets, then ${ x }$ would be the intersection of countably many open neighborhoods. A basis of open neighborhoods is given by the family of sets $bigcap_{i=1}^n p_{alpha_i} ^{-1}(U_i)$, where $p_alpha : P to I$ is the projection onto the coordinate $alpha$, $U_i subset I$ an open neighborhood of $x_i = p_{alpha_i}(x)$ and $n in mathbb{N}$ is arbitrary. But it is easy to see that any countable intersection of such sets is still an uncountable set since only countably many coordinates are subject to restrictions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 22:08

























answered Dec 19 '18 at 18:13









Paul FrostPaul Frost

10.9k3934




10.9k3934












  • $begingroup$
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    – Abdul Gaffar Khan
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  • $begingroup$
    Thanks for help
    $endgroup$
    – Abdul Gaffar Khan
    Dec 20 '18 at 19:40
















$begingroup$
Thanks for help
$endgroup$
– Abdul Gaffar Khan
Dec 20 '18 at 19:40




$begingroup$
Thanks for help
$endgroup$
– Abdul Gaffar Khan
Dec 20 '18 at 19:40


















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