A question about zero-sets












2












$begingroup$


The set $C(X)$ of all continuous, real-value functions no a topological space $X$ will be provided with an algebraic structure and order structure.



zero-set means:



$Z(f) = Z_{X} (f) = { x in X : f(x) = 0 } quad ( f in C(X) )$



$Z(X)= Z[C(X)]={ Z(f) : f in C(X) }$



I know that $Z(X)$ closed under countable intersection and also $Z(X)$ need not be closed under infinite union.




1: In a general space, Do countable or finite union of zero-sets be a
zero-set?Why?



2:Is $Z(X)$ closed under arbitrary intersection?Why?







general-topology topological-groups







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is $Z(X)$? What does it mean for $Z(X)$ to be, for instance, "closed under countable intersection"?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:15












  • $begingroup$
    @BenW $Z(X)$ is the set of all zero-sets of $X$.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:22










  • $begingroup$
    Hint: Find an example of a closed set that is not a zero set.
    $endgroup$
    – GEdgar
    Dec 26 '18 at 18:27










  • $begingroup$
    And what does it mean for $Z(X)$ to be closed under countable intersection?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:28
















2












$begingroup$


The set $C(X)$ of all continuous, real-value functions no a topological space $X$ will be provided with an algebraic structure and order structure.



zero-set means:



$Z(f) = Z_{X} (f) = { x in X : f(x) = 0 } quad ( f in C(X) )$



$Z(X)= Z[C(X)]={ Z(f) : f in C(X) }$



I know that $Z(X)$ closed under countable intersection and also $Z(X)$ need not be closed under infinite union.




1: In a general space, Do countable or finite union of zero-sets be a
zero-set?Why?



2:Is $Z(X)$ closed under arbitrary intersection?Why?







general-topology topological-groups







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is $Z(X)$? What does it mean for $Z(X)$ to be, for instance, "closed under countable intersection"?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:15












  • $begingroup$
    @BenW $Z(X)$ is the set of all zero-sets of $X$.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:22










  • $begingroup$
    Hint: Find an example of a closed set that is not a zero set.
    $endgroup$
    – GEdgar
    Dec 26 '18 at 18:27










  • $begingroup$
    And what does it mean for $Z(X)$ to be closed under countable intersection?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:28














2












2








2





$begingroup$


The set $C(X)$ of all continuous, real-value functions no a topological space $X$ will be provided with an algebraic structure and order structure.



zero-set means:



$Z(f) = Z_{X} (f) = { x in X : f(x) = 0 } quad ( f in C(X) )$



$Z(X)= Z[C(X)]={ Z(f) : f in C(X) }$



I know that $Z(X)$ closed under countable intersection and also $Z(X)$ need not be closed under infinite union.




1: In a general space, Do countable or finite union of zero-sets be a
zero-set?Why?



2:Is $Z(X)$ closed under arbitrary intersection?Why?







general-topology topological-groups







share|cite|improve this question











$endgroup$




The set $C(X)$ of all continuous, real-value functions no a topological space $X$ will be provided with an algebraic structure and order structure.



zero-set means:



$Z(f) = Z_{X} (f) = { x in X : f(x) = 0 } quad ( f in C(X) )$



$Z(X)= Z[C(X)]={ Z(f) : f in C(X) }$



I know that $Z(X)$ closed under countable intersection and also $Z(X)$ need not be closed under infinite union.




1: In a general space, Do countable or finite union of zero-sets be a
zero-set?Why?



2:Is $Z(X)$ closed under arbitrary intersection?Why?







general-topology topological-groups




general-topology topological-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 18:24

























asked Dec 26 '18 at 18:12







user387219















  • 2




    $begingroup$
    What is $Z(X)$? What does it mean for $Z(X)$ to be, for instance, "closed under countable intersection"?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:15












  • $begingroup$
    @BenW $Z(X)$ is the set of all zero-sets of $X$.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:22










  • $begingroup$
    Hint: Find an example of a closed set that is not a zero set.
    $endgroup$
    – GEdgar
    Dec 26 '18 at 18:27










  • $begingroup$
    And what does it mean for $Z(X)$ to be closed under countable intersection?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:28














  • 2




    $begingroup$
    What is $Z(X)$? What does it mean for $Z(X)$ to be, for instance, "closed under countable intersection"?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:15












  • $begingroup$
    @BenW $Z(X)$ is the set of all zero-sets of $X$.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:22










  • $begingroup$
    Hint: Find an example of a closed set that is not a zero set.
    $endgroup$
    – GEdgar
    Dec 26 '18 at 18:27










  • $begingroup$
    And what does it mean for $Z(X)$ to be closed under countable intersection?
    $endgroup$
    – Ben W
    Dec 26 '18 at 18:28








2




2




$begingroup$
What is $Z(X)$? What does it mean for $Z(X)$ to be, for instance, "closed under countable intersection"?
$endgroup$
– Ben W
Dec 26 '18 at 18:15






$begingroup$
What is $Z(X)$? What does it mean for $Z(X)$ to be, for instance, "closed under countable intersection"?
$endgroup$
– Ben W
Dec 26 '18 at 18:15














$begingroup$
@BenW $Z(X)$ is the set of all zero-sets of $X$.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 18:22




$begingroup$
@BenW $Z(X)$ is the set of all zero-sets of $X$.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 18:22












$begingroup$
Hint: Find an example of a closed set that is not a zero set.
$endgroup$
– GEdgar
Dec 26 '18 at 18:27




$begingroup$
Hint: Find an example of a closed set that is not a zero set.
$endgroup$
– GEdgar
Dec 26 '18 at 18:27












$begingroup$
And what does it mean for $Z(X)$ to be closed under countable intersection?
$endgroup$
– Ben W
Dec 26 '18 at 18:28




$begingroup$
And what does it mean for $Z(X)$ to be closed under countable intersection?
$endgroup$
– Ben W
Dec 26 '18 at 18:28










1 Answer
1






active

oldest

votes


















0












$begingroup$

$Z(f)$ is closed under finite unions : $Z(f) cup Z(g) = Z(fg)$ plus induction.



In general not closed under countable unions: singletons are in $Z(mathbb{R})$ but $mathbb{Q}$ is not.



Also $Z(X)$ will not be closed under arbitary intersections, but you need some non-metrisable counterexample (in metric spaces $Z(X)$ is just the collection of all closed sets which is closed under all intersections). Think (or search) about a counterexample with some closed set that is not a zero-set.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:38










  • $begingroup$
    @joe the rationals are a countable union of zerosets that’s not a zeroset
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:40










  • $begingroup$
    In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:43








  • 2




    $begingroup$
    @joe so find a non metric $X$ as an example.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:00






  • 2




    $begingroup$
    @joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:04











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

$Z(f)$ is closed under finite unions : $Z(f) cup Z(g) = Z(fg)$ plus induction.



In general not closed under countable unions: singletons are in $Z(mathbb{R})$ but $mathbb{Q}$ is not.



Also $Z(X)$ will not be closed under arbitary intersections, but you need some non-metrisable counterexample (in metric spaces $Z(X)$ is just the collection of all closed sets which is closed under all intersections). Think (or search) about a counterexample with some closed set that is not a zero-set.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:38










  • $begingroup$
    @joe the rationals are a countable union of zerosets that’s not a zeroset
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:40










  • $begingroup$
    In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:43








  • 2




    $begingroup$
    @joe so find a non metric $X$ as an example.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:00






  • 2




    $begingroup$
    @joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:04
















0












$begingroup$

$Z(f)$ is closed under finite unions : $Z(f) cup Z(g) = Z(fg)$ plus induction.



In general not closed under countable unions: singletons are in $Z(mathbb{R})$ but $mathbb{Q}$ is not.



Also $Z(X)$ will not be closed under arbitary intersections, but you need some non-metrisable counterexample (in metric spaces $Z(X)$ is just the collection of all closed sets which is closed under all intersections). Think (or search) about a counterexample with some closed set that is not a zero-set.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:38










  • $begingroup$
    @joe the rationals are a countable union of zerosets that’s not a zeroset
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:40










  • $begingroup$
    In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:43








  • 2




    $begingroup$
    @joe so find a non metric $X$ as an example.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:00






  • 2




    $begingroup$
    @joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:04














0












0








0





$begingroup$

$Z(f)$ is closed under finite unions : $Z(f) cup Z(g) = Z(fg)$ plus induction.



In general not closed under countable unions: singletons are in $Z(mathbb{R})$ but $mathbb{Q}$ is not.



Also $Z(X)$ will not be closed under arbitary intersections, but you need some non-metrisable counterexample (in metric spaces $Z(X)$ is just the collection of all closed sets which is closed under all intersections). Think (or search) about a counterexample with some closed set that is not a zero-set.






share|cite|improve this answer









$endgroup$



$Z(f)$ is closed under finite unions : $Z(f) cup Z(g) = Z(fg)$ plus induction.



In general not closed under countable unions: singletons are in $Z(mathbb{R})$ but $mathbb{Q}$ is not.



Also $Z(X)$ will not be closed under arbitary intersections, but you need some non-metrisable counterexample (in metric spaces $Z(X)$ is just the collection of all closed sets which is closed under all intersections). Think (or search) about a counterexample with some closed set that is not a zero-set.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 18:28









Henno BrandsmaHenno Brandsma

112k348121




112k348121












  • $begingroup$
    What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:38










  • $begingroup$
    @joe the rationals are a countable union of zerosets that’s not a zeroset
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:40










  • $begingroup$
    In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:43








  • 2




    $begingroup$
    @joe so find a non metric $X$ as an example.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:00






  • 2




    $begingroup$
    @joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:04


















  • $begingroup$
    What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:38










  • $begingroup$
    @joe the rationals are a countable union of zerosets that’s not a zeroset
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 18:40










  • $begingroup$
    In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
    $endgroup$
    – user387219
    Dec 26 '18 at 18:43








  • 2




    $begingroup$
    @joe so find a non metric $X$ as an example.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:00






  • 2




    $begingroup$
    @joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
    $endgroup$
    – Henno Brandsma
    Dec 26 '18 at 19:04
















$begingroup$
What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
$endgroup$
– user387219
Dec 26 '18 at 18:38




$begingroup$
What dose it mean? singletons are in $Z(mathbb{R})$ but $ mathbb{Q}$ is not.
$endgroup$
– user387219
Dec 26 '18 at 18:38












$begingroup$
@joe the rationals are a countable union of zerosets that’s not a zeroset
$endgroup$
– Henno Brandsma
Dec 26 '18 at 18:40




$begingroup$
@joe the rationals are a countable union of zerosets that’s not a zeroset
$endgroup$
– Henno Brandsma
Dec 26 '18 at 18:40












$begingroup$
In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
$endgroup$
– user387219
Dec 26 '18 at 18:43






$begingroup$
In metric space every closed set is zero-set, buy you say "search about a counterexample with some closed set that is not a zero-set". I can not see.
$endgroup$
– user387219
Dec 26 '18 at 18:43






2




2




$begingroup$
@joe so find a non metric $X$ as an example.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 19:00




$begingroup$
@joe so find a non metric $X$ as an example.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 19:00




2




2




$begingroup$
@joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 19:04




$begingroup$
@joe you can use that in a Tychonoff space all closed sets are intersections of zero-sets. Also, a zero-set is a $G_delta$ and the reverse holds in normal spaces.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 19:04


















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