Set-theoretic difficulties concerning normality












4












$begingroup$


In her "Lectures on Set Theoretic Topology" Mary Ellen Rudin states at the end of page 5 that "In the case on nomality this is made doubly difficult by the fact that normality is such a second order property that it can often not be decided whether a given topological space is normal or not within the usual axioms for set theory."



What does it means "nomality is such a second order property"? And why this is a particularity of normality? The regularity isn't a second order property?



Is it possible to define a topological space in ZFC such that it is undecidable wheter this space is normal or not?






general-topology logic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    maybe what she means is that it universally quantifies twice over the set of all closed subsets (for all closed C, for all closed D, there exist disjoint open sets etc). regular just quantifies over all closed subsets and all points. but i don't really know.
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 0:47








  • 1




    $begingroup$
    Normality is a property about sets. Consider $forall xforall y P(x,y)$. In first order logic, the variables range over the elements of some set. In second order logic, the variables can range over sets. See en.wikipedia.org/wiki/Second-order_logic for more information.
    $endgroup$
    – John Douma
    Dec 31 '18 at 0:58
















4












$begingroup$


In her "Lectures on Set Theoretic Topology" Mary Ellen Rudin states at the end of page 5 that "In the case on nomality this is made doubly difficult by the fact that normality is such a second order property that it can often not be decided whether a given topological space is normal or not within the usual axioms for set theory."



What does it means "nomality is such a second order property"? And why this is a particularity of normality? The regularity isn't a second order property?



Is it possible to define a topological space in ZFC such that it is undecidable wheter this space is normal or not?






general-topology logic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    maybe what she means is that it universally quantifies twice over the set of all closed subsets (for all closed C, for all closed D, there exist disjoint open sets etc). regular just quantifies over all closed subsets and all points. but i don't really know.
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 0:47








  • 1




    $begingroup$
    Normality is a property about sets. Consider $forall xforall y P(x,y)$. In first order logic, the variables range over the elements of some set. In second order logic, the variables can range over sets. See en.wikipedia.org/wiki/Second-order_logic for more information.
    $endgroup$
    – John Douma
    Dec 31 '18 at 0:58














4












4








4





$begingroup$


In her "Lectures on Set Theoretic Topology" Mary Ellen Rudin states at the end of page 5 that "In the case on nomality this is made doubly difficult by the fact that normality is such a second order property that it can often not be decided whether a given topological space is normal or not within the usual axioms for set theory."



What does it means "nomality is such a second order property"? And why this is a particularity of normality? The regularity isn't a second order property?



Is it possible to define a topological space in ZFC such that it is undecidable wheter this space is normal or not?






general-topology logic







share|cite|improve this question











$endgroup$




In her "Lectures on Set Theoretic Topology" Mary Ellen Rudin states at the end of page 5 that "In the case on nomality this is made doubly difficult by the fact that normality is such a second order property that it can often not be decided whether a given topological space is normal or not within the usual axioms for set theory."



What does it means "nomality is such a second order property"? And why this is a particularity of normality? The regularity isn't a second order property?



Is it possible to define a topological space in ZFC such that it is undecidable wheter this space is normal or not?






general-topology logic




general-topology logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 13:38









bof

52.2k558121




52.2k558121










asked Dec 31 '18 at 0:35









G. OttavianoG. Ottaviano

219110




219110








  • 1




    $begingroup$
    maybe what she means is that it universally quantifies twice over the set of all closed subsets (for all closed C, for all closed D, there exist disjoint open sets etc). regular just quantifies over all closed subsets and all points. but i don't really know.
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 0:47








  • 1




    $begingroup$
    Normality is a property about sets. Consider $forall xforall y P(x,y)$. In first order logic, the variables range over the elements of some set. In second order logic, the variables can range over sets. See en.wikipedia.org/wiki/Second-order_logic for more information.
    $endgroup$
    – John Douma
    Dec 31 '18 at 0:58














  • 1




    $begingroup$
    maybe what she means is that it universally quantifies twice over the set of all closed subsets (for all closed C, for all closed D, there exist disjoint open sets etc). regular just quantifies over all closed subsets and all points. but i don't really know.
    $endgroup$
    – Tim kinsella
    Dec 31 '18 at 0:47








  • 1




    $begingroup$
    Normality is a property about sets. Consider $forall xforall y P(x,y)$. In first order logic, the variables range over the elements of some set. In second order logic, the variables can range over sets. See en.wikipedia.org/wiki/Second-order_logic for more information.
    $endgroup$
    – John Douma
    Dec 31 '18 at 0:58








1




1




$begingroup$
maybe what she means is that it universally quantifies twice over the set of all closed subsets (for all closed C, for all closed D, there exist disjoint open sets etc). regular just quantifies over all closed subsets and all points. but i don't really know.
$endgroup$
– Tim kinsella
Dec 31 '18 at 0:47






$begingroup$
maybe what she means is that it universally quantifies twice over the set of all closed subsets (for all closed C, for all closed D, there exist disjoint open sets etc). regular just quantifies over all closed subsets and all points. but i don't really know.
$endgroup$
– Tim kinsella
Dec 31 '18 at 0:47






1




1




$begingroup$
Normality is a property about sets. Consider $forall xforall y P(x,y)$. In first order logic, the variables range over the elements of some set. In second order logic, the variables can range over sets. See en.wikipedia.org/wiki/Second-order_logic for more information.
$endgroup$
– John Douma
Dec 31 '18 at 0:58




$begingroup$
Normality is a property about sets. Consider $forall xforall y P(x,y)$. In first order logic, the variables range over the elements of some set. In second order logic, the variables can range over sets. See en.wikipedia.org/wiki/Second-order_logic for more information.
$endgroup$
– John Douma
Dec 31 '18 at 0:58










1 Answer
1






active

oldest

votes


















3












$begingroup$


What does it means "normality is such a second order property"?




It means that the statement "$X$ is a normal space" requires non-trivial quantification over the subsets of some set. In particular, for a topological space $(X,tau)$ with base $mathcal{B} subset tau$, the assertion that $X$ is normal, is equivalent to the following, second-order, statement,



$$forall A,B in mathcal{P}(mathcal{B}), exists U, V in mathcal{P}(mathcal{B}): text{ either } (cup A) cup (cup B) neq X, text{ or } quadquadquadquadquad$$
$$quadquadquadquadquadquad (Xbackslash cup A) backslash cup U= emptyset,,, (Xbackslash cup B)backslash cup V = emptyset, text{ and } (cup U) cap (cup V) = emptyset.$$




Is it possible to define a topological space in ZFC such that it is undecidable whether this space is normal or not?




Yes.



A nice way to do this is to consider the spaces associated with ladder systems on $omega_1$.



Definition:




  1. A Ladder System on a stationary set $S subset lim(omega_1)$ is an $S$-indexed sequence $M=langle M_gamma: gamma in Srangle$ of countable subsets of $omega_1$, such that, $operatorname{ot}(M_gamma) = omega,$ and $,sup (M_gamma) = gamma $.


  2. For each ladder system $M=langle M_gamma: gamma in Srangle$, we let $X_M$ denote the topological space defined on the set $A=omega_1 times {0} cup S times {1}$, by letting each $(gamma, 0) in A$ be isolated and taking as basic open neighborhoods of each $(gamma,1) in A$, sets of the form ${ (gamma,1) } cup B$ where $B subset M_gamma times {0}$ co-finite.



As it turns out, whether any of the spaces $X_M$ are normal, is independent of $mathsf{ZFC}$; for a nice discussion of this result, see the paper,




Balogh, Zoltán; Eisworth, Todd; Gruenhage, Gary; Pavlov, Oleg; Szeptycki, Paul, Uniformization and anti-uniformization properties of ladder systems, Fundam. Math. 181, No. 3, 189-213 (2004). ZBL1051.03034.,







share|cite|improve this answer











$endgroup$













  • $begingroup$
    But isn't also the regularity a second order property?
    $endgroup$
    – G. Ottaviano
    Jan 10 at 19:55






  • 1




    $begingroup$
    @Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
    $endgroup$
    – Not Mike
    Jan 10 at 21:59










  • $begingroup$
    Thank you! Very clear
    $endgroup$
    – G. Ottaviano
    Jan 10 at 23:47











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$begingroup$


What does it means "normality is such a second order property"?




It means that the statement "$X$ is a normal space" requires non-trivial quantification over the subsets of some set. In particular, for a topological space $(X,tau)$ with base $mathcal{B} subset tau$, the assertion that $X$ is normal, is equivalent to the following, second-order, statement,



$$forall A,B in mathcal{P}(mathcal{B}), exists U, V in mathcal{P}(mathcal{B}): text{ either } (cup A) cup (cup B) neq X, text{ or } quadquadquadquadquad$$
$$quadquadquadquadquadquad (Xbackslash cup A) backslash cup U= emptyset,,, (Xbackslash cup B)backslash cup V = emptyset, text{ and } (cup U) cap (cup V) = emptyset.$$




Is it possible to define a topological space in ZFC such that it is undecidable whether this space is normal or not?




Yes.



A nice way to do this is to consider the spaces associated with ladder systems on $omega_1$.



Definition:




  1. A Ladder System on a stationary set $S subset lim(omega_1)$ is an $S$-indexed sequence $M=langle M_gamma: gamma in Srangle$ of countable subsets of $omega_1$, such that, $operatorname{ot}(M_gamma) = omega,$ and $,sup (M_gamma) = gamma $.


  2. For each ladder system $M=langle M_gamma: gamma in Srangle$, we let $X_M$ denote the topological space defined on the set $A=omega_1 times {0} cup S times {1}$, by letting each $(gamma, 0) in A$ be isolated and taking as basic open neighborhoods of each $(gamma,1) in A$, sets of the form ${ (gamma,1) } cup B$ where $B subset M_gamma times {0}$ co-finite.



As it turns out, whether any of the spaces $X_M$ are normal, is independent of $mathsf{ZFC}$; for a nice discussion of this result, see the paper,




Balogh, Zoltán; Eisworth, Todd; Gruenhage, Gary; Pavlov, Oleg; Szeptycki, Paul, Uniformization and anti-uniformization properties of ladder systems, Fundam. Math. 181, No. 3, 189-213 (2004). ZBL1051.03034.,







share|cite|improve this answer











$endgroup$













  • $begingroup$
    But isn't also the regularity a second order property?
    $endgroup$
    – G. Ottaviano
    Jan 10 at 19:55






  • 1




    $begingroup$
    @Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
    $endgroup$
    – Not Mike
    Jan 10 at 21:59










  • $begingroup$
    Thank you! Very clear
    $endgroup$
    – G. Ottaviano
    Jan 10 at 23:47
















3












$begingroup$


What does it means "normality is such a second order property"?




It means that the statement "$X$ is a normal space" requires non-trivial quantification over the subsets of some set. In particular, for a topological space $(X,tau)$ with base $mathcal{B} subset tau$, the assertion that $X$ is normal, is equivalent to the following, second-order, statement,



$$forall A,B in mathcal{P}(mathcal{B}), exists U, V in mathcal{P}(mathcal{B}): text{ either } (cup A) cup (cup B) neq X, text{ or } quadquadquadquadquad$$
$$quadquadquadquadquadquad (Xbackslash cup A) backslash cup U= emptyset,,, (Xbackslash cup B)backslash cup V = emptyset, text{ and } (cup U) cap (cup V) = emptyset.$$




Is it possible to define a topological space in ZFC such that it is undecidable whether this space is normal or not?




Yes.



A nice way to do this is to consider the spaces associated with ladder systems on $omega_1$.



Definition:




  1. A Ladder System on a stationary set $S subset lim(omega_1)$ is an $S$-indexed sequence $M=langle M_gamma: gamma in Srangle$ of countable subsets of $omega_1$, such that, $operatorname{ot}(M_gamma) = omega,$ and $,sup (M_gamma) = gamma $.


  2. For each ladder system $M=langle M_gamma: gamma in Srangle$, we let $X_M$ denote the topological space defined on the set $A=omega_1 times {0} cup S times {1}$, by letting each $(gamma, 0) in A$ be isolated and taking as basic open neighborhoods of each $(gamma,1) in A$, sets of the form ${ (gamma,1) } cup B$ where $B subset M_gamma times {0}$ co-finite.



As it turns out, whether any of the spaces $X_M$ are normal, is independent of $mathsf{ZFC}$; for a nice discussion of this result, see the paper,




Balogh, Zoltán; Eisworth, Todd; Gruenhage, Gary; Pavlov, Oleg; Szeptycki, Paul, Uniformization and anti-uniformization properties of ladder systems, Fundam. Math. 181, No. 3, 189-213 (2004). ZBL1051.03034.,







share|cite|improve this answer











$endgroup$













  • $begingroup$
    But isn't also the regularity a second order property?
    $endgroup$
    – G. Ottaviano
    Jan 10 at 19:55






  • 1




    $begingroup$
    @Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
    $endgroup$
    – Not Mike
    Jan 10 at 21:59










  • $begingroup$
    Thank you! Very clear
    $endgroup$
    – G. Ottaviano
    Jan 10 at 23:47














3












3








3





$begingroup$


What does it means "normality is such a second order property"?




It means that the statement "$X$ is a normal space" requires non-trivial quantification over the subsets of some set. In particular, for a topological space $(X,tau)$ with base $mathcal{B} subset tau$, the assertion that $X$ is normal, is equivalent to the following, second-order, statement,



$$forall A,B in mathcal{P}(mathcal{B}), exists U, V in mathcal{P}(mathcal{B}): text{ either } (cup A) cup (cup B) neq X, text{ or } quadquadquadquadquad$$
$$quadquadquadquadquadquad (Xbackslash cup A) backslash cup U= emptyset,,, (Xbackslash cup B)backslash cup V = emptyset, text{ and } (cup U) cap (cup V) = emptyset.$$




Is it possible to define a topological space in ZFC such that it is undecidable whether this space is normal or not?




Yes.



A nice way to do this is to consider the spaces associated with ladder systems on $omega_1$.



Definition:




  1. A Ladder System on a stationary set $S subset lim(omega_1)$ is an $S$-indexed sequence $M=langle M_gamma: gamma in Srangle$ of countable subsets of $omega_1$, such that, $operatorname{ot}(M_gamma) = omega,$ and $,sup (M_gamma) = gamma $.


  2. For each ladder system $M=langle M_gamma: gamma in Srangle$, we let $X_M$ denote the topological space defined on the set $A=omega_1 times {0} cup S times {1}$, by letting each $(gamma, 0) in A$ be isolated and taking as basic open neighborhoods of each $(gamma,1) in A$, sets of the form ${ (gamma,1) } cup B$ where $B subset M_gamma times {0}$ co-finite.



As it turns out, whether any of the spaces $X_M$ are normal, is independent of $mathsf{ZFC}$; for a nice discussion of this result, see the paper,




Balogh, Zoltán; Eisworth, Todd; Gruenhage, Gary; Pavlov, Oleg; Szeptycki, Paul, Uniformization and anti-uniformization properties of ladder systems, Fundam. Math. 181, No. 3, 189-213 (2004). ZBL1051.03034.,







share|cite|improve this answer











$endgroup$




What does it means "normality is such a second order property"?




It means that the statement "$X$ is a normal space" requires non-trivial quantification over the subsets of some set. In particular, for a topological space $(X,tau)$ with base $mathcal{B} subset tau$, the assertion that $X$ is normal, is equivalent to the following, second-order, statement,



$$forall A,B in mathcal{P}(mathcal{B}), exists U, V in mathcal{P}(mathcal{B}): text{ either } (cup A) cup (cup B) neq X, text{ or } quadquadquadquadquad$$
$$quadquadquadquadquadquad (Xbackslash cup A) backslash cup U= emptyset,,, (Xbackslash cup B)backslash cup V = emptyset, text{ and } (cup U) cap (cup V) = emptyset.$$




Is it possible to define a topological space in ZFC such that it is undecidable whether this space is normal or not?




Yes.



A nice way to do this is to consider the spaces associated with ladder systems on $omega_1$.



Definition:




  1. A Ladder System on a stationary set $S subset lim(omega_1)$ is an $S$-indexed sequence $M=langle M_gamma: gamma in Srangle$ of countable subsets of $omega_1$, such that, $operatorname{ot}(M_gamma) = omega,$ and $,sup (M_gamma) = gamma $.


  2. For each ladder system $M=langle M_gamma: gamma in Srangle$, we let $X_M$ denote the topological space defined on the set $A=omega_1 times {0} cup S times {1}$, by letting each $(gamma, 0) in A$ be isolated and taking as basic open neighborhoods of each $(gamma,1) in A$, sets of the form ${ (gamma,1) } cup B$ where $B subset M_gamma times {0}$ co-finite.



As it turns out, whether any of the spaces $X_M$ are normal, is independent of $mathsf{ZFC}$; for a nice discussion of this result, see the paper,




Balogh, Zoltán; Eisworth, Todd; Gruenhage, Gary; Pavlov, Oleg; Szeptycki, Paul, Uniformization and anti-uniformization properties of ladder systems, Fundam. Math. 181, No. 3, 189-213 (2004). ZBL1051.03034.,








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 8:42

























answered Jan 2 at 12:52









Not MikeNot Mike

36329




36329












  • $begingroup$
    But isn't also the regularity a second order property?
    $endgroup$
    – G. Ottaviano
    Jan 10 at 19:55






  • 1




    $begingroup$
    @Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
    $endgroup$
    – Not Mike
    Jan 10 at 21:59










  • $begingroup$
    Thank you! Very clear
    $endgroup$
    – G. Ottaviano
    Jan 10 at 23:47


















  • $begingroup$
    But isn't also the regularity a second order property?
    $endgroup$
    – G. Ottaviano
    Jan 10 at 19:55






  • 1




    $begingroup$
    @Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
    $endgroup$
    – Not Mike
    Jan 10 at 21:59










  • $begingroup$
    Thank you! Very clear
    $endgroup$
    – G. Ottaviano
    Jan 10 at 23:47
















$begingroup$
But isn't also the regularity a second order property?
$endgroup$
– G. Ottaviano
Jan 10 at 19:55




$begingroup$
But isn't also the regularity a second order property?
$endgroup$
– G. Ottaviano
Jan 10 at 19:55




1




1




$begingroup$
@Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
$endgroup$
– Not Mike
Jan 10 at 21:59




$begingroup$
@Ottaviano It turns out that regularity is equivalent to a much simpler statement of the form "there is some base with a certain first-order property" which means the property behaves much nicer.
$endgroup$
– Not Mike
Jan 10 at 21:59












$begingroup$
Thank you! Very clear
$endgroup$
– G. Ottaviano
Jan 10 at 23:47




$begingroup$
Thank you! Very clear
$endgroup$
– G. Ottaviano
Jan 10 at 23:47


















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