Compact spaces that depend on choice?
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I've often encountered topological spaces like $2^omega$, where there is a short proof of compactness using Tychonoff's theorem and a long proof of compactness not using Tychonoff's theorem. Are there situations where this is not the case, i.e. the compactness of the space depends on Tychonoff's theorem?
Essentially, I'm looking for any examples of topological spaces $(X,tau)$ such that
1. $(X,tau)$ exists in ZFC.
2. $(X,tau)$, or something more or less equivalent, exists in some model of ZF without choice.
3. $(X,tau)$ is compact in ZFC but not in the other model.
By "more or less equivalent" I mean that they are constructed in the same way, or have similar definitions, or have the same name, or anything else that would make them analogous; my knowledge of set theory is not particularly deep, so the notion is a little fuzzy.
general-topology set-theory compactness axiom-of-choice
asked Aug 9 at 20:50
namsos
806
add a comment |Â
up vote
3
down vote
favorite
I've often encountered topological spaces like $2^omega$, where there is a short proof of compactness using Tychonoff's theorem and a long proof of compactness not using Tychonoff's theorem. Are there situations where this is not the case, i.e. the compactness of the space depends on Tychonoff's theorem?
Essentially, I'm looking for any examples of topological spaces $(X,tau)$ such that
1. $(X,tau)$ exists in ZFC.
2. $(X,tau)$, or something more or less equivalent, exists in some model of ZF without choice.
3. $(X,tau)$ is compact in ZFC but not in the other model.
By "more or less equivalent" I mean that they are constructed in the same way, or have similar definitions, or have the same name, or anything else that would make them analogous; my knowledge of set theory is not particularly deep, so the notion is a little fuzzy.
general-topology set-theory compactness axiom-of-choice
asked Aug 9 at 20:50
namsos
806
3
A common proof that Tychonoff's theorem is equivalent to AC takes an indexed family of sets $ X_i mid i in I $ and constructs a space which is compact if and only if $prod_iin I X_i ne emptyset$.
– Daniel Schepler
Aug 9 at 20:55
3
So for example, if you take the family $ x + mathbbQ mid x in mathbbR $ used in constructing a nonmeasurable set, then that won't have a "choice function" in a model of ZF where every subset of $mathbbR$ is measurable...
– Daniel Schepler
Aug 9 at 20:57
3
As @Daniel suggests, this amounts to two parts: (1) analysis of the proof that Tychonoff's theorem implies choice; (2) finding a family of sets which does not provably admit a choice function. The first one is "easy" (or rather, stratightforward). The second part requires some more knowledge of set theory. But here's a key point: (a) $X$ can be well-ordered if and only if $mathcal P(X)setminusvarnothing$ admits a choice function; and (b) $Bbb R$ is not provably well-orderable in ZF.
– Asaf Karagila♦
Aug 9 at 21:03
1
How about $[0,1]cupxinmathbb R:text the axiom of choice is false$?
– bof
Aug 9 at 22:53
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I've often encountered topological spaces like $2^omega$, where there is a short proof of compactness using Tychonoff's theorem and a long proof of compactness not using Tychonoff's theorem. Are there situations where this is not the case, i.e. the compactness of the space depends on Tychonoff's theorem?
Essentially, I'm looking for any examples of topological spaces $(X,tau)$ such that
1. $(X,tau)$ exists in ZFC.
2. $(X,tau)$, or something more or less equivalent, exists in some model of ZF without choice.
3. $(X,tau)$ is compact in ZFC but not in the other model.
By "more or less equivalent" I mean that they are constructed in the same way, or have similar definitions, or have the same name, or anything else that would make them analogous; my knowledge of set theory is not particularly deep, so the notion is a little fuzzy.
general-topology set-theory compactness axiom-of-choice
asked Aug 9 at 20:50
namsos
806
I've often encountered topological spaces like $2^omega$, where there is a short proof of compactness using Tychonoff's theorem and a long proof of compactness not using Tychonoff's theorem. Are there situations where this is not the case, i.e. the compactness of the space depends on Tychonoff's theorem?
Essentially, I'm looking for any examples of topological spaces $(X,tau)$ such that
1. $(X,tau)$ exists in ZFC.
2. $(X,tau)$, or something more or less equivalent, exists in some model of ZF without choice.
3. $(X,tau)$ is compact in ZFC but not in the other model.
By "more or less equivalent" I mean that they are constructed in the same way, or have similar definitions, or have the same name, or anything else that would make them analogous; my knowledge of set theory is not particularly deep, so the notion is a little fuzzy.
general-topology set-theory compactness axiom-of-choice
asked Aug 9 at 20:50
namsos
806
asked Aug 9 at 20:50
namsos
806
asked Aug 9 at 20:50
namsos
806
asked Aug 9 at 20:50
namsos
806
806
3
A common proof that Tychonoff's theorem is equivalent to AC takes an indexed family of sets $ X_i mid i in I $ and constructs a space which is compact if and only if $prod_iin I X_i ne emptyset$.
– Daniel Schepler
Aug 9 at 20:55
3
So for example, if you take the family $ x + mathbbQ mid x in mathbbR $ used in constructing a nonmeasurable set, then that won't have a "choice function" in a model of ZF where every subset of $mathbbR$ is measurable...
– Daniel Schepler
Aug 9 at 20:57
3
As @Daniel suggests, this amounts to two parts: (1) analysis of the proof that Tychonoff's theorem implies choice; (2) finding a family of sets which does not provably admit a choice function. The first one is "easy" (or rather, stratightforward). The second part requires some more knowledge of set theory. But here's a key point: (a) $X$ can be well-ordered if and only if $mathcal P(X)setminusvarnothing$ admits a choice function; and (b) $Bbb R$ is not provably well-orderable in ZF.
– Asaf Karagila♦
Aug 9 at 21:03
1
How about $[0,1]cupxinmathbb R:text the axiom of choice is false$?
– bof
Aug 9 at 22:53
add a comment |Â
3
A common proof that Tychonoff's theorem is equivalent to AC takes an indexed family of sets $ X_i mid i in I $ and constructs a space which is compact if and only if $prod_iin I X_i ne emptyset$.
– Daniel Schepler
Aug 9 at 20:55
3
So for example, if you take the family $ x + mathbbQ mid x in mathbbR $ used in constructing a nonmeasurable set, then that won't have a "choice function" in a model of ZF where every subset of $mathbbR$ is measurable...
– Daniel Schepler
Aug 9 at 20:57
3
As @Daniel suggests, this amounts to two parts: (1) analysis of the proof that Tychonoff's theorem implies choice; (2) finding a family of sets which does not provably admit a choice function. The first one is "easy" (or rather, stratightforward). The second part requires some more knowledge of set theory. But here's a key point: (a) $X$ can be well-ordered if and only if $mathcal P(X)setminusvarnothing$ admits a choice function; and (b) $Bbb R$ is not provably well-orderable in ZF.
– Asaf Karagila♦
Aug 9 at 21:03
1
How about $[0,1]cupxinmathbb R:text the axiom of choice is false$?
– bof
Aug 9 at 22:53
3
3
A common proof that Tychonoff's theorem is equivalent to AC takes an indexed family of sets $ X_i mid i in I $ and constructs a space which is compact if and only if $prod_iin I X_i ne emptyset$.
– Daniel Schepler
Aug 9 at 20:55
A common proof that Tychonoff's theorem is equivalent to AC takes an indexed family of sets $ X_i mid i in I $ and constructs a space which is compact if and only if $prod_iin I X_i ne emptyset$.
– Daniel Schepler
Aug 9 at 20:55
3
3
So for example, if you take the family $ x + mathbbQ mid x in mathbbR $ used in constructing a nonmeasurable set, then that won't have a "choice function" in a model of ZF where every subset of $mathbbR$ is measurable...
– Daniel Schepler
Aug 9 at 20:57
So for example, if you take the family $ x + mathbbQ mid x in mathbbR $ used in constructing a nonmeasurable set, then that won't have a "choice function" in a model of ZF where every subset of $mathbbR$ is measurable...
– Daniel Schepler
Aug 9 at 20:57
3
3
As @Daniel suggests, this amounts to two parts: (1) analysis of the proof that Tychonoff's theorem implies choice; (2) finding a family of sets which does not provably admit a choice function. The first one is "easy" (or rather, stratightforward). The second part requires some more knowledge of set theory. But here's a key point: (a) $X$ can be well-ordered if and only if $mathcal P(X)setminusvarnothing$ admits a choice function; and (b) $Bbb R$ is not provably well-orderable in ZF.
– Asaf Karagila♦
Aug 9 at 21:03
As @Daniel suggests, this amounts to two parts: (1) analysis of the proof that Tychonoff's theorem implies choice; (2) finding a family of sets which does not provably admit a choice function. The first one is "easy" (or rather, stratightforward). The second part requires some more knowledge of set theory. But here's a key point: (a) $X$ can be well-ordered if and only if $mathcal P(X)setminusvarnothing$ admits a choice function; and (b) $Bbb R$ is not provably well-orderable in ZF.
– Asaf Karagila♦
Aug 9 at 21:03
1
1
How about $[0,1]cupxinmathbb R:text the axiom of choice is false$?
– bof
Aug 9 at 22:53
How about $[0,1]cupxinmathbb R:text the axiom of choice is false$?
– bof
Aug 9 at 22:53
add a comment |Â
1 Answer
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oldest
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up vote
5
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accepted
As mentioned in the comments, you can easily get lots of examples by reverse-engineering from the proof that Tychonoff's theorem implies AC. There are more natural examples, though.
In particular, for instance, $X=0,1^mathcalP(mathbbN)$ (with the product topology) cannot be proven to be compact in ZF. To prove this consider for each $ninmathbbN$ the function $F_n:mathcalP(mathbbN)to0,1$ defined by $F_n(S)=1$ if $nin S$ and $F_n(S)=0$ if $nnotin S$. Then it is easy to verify that $A=F_n:ninmathbbN$ is an infinite discrete subset of $X$. If $X$ is compact, then $A$ has an accumulation point. But an accumulation point of $A$ can be shown to be the same thing as the characteristic function of a nonprincipal ultrafilter on $mathbbN$. So, in a model of ZF where no such nonprincipal ultrafilters exist, $A$ will have no accumulation points, and $X$ will not be compact.
It follows that if $Y$ is any $T_1$ space with more than one point and $I$ is any set with cardinality at least $mathfrakc$, then $Y^I$ cannot be proven to be compact in ZF (since $X$ embeds in it as a closed subspace).
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
As mentioned in the comments, you can easily get lots of examples by reverse-engineering from the proof that Tychonoff's theorem implies AC. There are more natural examples, though.
In particular, for instance, $X=0,1^mathcalP(mathbbN)$ (with the product topology) cannot be proven to be compact in ZF. To prove this consider for each $ninmathbbN$ the function $F_n:mathcalP(mathbbN)to0,1$ defined by $F_n(S)=1$ if $nin S$ and $F_n(S)=0$ if $nnotin S$. Then it is easy to verify that $A=F_n:ninmathbbN$ is an infinite discrete subset of $X$. If $X$ is compact, then $A$ has an accumulation point. But an accumulation point of $A$ can be shown to be the same thing as the characteristic function of a nonprincipal ultrafilter on $mathbbN$. So, in a model of ZF where no such nonprincipal ultrafilters exist, $A$ will have no accumulation points, and $X$ will not be compact.
It follows that if $Y$ is any $T_1$ space with more than one point and $I$ is any set with cardinality at least $mathfrakc$, then $Y^I$ cannot be proven to be compact in ZF (since $X$ embeds in it as a closed subspace).
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
add a comment |Â
up vote
5
down vote
accepted
As mentioned in the comments, you can easily get lots of examples by reverse-engineering from the proof that Tychonoff's theorem implies AC. There are more natural examples, though.
In particular, for instance, $X=0,1^mathcalP(mathbbN)$ (with the product topology) cannot be proven to be compact in ZF. To prove this consider for each $ninmathbbN$ the function $F_n:mathcalP(mathbbN)to0,1$ defined by $F_n(S)=1$ if $nin S$ and $F_n(S)=0$ if $nnotin S$. Then it is easy to verify that $A=F_n:ninmathbbN$ is an infinite discrete subset of $X$. If $X$ is compact, then $A$ has an accumulation point. But an accumulation point of $A$ can be shown to be the same thing as the characteristic function of a nonprincipal ultrafilter on $mathbbN$. So, in a model of ZF where no such nonprincipal ultrafilters exist, $A$ will have no accumulation points, and $X$ will not be compact.
It follows that if $Y$ is any $T_1$ space with more than one point and $I$ is any set with cardinality at least $mathfrakc$, then $Y^I$ cannot be proven to be compact in ZF (since $X$ embeds in it as a closed subspace).
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
As mentioned in the comments, you can easily get lots of examples by reverse-engineering from the proof that Tychonoff's theorem implies AC. There are more natural examples, though.
In particular, for instance, $X=0,1^mathcalP(mathbbN)$ (with the product topology) cannot be proven to be compact in ZF. To prove this consider for each $ninmathbbN$ the function $F_n:mathcalP(mathbbN)to0,1$ defined by $F_n(S)=1$ if $nin S$ and $F_n(S)=0$ if $nnotin S$. Then it is easy to verify that $A=F_n:ninmathbbN$ is an infinite discrete subset of $X$. If $X$ is compact, then $A$ has an accumulation point. But an accumulation point of $A$ can be shown to be the same thing as the characteristic function of a nonprincipal ultrafilter on $mathbbN$. So, in a model of ZF where no such nonprincipal ultrafilters exist, $A$ will have no accumulation points, and $X$ will not be compact.
It follows that if $Y$ is any $T_1$ space with more than one point and $I$ is any set with cardinality at least $mathfrakc$, then $Y^I$ cannot be proven to be compact in ZF (since $X$ embeds in it as a closed subspace).
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
As mentioned in the comments, you can easily get lots of examples by reverse-engineering from the proof that Tychonoff's theorem implies AC. There are more natural examples, though.
In particular, for instance, $X=0,1^mathcalP(mathbbN)$ (with the product topology) cannot be proven to be compact in ZF. To prove this consider for each $ninmathbbN$ the function $F_n:mathcalP(mathbbN)to0,1$ defined by $F_n(S)=1$ if $nin S$ and $F_n(S)=0$ if $nnotin S$. Then it is easy to verify that $A=F_n:ninmathbbN$ is an infinite discrete subset of $X$. If $X$ is compact, then $A$ has an accumulation point. But an accumulation point of $A$ can be shown to be the same thing as the characteristic function of a nonprincipal ultrafilter on $mathbbN$. So, in a model of ZF where no such nonprincipal ultrafilters exist, $A$ will have no accumulation points, and $X$ will not be compact.
It follows that if $Y$ is any $T_1$ space with more than one point and $I$ is any set with cardinality at least $mathfrakc$, then $Y^I$ cannot be proven to be compact in ZF (since $X$ embeds in it as a closed subspace).
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
answered Aug 9 at 22:36
Eric Wofsey
163k12190301
163k12190301
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
add a comment |Â
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
See also the last exercise to Section 4.8 in Herrlich's Axiom of Choice book.
– Asaf Karagila♦
Aug 10 at 8:20
add a comment |Â
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3
A common proof that Tychonoff's theorem is equivalent to AC takes an indexed family of sets $ X_i mid i in I $ and constructs a space which is compact if and only if $prod_iin I X_i ne emptyset$.
– Daniel Schepler
Aug 9 at 20:55
3
So for example, if you take the family $ x + mathbbQ mid x in mathbbR $ used in constructing a nonmeasurable set, then that won't have a "choice function" in a model of ZF where every subset of $mathbbR$ is measurable...
– Daniel Schepler
Aug 9 at 20:57
3
As @Daniel suggests, this amounts to two parts: (1) analysis of the proof that Tychonoff's theorem implies choice; (2) finding a family of sets which does not provably admit a choice function. The first one is "easy" (or rather, stratightforward). The second part requires some more knowledge of set theory. But here's a key point: (a) $X$ can be well-ordered if and only if $mathcal P(X)setminusvarnothing$ admits a choice function; and (b) $Bbb R$ is not provably well-orderable in ZF.
– Asaf Karagila♦
Aug 9 at 21:03
1
How about $[0,1]cupxinmathbb R:text the axiom of choice is false$?
– bof
Aug 9 at 22:53