When Can You State that a Given Class is a Set?












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Just wondering if the following is of any interest (I am an amateur in these areas, so this might be so much malarkey).



Let $mathcal X$ be a class in ZFC set theory. Let $mathbb Phi:mathcal X to mathcal X $ be a defined bijective correspondence with the property that for every $X in mathcal X$, $mathbb Phi(X) ne X$. Then $mathcal X$ is a set.




Question 1: Can the above statement even be formulated in ZFC?



Question 2: If if does make sense, is it true or could it be used in
an axiomatc framework?







set-theory axioms







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Incidentally, by far the simplest criterion for set-hood is: a class $mathcal{X}$ is a set iff it contains only elements of bounded rank in the cumulative hierarchy, that is, iff there is some ordinal $alpha$ such that every element of $mathcal{X}$ is in $V_alpha$. One direction is obvious by separation, and the other direction follows since $(1)$ the class of ranks of elements of a set is always a set and $(2)$ no unbounded class of ordinals is a set (take the union).
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:47








  • 1




    $begingroup$
    What's nice about this is that it establishes the class $Ord$ as the minimal proper class, in a precise sense. If $mathcal{X}$ is a proper class, then we can define a surjection from $mathcal{X}$ to $Ord$: send $ainmathcal{X}$ to the Mostowski collapse of the set of ranks of elements of $mathcal{X}$ less than the rank of $a$ itself. Meanwhile, there are models of ZFC in which there is no definable (even with parameters) surjection from $Ord$ to $V$, so this is a nontrivial fact. On the other hand, the ideal result - that $Ord$ in fact injects into any proper class - doesn't always hold.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:50








  • 1




    $begingroup$
    It's also worth noting that in ZF alone there may be sets to which $Ord$ does not definably surject - these will be exactly the non-well-orderable sets (if $f:Ordrightarrow x$ is onto, for $a,bin x$ we can set $atriangleleft b$ if $min(f^{-1}(a))<min(f^{-1}(b))$; note that this uses the fact that $Ord$ is well-ordered). So the axiom of choice is equivalent to the existence of a surjection from $Ord$ to every set. Okay fine this isn't expressible in set theory, but an appropriate version of it is, namely "For every set $x$ there is a set $a$ of ordinals and a surjection from $a$ to $x$."
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:54












  • $begingroup$
    @noah Funny that the simplest criterion doesn't sound that weird! It seem intuitive, and in these matters you can expect some 'way out' arguments.
    $endgroup$
    – CopyPasteIt
    Dec 24 '18 at 19:55






  • 1




    $begingroup$
    Yes, one of the nice features of ZFC (or indeed ZF, or indeed Z+[a bit more]) is how much it simplifies the set/class issue. Telling whether a class is a set is much harder in, say, Quine's set theory NF (or its variants), but in ZF-style set theory everything boils down to just rank.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:57
















0












$begingroup$


Just wondering if the following is of any interest (I am an amateur in these areas, so this might be so much malarkey).



Let $mathcal X$ be a class in ZFC set theory. Let $mathbb Phi:mathcal X to mathcal X $ be a defined bijective correspondence with the property that for every $X in mathcal X$, $mathbb Phi(X) ne X$. Then $mathcal X$ is a set.




Question 1: Can the above statement even be formulated in ZFC?



Question 2: If if does make sense, is it true or could it be used in
an axiomatc framework?







set-theory axioms







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Incidentally, by far the simplest criterion for set-hood is: a class $mathcal{X}$ is a set iff it contains only elements of bounded rank in the cumulative hierarchy, that is, iff there is some ordinal $alpha$ such that every element of $mathcal{X}$ is in $V_alpha$. One direction is obvious by separation, and the other direction follows since $(1)$ the class of ranks of elements of a set is always a set and $(2)$ no unbounded class of ordinals is a set (take the union).
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:47








  • 1




    $begingroup$
    What's nice about this is that it establishes the class $Ord$ as the minimal proper class, in a precise sense. If $mathcal{X}$ is a proper class, then we can define a surjection from $mathcal{X}$ to $Ord$: send $ainmathcal{X}$ to the Mostowski collapse of the set of ranks of elements of $mathcal{X}$ less than the rank of $a$ itself. Meanwhile, there are models of ZFC in which there is no definable (even with parameters) surjection from $Ord$ to $V$, so this is a nontrivial fact. On the other hand, the ideal result - that $Ord$ in fact injects into any proper class - doesn't always hold.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:50








  • 1




    $begingroup$
    It's also worth noting that in ZF alone there may be sets to which $Ord$ does not definably surject - these will be exactly the non-well-orderable sets (if $f:Ordrightarrow x$ is onto, for $a,bin x$ we can set $atriangleleft b$ if $min(f^{-1}(a))<min(f^{-1}(b))$; note that this uses the fact that $Ord$ is well-ordered). So the axiom of choice is equivalent to the existence of a surjection from $Ord$ to every set. Okay fine this isn't expressible in set theory, but an appropriate version of it is, namely "For every set $x$ there is a set $a$ of ordinals and a surjection from $a$ to $x$."
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:54












  • $begingroup$
    @noah Funny that the simplest criterion doesn't sound that weird! It seem intuitive, and in these matters you can expect some 'way out' arguments.
    $endgroup$
    – CopyPasteIt
    Dec 24 '18 at 19:55






  • 1




    $begingroup$
    Yes, one of the nice features of ZFC (or indeed ZF, or indeed Z+[a bit more]) is how much it simplifies the set/class issue. Telling whether a class is a set is much harder in, say, Quine's set theory NF (or its variants), but in ZF-style set theory everything boils down to just rank.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:57














0












0








0





$begingroup$


Just wondering if the following is of any interest (I am an amateur in these areas, so this might be so much malarkey).



Let $mathcal X$ be a class in ZFC set theory. Let $mathbb Phi:mathcal X to mathcal X $ be a defined bijective correspondence with the property that for every $X in mathcal X$, $mathbb Phi(X) ne X$. Then $mathcal X$ is a set.




Question 1: Can the above statement even be formulated in ZFC?



Question 2: If if does make sense, is it true or could it be used in
an axiomatc framework?







set-theory axioms







share|cite|improve this question









$endgroup$




Just wondering if the following is of any interest (I am an amateur in these areas, so this might be so much malarkey).



Let $mathcal X$ be a class in ZFC set theory. Let $mathbb Phi:mathcal X to mathcal X $ be a defined bijective correspondence with the property that for every $X in mathcal X$, $mathbb Phi(X) ne X$. Then $mathcal X$ is a set.




Question 1: Can the above statement even be formulated in ZFC?



Question 2: If if does make sense, is it true or could it be used in
an axiomatc framework?







set-theory axioms




set-theory axioms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 18:51









CopyPasteItCopyPasteIt

4,4321828




4,4321828








  • 1




    $begingroup$
    Incidentally, by far the simplest criterion for set-hood is: a class $mathcal{X}$ is a set iff it contains only elements of bounded rank in the cumulative hierarchy, that is, iff there is some ordinal $alpha$ such that every element of $mathcal{X}$ is in $V_alpha$. One direction is obvious by separation, and the other direction follows since $(1)$ the class of ranks of elements of a set is always a set and $(2)$ no unbounded class of ordinals is a set (take the union).
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:47








  • 1




    $begingroup$
    What's nice about this is that it establishes the class $Ord$ as the minimal proper class, in a precise sense. If $mathcal{X}$ is a proper class, then we can define a surjection from $mathcal{X}$ to $Ord$: send $ainmathcal{X}$ to the Mostowski collapse of the set of ranks of elements of $mathcal{X}$ less than the rank of $a$ itself. Meanwhile, there are models of ZFC in which there is no definable (even with parameters) surjection from $Ord$ to $V$, so this is a nontrivial fact. On the other hand, the ideal result - that $Ord$ in fact injects into any proper class - doesn't always hold.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:50








  • 1




    $begingroup$
    It's also worth noting that in ZF alone there may be sets to which $Ord$ does not definably surject - these will be exactly the non-well-orderable sets (if $f:Ordrightarrow x$ is onto, for $a,bin x$ we can set $atriangleleft b$ if $min(f^{-1}(a))<min(f^{-1}(b))$; note that this uses the fact that $Ord$ is well-ordered). So the axiom of choice is equivalent to the existence of a surjection from $Ord$ to every set. Okay fine this isn't expressible in set theory, but an appropriate version of it is, namely "For every set $x$ there is a set $a$ of ordinals and a surjection from $a$ to $x$."
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:54












  • $begingroup$
    @noah Funny that the simplest criterion doesn't sound that weird! It seem intuitive, and in these matters you can expect some 'way out' arguments.
    $endgroup$
    – CopyPasteIt
    Dec 24 '18 at 19:55






  • 1




    $begingroup$
    Yes, one of the nice features of ZFC (or indeed ZF, or indeed Z+[a bit more]) is how much it simplifies the set/class issue. Telling whether a class is a set is much harder in, say, Quine's set theory NF (or its variants), but in ZF-style set theory everything boils down to just rank.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:57














  • 1




    $begingroup$
    Incidentally, by far the simplest criterion for set-hood is: a class $mathcal{X}$ is a set iff it contains only elements of bounded rank in the cumulative hierarchy, that is, iff there is some ordinal $alpha$ such that every element of $mathcal{X}$ is in $V_alpha$. One direction is obvious by separation, and the other direction follows since $(1)$ the class of ranks of elements of a set is always a set and $(2)$ no unbounded class of ordinals is a set (take the union).
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:47








  • 1




    $begingroup$
    What's nice about this is that it establishes the class $Ord$ as the minimal proper class, in a precise sense. If $mathcal{X}$ is a proper class, then we can define a surjection from $mathcal{X}$ to $Ord$: send $ainmathcal{X}$ to the Mostowski collapse of the set of ranks of elements of $mathcal{X}$ less than the rank of $a$ itself. Meanwhile, there are models of ZFC in which there is no definable (even with parameters) surjection from $Ord$ to $V$, so this is a nontrivial fact. On the other hand, the ideal result - that $Ord$ in fact injects into any proper class - doesn't always hold.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:50








  • 1




    $begingroup$
    It's also worth noting that in ZF alone there may be sets to which $Ord$ does not definably surject - these will be exactly the non-well-orderable sets (if $f:Ordrightarrow x$ is onto, for $a,bin x$ we can set $atriangleleft b$ if $min(f^{-1}(a))<min(f^{-1}(b))$; note that this uses the fact that $Ord$ is well-ordered). So the axiom of choice is equivalent to the existence of a surjection from $Ord$ to every set. Okay fine this isn't expressible in set theory, but an appropriate version of it is, namely "For every set $x$ there is a set $a$ of ordinals and a surjection from $a$ to $x$."
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:54












  • $begingroup$
    @noah Funny that the simplest criterion doesn't sound that weird! It seem intuitive, and in these matters you can expect some 'way out' arguments.
    $endgroup$
    – CopyPasteIt
    Dec 24 '18 at 19:55






  • 1




    $begingroup$
    Yes, one of the nice features of ZFC (or indeed ZF, or indeed Z+[a bit more]) is how much it simplifies the set/class issue. Telling whether a class is a set is much harder in, say, Quine's set theory NF (or its variants), but in ZF-style set theory everything boils down to just rank.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 19:57








1




1




$begingroup$
Incidentally, by far the simplest criterion for set-hood is: a class $mathcal{X}$ is a set iff it contains only elements of bounded rank in the cumulative hierarchy, that is, iff there is some ordinal $alpha$ such that every element of $mathcal{X}$ is in $V_alpha$. One direction is obvious by separation, and the other direction follows since $(1)$ the class of ranks of elements of a set is always a set and $(2)$ no unbounded class of ordinals is a set (take the union).
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:47






$begingroup$
Incidentally, by far the simplest criterion for set-hood is: a class $mathcal{X}$ is a set iff it contains only elements of bounded rank in the cumulative hierarchy, that is, iff there is some ordinal $alpha$ such that every element of $mathcal{X}$ is in $V_alpha$. One direction is obvious by separation, and the other direction follows since $(1)$ the class of ranks of elements of a set is always a set and $(2)$ no unbounded class of ordinals is a set (take the union).
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:47






1




1




$begingroup$
What's nice about this is that it establishes the class $Ord$ as the minimal proper class, in a precise sense. If $mathcal{X}$ is a proper class, then we can define a surjection from $mathcal{X}$ to $Ord$: send $ainmathcal{X}$ to the Mostowski collapse of the set of ranks of elements of $mathcal{X}$ less than the rank of $a$ itself. Meanwhile, there are models of ZFC in which there is no definable (even with parameters) surjection from $Ord$ to $V$, so this is a nontrivial fact. On the other hand, the ideal result - that $Ord$ in fact injects into any proper class - doesn't always hold.
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:50






$begingroup$
What's nice about this is that it establishes the class $Ord$ as the minimal proper class, in a precise sense. If $mathcal{X}$ is a proper class, then we can define a surjection from $mathcal{X}$ to $Ord$: send $ainmathcal{X}$ to the Mostowski collapse of the set of ranks of elements of $mathcal{X}$ less than the rank of $a$ itself. Meanwhile, there are models of ZFC in which there is no definable (even with parameters) surjection from $Ord$ to $V$, so this is a nontrivial fact. On the other hand, the ideal result - that $Ord$ in fact injects into any proper class - doesn't always hold.
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:50






1




1




$begingroup$
It's also worth noting that in ZF alone there may be sets to which $Ord$ does not definably surject - these will be exactly the non-well-orderable sets (if $f:Ordrightarrow x$ is onto, for $a,bin x$ we can set $atriangleleft b$ if $min(f^{-1}(a))<min(f^{-1}(b))$; note that this uses the fact that $Ord$ is well-ordered). So the axiom of choice is equivalent to the existence of a surjection from $Ord$ to every set. Okay fine this isn't expressible in set theory, but an appropriate version of it is, namely "For every set $x$ there is a set $a$ of ordinals and a surjection from $a$ to $x$."
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:54






$begingroup$
It's also worth noting that in ZF alone there may be sets to which $Ord$ does not definably surject - these will be exactly the non-well-orderable sets (if $f:Ordrightarrow x$ is onto, for $a,bin x$ we can set $atriangleleft b$ if $min(f^{-1}(a))<min(f^{-1}(b))$; note that this uses the fact that $Ord$ is well-ordered). So the axiom of choice is equivalent to the existence of a surjection from $Ord$ to every set. Okay fine this isn't expressible in set theory, but an appropriate version of it is, namely "For every set $x$ there is a set $a$ of ordinals and a surjection from $a$ to $x$."
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:54














$begingroup$
@noah Funny that the simplest criterion doesn't sound that weird! It seem intuitive, and in these matters you can expect some 'way out' arguments.
$endgroup$
– CopyPasteIt
Dec 24 '18 at 19:55




$begingroup$
@noah Funny that the simplest criterion doesn't sound that weird! It seem intuitive, and in these matters you can expect some 'way out' arguments.
$endgroup$
– CopyPasteIt
Dec 24 '18 at 19:55




1




1




$begingroup$
Yes, one of the nice features of ZFC (or indeed ZF, or indeed Z+[a bit more]) is how much it simplifies the set/class issue. Telling whether a class is a set is much harder in, say, Quine's set theory NF (or its variants), but in ZF-style set theory everything boils down to just rank.
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:57




$begingroup$
Yes, one of the nice features of ZFC (or indeed ZF, or indeed Z+[a bit more]) is how much it simplifies the set/class issue. Telling whether a class is a set is much harder in, say, Quine's set theory NF (or its variants), but in ZF-style set theory everything boils down to just rank.
$endgroup$
– Noah Schweber
Dec 24 '18 at 19:57










3 Answers
3






active

oldest

votes


















3












$begingroup$

First, note that the principle you're considering is false: consider the class $mathcal{X}$ of ordinals which are either limits or successors of limits. Then there is an obvious self-bijection of $mathcal{X}$ with no fixed points - namely, for $alphainmathcal{X}$ we send $alpha$ to $alpha+1$ if $alpha$ is a limit and we send $alpha$ to the predecessor of $alpha$ if $alpha$ is not a limit - but $mathcal{X}$ is a proper class since no unbounded class of ordinals is a set.



Another example which may be easier to think about at first: let $mathcal{A}$ be the class of all sets of the form $xtimes{0}$ and let $mathcal{B}$ be the class of all sets of the form $xtimes{1}$. Their union $mathcal{C}:=mathcal{A}cupmathcal{B}$ is clearly a proper class, but by swapping $0$ and $1$ we get a self-bijection of $mathcal{C}$ with no fixed points.




As to expressing it appropriately, there is a serious problem: you can't quantify over class functions in set theory (that is, you can't say "for some $Phi$"). What you can do is express the principle as a scheme: for each formula $Phi$ defining a class function and each formula $chi$ defining a class, we can write a sentence which says "$Phi$ defines a fixed-point-free self-bijection of the class defined by $chi$." However, this is enough to express what you want (ignoring its falsity for the moment), so it's not a huge problem.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
    $endgroup$
    – DanielWainfleet
    Dec 24 '18 at 23:31










  • $begingroup$
    So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
    $endgroup$
    – DanielWainfleet
    Dec 25 '18 at 0:47










  • $begingroup$
    @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
    $endgroup$
    – Noah Schweber
    Dec 25 '18 at 4:37












  • $begingroup$
    My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
    $endgroup$
    – DanielWainfleet
    Dec 26 '18 at 0:04






  • 1




    $begingroup$
    @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
    $endgroup$
    – Noah Schweber
    Dec 26 '18 at 1:04





















3












$begingroup$

The claim is that if $mathcal X$ can be bijected to itself with no fixed points, $mathcal X$ is a set. The closest we can get to formulating this in ZFC is to introduce a unary predicate $chi(x)$ meaning $xinmathcal X$ and a binary predicate $phi(x,,y)$ meaning $Phi(x)=y$, and write (I hope I don't mess up the details!)$$(forall x(chi(x)toforall y,,z(phi(x,,y)landphi(x,,z)to y=zne x))land exists y (chi(y)land phi(y,,x)))to(exists zforall x(chi(x)leftrightarrow xin z)).$$(Since ZFC's objects are all sets, "this is a set" means "there's a set with the members this has".) Note each choice of the predicates $chi,,phi$ has its own version of this statement; we can't quantify over them in one statement, because the logic is first-order.



So that's question 1 answered. The conjecture isn't in general true, though. For example, let's biject the class of ordinals to itself according to the following rule: write an ordinal $alpha$ as $gamma+n$ with $n$ finite and $gamma$ a limit ordinal; then $Phi(gamma+n)=gamma+n+(-1)^n$. This doesn't have fixed points, but the ordinals don't form a set.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I don't know if it would be consistent with ZFC, but you can to the best of my knowledge formulate said condition in the language of sets (or in any other language), but it would require infinitely many axioms.



    A class of sets is basically a family of sets given by a formula (possibly with parameters). You basically want to consider the following axiom for any pair or two formulas (possibly involving parameters) $phi(x)$ and $psi(x,y,bar{z})$, the first one having $1$ free variable and the second one $2+n$.



    For every $bar{a}$ (array of $n$ parameters), if $psi(x,y,bar{a})$ defines a bijection $mathcal{X}rightarrowmathcal{X}$ with no fixed points, where $mathcal{X}$ is the class defined by $phi(x)$, then $mathcal{X}$ is a set.



    You can do this for every pair of formulas to formulate your condition.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      First, note that the principle you're considering is false: consider the class $mathcal{X}$ of ordinals which are either limits or successors of limits. Then there is an obvious self-bijection of $mathcal{X}$ with no fixed points - namely, for $alphainmathcal{X}$ we send $alpha$ to $alpha+1$ if $alpha$ is a limit and we send $alpha$ to the predecessor of $alpha$ if $alpha$ is not a limit - but $mathcal{X}$ is a proper class since no unbounded class of ordinals is a set.



      Another example which may be easier to think about at first: let $mathcal{A}$ be the class of all sets of the form $xtimes{0}$ and let $mathcal{B}$ be the class of all sets of the form $xtimes{1}$. Their union $mathcal{C}:=mathcal{A}cupmathcal{B}$ is clearly a proper class, but by swapping $0$ and $1$ we get a self-bijection of $mathcal{C}$ with no fixed points.




      As to expressing it appropriately, there is a serious problem: you can't quantify over class functions in set theory (that is, you can't say "for some $Phi$"). What you can do is express the principle as a scheme: for each formula $Phi$ defining a class function and each formula $chi$ defining a class, we can write a sentence which says "$Phi$ defines a fixed-point-free self-bijection of the class defined by $chi$." However, this is enough to express what you want (ignoring its falsity for the moment), so it's not a huge problem.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
        $endgroup$
        – DanielWainfleet
        Dec 24 '18 at 23:31










      • $begingroup$
        So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
        $endgroup$
        – DanielWainfleet
        Dec 25 '18 at 0:47










      • $begingroup$
        @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
        $endgroup$
        – Noah Schweber
        Dec 25 '18 at 4:37












      • $begingroup$
        My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
        $endgroup$
        – DanielWainfleet
        Dec 26 '18 at 0:04






      • 1




        $begingroup$
        @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
        $endgroup$
        – Noah Schweber
        Dec 26 '18 at 1:04


















      3












      $begingroup$

      First, note that the principle you're considering is false: consider the class $mathcal{X}$ of ordinals which are either limits or successors of limits. Then there is an obvious self-bijection of $mathcal{X}$ with no fixed points - namely, for $alphainmathcal{X}$ we send $alpha$ to $alpha+1$ if $alpha$ is a limit and we send $alpha$ to the predecessor of $alpha$ if $alpha$ is not a limit - but $mathcal{X}$ is a proper class since no unbounded class of ordinals is a set.



      Another example which may be easier to think about at first: let $mathcal{A}$ be the class of all sets of the form $xtimes{0}$ and let $mathcal{B}$ be the class of all sets of the form $xtimes{1}$. Their union $mathcal{C}:=mathcal{A}cupmathcal{B}$ is clearly a proper class, but by swapping $0$ and $1$ we get a self-bijection of $mathcal{C}$ with no fixed points.




      As to expressing it appropriately, there is a serious problem: you can't quantify over class functions in set theory (that is, you can't say "for some $Phi$"). What you can do is express the principle as a scheme: for each formula $Phi$ defining a class function and each formula $chi$ defining a class, we can write a sentence which says "$Phi$ defines a fixed-point-free self-bijection of the class defined by $chi$." However, this is enough to express what you want (ignoring its falsity for the moment), so it's not a huge problem.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
        $endgroup$
        – DanielWainfleet
        Dec 24 '18 at 23:31










      • $begingroup$
        So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
        $endgroup$
        – DanielWainfleet
        Dec 25 '18 at 0:47










      • $begingroup$
        @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
        $endgroup$
        – Noah Schweber
        Dec 25 '18 at 4:37












      • $begingroup$
        My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
        $endgroup$
        – DanielWainfleet
        Dec 26 '18 at 0:04






      • 1




        $begingroup$
        @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
        $endgroup$
        – Noah Schweber
        Dec 26 '18 at 1:04
















      3












      3








      3





      $begingroup$

      First, note that the principle you're considering is false: consider the class $mathcal{X}$ of ordinals which are either limits or successors of limits. Then there is an obvious self-bijection of $mathcal{X}$ with no fixed points - namely, for $alphainmathcal{X}$ we send $alpha$ to $alpha+1$ if $alpha$ is a limit and we send $alpha$ to the predecessor of $alpha$ if $alpha$ is not a limit - but $mathcal{X}$ is a proper class since no unbounded class of ordinals is a set.



      Another example which may be easier to think about at first: let $mathcal{A}$ be the class of all sets of the form $xtimes{0}$ and let $mathcal{B}$ be the class of all sets of the form $xtimes{1}$. Their union $mathcal{C}:=mathcal{A}cupmathcal{B}$ is clearly a proper class, but by swapping $0$ and $1$ we get a self-bijection of $mathcal{C}$ with no fixed points.




      As to expressing it appropriately, there is a serious problem: you can't quantify over class functions in set theory (that is, you can't say "for some $Phi$"). What you can do is express the principle as a scheme: for each formula $Phi$ defining a class function and each formula $chi$ defining a class, we can write a sentence which says "$Phi$ defines a fixed-point-free self-bijection of the class defined by $chi$." However, this is enough to express what you want (ignoring its falsity for the moment), so it's not a huge problem.






      share|cite|improve this answer











      $endgroup$



      First, note that the principle you're considering is false: consider the class $mathcal{X}$ of ordinals which are either limits or successors of limits. Then there is an obvious self-bijection of $mathcal{X}$ with no fixed points - namely, for $alphainmathcal{X}$ we send $alpha$ to $alpha+1$ if $alpha$ is a limit and we send $alpha$ to the predecessor of $alpha$ if $alpha$ is not a limit - but $mathcal{X}$ is a proper class since no unbounded class of ordinals is a set.



      Another example which may be easier to think about at first: let $mathcal{A}$ be the class of all sets of the form $xtimes{0}$ and let $mathcal{B}$ be the class of all sets of the form $xtimes{1}$. Their union $mathcal{C}:=mathcal{A}cupmathcal{B}$ is clearly a proper class, but by swapping $0$ and $1$ we get a self-bijection of $mathcal{C}$ with no fixed points.




      As to expressing it appropriately, there is a serious problem: you can't quantify over class functions in set theory (that is, you can't say "for some $Phi$"). What you can do is express the principle as a scheme: for each formula $Phi$ defining a class function and each formula $chi$ defining a class, we can write a sentence which says "$Phi$ defines a fixed-point-free self-bijection of the class defined by $chi$." However, this is enough to express what you want (ignoring its falsity for the moment), so it's not a huge problem.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 24 '18 at 19:14

























      answered Dec 24 '18 at 19:09









      Noah SchweberNoah Schweber

      129k10154295




      129k10154295












      • $begingroup$
        I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
        $endgroup$
        – DanielWainfleet
        Dec 24 '18 at 23:31










      • $begingroup$
        So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
        $endgroup$
        – DanielWainfleet
        Dec 25 '18 at 0:47










      • $begingroup$
        @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
        $endgroup$
        – Noah Schweber
        Dec 25 '18 at 4:37












      • $begingroup$
        My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
        $endgroup$
        – DanielWainfleet
        Dec 26 '18 at 0:04






      • 1




        $begingroup$
        @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
        $endgroup$
        – Noah Schweber
        Dec 26 '18 at 1:04




















      • $begingroup$
        I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
        $endgroup$
        – DanielWainfleet
        Dec 24 '18 at 23:31










      • $begingroup$
        So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
        $endgroup$
        – DanielWainfleet
        Dec 25 '18 at 0:47










      • $begingroup$
        @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
        $endgroup$
        – Noah Schweber
        Dec 25 '18 at 4:37












      • $begingroup$
        My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
        $endgroup$
        – DanielWainfleet
        Dec 26 '18 at 0:04






      • 1




        $begingroup$
        @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
        $endgroup$
        – Noah Schweber
        Dec 26 '18 at 1:04


















      $begingroup$
      I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
      $endgroup$
      – DanielWainfleet
      Dec 24 '18 at 23:31




      $begingroup$
      I was going to give a more complicated counter-example with X=L (Godel's class) but yours is just obvious.
      $endgroup$
      – DanielWainfleet
      Dec 24 '18 at 23:31












      $begingroup$
      So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
      $endgroup$
      – DanielWainfleet
      Dec 25 '18 at 0:47




      $begingroup$
      So in ZFC can we give an explicit formula for a bijection $f:Vto V$ with no fixed point?
      $endgroup$
      – DanielWainfleet
      Dec 25 '18 at 0:47












      $begingroup$
      @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
      $endgroup$
      – Noah Schweber
      Dec 25 '18 at 4:37






      $begingroup$
      @DanielWainfleet Sorry, my previous comment was silly (I just got non-identity, not no-fixed-points). But the answer to your question is definitely yes. Here's one way (being a bit sloppy): say that a set $xnotin V_omega$ has depth $n$ if $x$ can be written as ${{...{a}...}}$ for some set $a$ with more than one element, using $n$ pairs of curly braces. E.g. ${{omega}}$ has depth $2$. For $xnotin V_omega$ let $hat{x}={x}$ if $x$ has even depth and ${hat{x}}=x$ if $x$ has odd depth. The map $xmapstohat{x}$ is a derangement of $Vsetminus V_omega$; now complete it.
      $endgroup$
      – Noah Schweber
      Dec 25 '18 at 4:37














      $begingroup$
      My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
      $endgroup$
      – DanielWainfleet
      Dec 26 '18 at 0:04




      $begingroup$
      My idea for a bijection $f:Lto L,$ with no fixed points,employs the "canonical" well-ordering of $L,$ but since this well-ordering can be used to define a bijection from $L$ to $On$ ( the Ordinals ) it is simpler, as you have done, to just take a bijection from $On$ to $On$ with no fixed points..... I don't understand your last comment. What is the depth of $omega_1$?
      $endgroup$
      – DanielWainfleet
      Dec 26 '18 at 0:04




      1




      1




      $begingroup$
      @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
      $endgroup$
      – Noah Schweber
      Dec 26 '18 at 1:04






      $begingroup$
      @DanielWainfleet $omega_1$ has depth $0$, so it would get swapped with ${omega_1}$, which has depth $1$. ${{omega_1}}$ would go to ${{{omega_1}}}$, and so on.
      $endgroup$
      – Noah Schweber
      Dec 26 '18 at 1:04













      3












      $begingroup$

      The claim is that if $mathcal X$ can be bijected to itself with no fixed points, $mathcal X$ is a set. The closest we can get to formulating this in ZFC is to introduce a unary predicate $chi(x)$ meaning $xinmathcal X$ and a binary predicate $phi(x,,y)$ meaning $Phi(x)=y$, and write (I hope I don't mess up the details!)$$(forall x(chi(x)toforall y,,z(phi(x,,y)landphi(x,,z)to y=zne x))land exists y (chi(y)land phi(y,,x)))to(exists zforall x(chi(x)leftrightarrow xin z)).$$(Since ZFC's objects are all sets, "this is a set" means "there's a set with the members this has".) Note each choice of the predicates $chi,,phi$ has its own version of this statement; we can't quantify over them in one statement, because the logic is first-order.



      So that's question 1 answered. The conjecture isn't in general true, though. For example, let's biject the class of ordinals to itself according to the following rule: write an ordinal $alpha$ as $gamma+n$ with $n$ finite and $gamma$ a limit ordinal; then $Phi(gamma+n)=gamma+n+(-1)^n$. This doesn't have fixed points, but the ordinals don't form a set.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The claim is that if $mathcal X$ can be bijected to itself with no fixed points, $mathcal X$ is a set. The closest we can get to formulating this in ZFC is to introduce a unary predicate $chi(x)$ meaning $xinmathcal X$ and a binary predicate $phi(x,,y)$ meaning $Phi(x)=y$, and write (I hope I don't mess up the details!)$$(forall x(chi(x)toforall y,,z(phi(x,,y)landphi(x,,z)to y=zne x))land exists y (chi(y)land phi(y,,x)))to(exists zforall x(chi(x)leftrightarrow xin z)).$$(Since ZFC's objects are all sets, "this is a set" means "there's a set with the members this has".) Note each choice of the predicates $chi,,phi$ has its own version of this statement; we can't quantify over them in one statement, because the logic is first-order.



        So that's question 1 answered. The conjecture isn't in general true, though. For example, let's biject the class of ordinals to itself according to the following rule: write an ordinal $alpha$ as $gamma+n$ with $n$ finite and $gamma$ a limit ordinal; then $Phi(gamma+n)=gamma+n+(-1)^n$. This doesn't have fixed points, but the ordinals don't form a set.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The claim is that if $mathcal X$ can be bijected to itself with no fixed points, $mathcal X$ is a set. The closest we can get to formulating this in ZFC is to introduce a unary predicate $chi(x)$ meaning $xinmathcal X$ and a binary predicate $phi(x,,y)$ meaning $Phi(x)=y$, and write (I hope I don't mess up the details!)$$(forall x(chi(x)toforall y,,z(phi(x,,y)landphi(x,,z)to y=zne x))land exists y (chi(y)land phi(y,,x)))to(exists zforall x(chi(x)leftrightarrow xin z)).$$(Since ZFC's objects are all sets, "this is a set" means "there's a set with the members this has".) Note each choice of the predicates $chi,,phi$ has its own version of this statement; we can't quantify over them in one statement, because the logic is first-order.



          So that's question 1 answered. The conjecture isn't in general true, though. For example, let's biject the class of ordinals to itself according to the following rule: write an ordinal $alpha$ as $gamma+n$ with $n$ finite and $gamma$ a limit ordinal; then $Phi(gamma+n)=gamma+n+(-1)^n$. This doesn't have fixed points, but the ordinals don't form a set.






          share|cite|improve this answer









          $endgroup$



          The claim is that if $mathcal X$ can be bijected to itself with no fixed points, $mathcal X$ is a set. The closest we can get to formulating this in ZFC is to introduce a unary predicate $chi(x)$ meaning $xinmathcal X$ and a binary predicate $phi(x,,y)$ meaning $Phi(x)=y$, and write (I hope I don't mess up the details!)$$(forall x(chi(x)toforall y,,z(phi(x,,y)landphi(x,,z)to y=zne x))land exists y (chi(y)land phi(y,,x)))to(exists zforall x(chi(x)leftrightarrow xin z)).$$(Since ZFC's objects are all sets, "this is a set" means "there's a set with the members this has".) Note each choice of the predicates $chi,,phi$ has its own version of this statement; we can't quantify over them in one statement, because the logic is first-order.



          So that's question 1 answered. The conjecture isn't in general true, though. For example, let's biject the class of ordinals to itself according to the following rule: write an ordinal $alpha$ as $gamma+n$ with $n$ finite and $gamma$ a limit ordinal; then $Phi(gamma+n)=gamma+n+(-1)^n$. This doesn't have fixed points, but the ordinals don't form a set.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 19:05









          J.G.J.G.

          34.4k23252




          34.4k23252























              0












              $begingroup$

              I don't know if it would be consistent with ZFC, but you can to the best of my knowledge formulate said condition in the language of sets (or in any other language), but it would require infinitely many axioms.



              A class of sets is basically a family of sets given by a formula (possibly with parameters). You basically want to consider the following axiom for any pair or two formulas (possibly involving parameters) $phi(x)$ and $psi(x,y,bar{z})$, the first one having $1$ free variable and the second one $2+n$.



              For every $bar{a}$ (array of $n$ parameters), if $psi(x,y,bar{a})$ defines a bijection $mathcal{X}rightarrowmathcal{X}$ with no fixed points, where $mathcal{X}$ is the class defined by $phi(x)$, then $mathcal{X}$ is a set.



              You can do this for every pair of formulas to formulate your condition.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I don't know if it would be consistent with ZFC, but you can to the best of my knowledge formulate said condition in the language of sets (or in any other language), but it would require infinitely many axioms.



                A class of sets is basically a family of sets given by a formula (possibly with parameters). You basically want to consider the following axiom for any pair or two formulas (possibly involving parameters) $phi(x)$ and $psi(x,y,bar{z})$, the first one having $1$ free variable and the second one $2+n$.



                For every $bar{a}$ (array of $n$ parameters), if $psi(x,y,bar{a})$ defines a bijection $mathcal{X}rightarrowmathcal{X}$ with no fixed points, where $mathcal{X}$ is the class defined by $phi(x)$, then $mathcal{X}$ is a set.



                You can do this for every pair of formulas to formulate your condition.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I don't know if it would be consistent with ZFC, but you can to the best of my knowledge formulate said condition in the language of sets (or in any other language), but it would require infinitely many axioms.



                  A class of sets is basically a family of sets given by a formula (possibly with parameters). You basically want to consider the following axiom for any pair or two formulas (possibly involving parameters) $phi(x)$ and $psi(x,y,bar{z})$, the first one having $1$ free variable and the second one $2+n$.



                  For every $bar{a}$ (array of $n$ parameters), if $psi(x,y,bar{a})$ defines a bijection $mathcal{X}rightarrowmathcal{X}$ with no fixed points, where $mathcal{X}$ is the class defined by $phi(x)$, then $mathcal{X}$ is a set.



                  You can do this for every pair of formulas to formulate your condition.






                  share|cite|improve this answer









                  $endgroup$



                  I don't know if it would be consistent with ZFC, but you can to the best of my knowledge formulate said condition in the language of sets (or in any other language), but it would require infinitely many axioms.



                  A class of sets is basically a family of sets given by a formula (possibly with parameters). You basically want to consider the following axiom for any pair or two formulas (possibly involving parameters) $phi(x)$ and $psi(x,y,bar{z})$, the first one having $1$ free variable and the second one $2+n$.



                  For every $bar{a}$ (array of $n$ parameters), if $psi(x,y,bar{a})$ defines a bijection $mathcal{X}rightarrowmathcal{X}$ with no fixed points, where $mathcal{X}$ is the class defined by $phi(x)$, then $mathcal{X}$ is a set.



                  You can do this for every pair of formulas to formulate your condition.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 19:17









                  AnguepaAnguepa

                  1,384819




                  1,384819






























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                      Brian Clough

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                      Brian Clough Datos personales Nombre completo Brian Howard Clough Apodo(s) Old Big 'Ead, Cloughie [ 1 ] ​ Nacimiento Middlesbrough, Inglaterra, Reino Unido 21 de marzo de 1935 Nacionalidad(es) Fallecimiento Derby, Reino Unido 20 de septiembre de 2004 (69 años) Altura 1,78 [ 1 ] ​ metros Carrera Deporte Fútbol Debut deportivo 1955 (Como jugador) 1965 (Como entrenador) (Middlesbrough F. C. (Como jugador) Hartlepool United F. C. (Como entrenador) ) Posición Delantero Goles en clubes 269 [ 2 ] ​ Retirada deportiva 1964 (Como jugador) 1993 (Como entrenador) (Sunderland A. F. C. (Como jugador) Nottingham Forest (Como entrenador) ) Carrera internacional Part. 2 [editar datos en Wikidata] Brian Howard Clough , OBE [ 3 ] ​ (Middlesbrough, Inglaterra, Reino Unido, 21 de marzo de 1935 – Derby, Inglaterra, Reino Unido, 20 de septiembre de 2004) fue un jugador y entrenador británico de fútbol. Es conocido por...
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                      Cáceres

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                      Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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