Prove that the topological closure of a set is definable if the set is definable












1












$begingroup$


Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18
















1












$begingroup$


Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18














1












1








1





$begingroup$


Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory







share|cite|improve this question











$endgroup$




Let $L$ be the language $L={<,=,+,-,cdot, 0,1}$, with standard interpretations, and let $mathcal{A}=langlemathbb{R}, <,=,+,-,cdot,0,1rangle$. Let $Ssubseteqmathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$bar{S}:={ainmathbb{R}^n:text{every open ball centered at $a$, contains a point of $S$}},$$ is also definable.



My attempt at a solution:



Clearly, the set $bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $phi^{mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.



The boundary is the set of all points $ainmathbb{R}^n$, so that the following holds:




  • $anotin S$


  • $forallepsilon>0exists sin S$ such that $d(a,s)<epsilon$, where $d$ is the distance function between two points


Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$phi^{mathcal{A}}cup(mathbb{R}^nsetminusphi^{mathcal{A}}cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.






logic first-order-logic predicate-logic model-theory




logic first-order-logic predicate-logic model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 19:57









Asaf Karagila♦

305k33436766




305k33436766










asked Dec 5 '18 at 19:11









quanticboltquanticbolt

769512




769512








  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18














  • 1




    $begingroup$
    What difficulty do you have defining $B$? Which part of it do you not know how to express?
    $endgroup$
    – Eric Wofsey
    Dec 5 '18 at 20:54










  • $begingroup$
    I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:04










  • $begingroup$
    You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
    $endgroup$
    – Rob Arthan
    Dec 5 '18 at 21:12












  • $begingroup$
    So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
    $endgroup$
    – quanticbolt
    Dec 5 '18 at 21:15








  • 1




    $begingroup$
    @quanticbolt What if $epsilon = 0$?
    $endgroup$
    – Alex Kruckman
    Dec 6 '18 at 2:18








1




1




$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54




$begingroup$
What difficulty do you have defining $B$? Which part of it do you not know how to express?
$endgroup$
– Eric Wofsey
Dec 5 '18 at 20:54












$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04




$begingroup$
I think I figured it out. I think the $psi$ which defines $B$ is given by $$psi:=exists x_1...exists x_n(phiwedgeforallepsilon((0<epsilon)to((x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<epsiloncdotepsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $sin S$ (i.e. satisfying $phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon.
$endgroup$
– quanticbolt
Dec 5 '18 at 21:04












$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12






$begingroup$
You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $mathbf{y} in S cup S'$ iff for any $epsilon > 0$, there is a $mathbb{x} in S$, with $d(mathbb{x}, mathbb{y}) < epsilon$. Now translate that into logical notation (the $forall$ should come outside the $exists$, not inside).
$endgroup$
– Rob Arthan
Dec 5 '18 at 21:12














$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15






$begingroup$
So can I say: $$forallepsilon((0<epsilon)wedgeexists x_1...exists x_n(phiwedge(x_1-y_1)cdot(x_1-y_1)+...+(x_n-y_n)cdot(x_n-y_n)<(epsiloncdotepsilon)))$$
$endgroup$
– quanticbolt
Dec 5 '18 at 21:15






1




1




$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18




$begingroup$
@quanticbolt What if $epsilon = 0$?
$endgroup$
– Alex Kruckman
Dec 6 '18 at 2:18










1 Answer
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$begingroup$

As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



$$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






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    $begingroup$

    As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



    $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



    You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



      $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



      You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



        $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



        You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.






        share|cite|improve this answer









        $endgroup$



        As you wrote in the question, the closure $overline{S}$ is definable by $(forall varepsilon > 0) (exists yin S), d(x,y)<varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $varphi(x_1,dots,x_n)$, we have:



        $$forall varepsilon, ((varepsilon > 0) rightarrow exists y_1dots exists y_n, (varphi(y_1,dots,y_n)land ((x_1-y_1)cdot (x_1-y_1) + dots + (x_n-y_n)cdot (x_n-y_n) < varepsilon cdot varepsilon))).$$



        You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 1:42









        Alex KruckmanAlex Kruckman

        27.6k32658




        27.6k32658






























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