Show that $(0,1)$ is a substructure of $(0,1]$ in the language $L={<}$ but is not an elementary...












0












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Show that the ordered set $(0,1)$ is a substructure of $(0,1]$ in the language $L={<}$ but is not an elementary substructure of it in the same language.




Since the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$ , it's also a substructure of it.



Now, $varphi: (exists x)(x=1)$ is satisfied in $(0,1]$ but not in $(0,1)$ so, $(0,1)$ is not an elementary substructure of $(0,1]$.



So, I wonder if saying "the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$" is enough to prove that $(0,1)$ is a substructure of $(0,1]$.






logic first-order-logic







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$endgroup$








  • 2




    $begingroup$
    $exists x (x=1)$ doesn’t work since the sentence needs to only have parameters in the substructure. (Otherwise nothing would be an elementary substructure of anything ever.) Replace with “there is a greatest element.”
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 20:48


















0












$begingroup$



Show that the ordered set $(0,1)$ is a substructure of $(0,1]$ in the language $L={<}$ but is not an elementary substructure of it in the same language.




Since the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$ , it's also a substructure of it.



Now, $varphi: (exists x)(x=1)$ is satisfied in $(0,1]$ but not in $(0,1)$ so, $(0,1)$ is not an elementary substructure of $(0,1]$.



So, I wonder if saying "the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$" is enough to prove that $(0,1)$ is a substructure of $(0,1]$.






logic first-order-logic







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $exists x (x=1)$ doesn’t work since the sentence needs to only have parameters in the substructure. (Otherwise nothing would be an elementary substructure of anything ever.) Replace with “there is a greatest element.”
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 20:48
















0












0








0





$begingroup$



Show that the ordered set $(0,1)$ is a substructure of $(0,1]$ in the language $L={<}$ but is not an elementary substructure of it in the same language.




Since the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$ , it's also a substructure of it.



Now, $varphi: (exists x)(x=1)$ is satisfied in $(0,1]$ but not in $(0,1)$ so, $(0,1)$ is not an elementary substructure of $(0,1]$.



So, I wonder if saying "the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$" is enough to prove that $(0,1)$ is a substructure of $(0,1]$.






logic first-order-logic







share|cite|improve this question









$endgroup$





Show that the ordered set $(0,1)$ is a substructure of $(0,1]$ in the language $L={<}$ but is not an elementary substructure of it in the same language.




Since the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$ , it's also a substructure of it.



Now, $varphi: (exists x)(x=1)$ is satisfied in $(0,1]$ but not in $(0,1)$ so, $(0,1)$ is not an elementary substructure of $(0,1]$.



So, I wonder if saying "the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$" is enough to prove that $(0,1)$ is a substructure of $(0,1]$.






logic first-order-logic




logic first-order-logic






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share|cite|improve this question










asked Dec 24 '18 at 20:29









Leyla AlkanLeyla Alkan

1,5401724




1,5401724








  • 2




    $begingroup$
    $exists x (x=1)$ doesn’t work since the sentence needs to only have parameters in the substructure. (Otherwise nothing would be an elementary substructure of anything ever.) Replace with “there is a greatest element.”
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 20:48
















  • 2




    $begingroup$
    $exists x (x=1)$ doesn’t work since the sentence needs to only have parameters in the substructure. (Otherwise nothing would be an elementary substructure of anything ever.) Replace with “there is a greatest element.”
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 20:48










2




2




$begingroup$
$exists x (x=1)$ doesn’t work since the sentence needs to only have parameters in the substructure. (Otherwise nothing would be an elementary substructure of anything ever.) Replace with “there is a greatest element.”
$endgroup$
– spaceisdarkgreen
Dec 24 '18 at 20:48






$begingroup$
$exists x (x=1)$ doesn’t work since the sentence needs to only have parameters in the substructure. (Otherwise nothing would be an elementary substructure of anything ever.) Replace with “there is a greatest element.”
$endgroup$
– spaceisdarkgreen
Dec 24 '18 at 20:48












1 Answer
1






active

oldest

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3












$begingroup$

As spaceisdarkgreen points out, what you've done isn't correct: the language $L$ of pure linear order doesn't have a symbol for $1$, so "$exists x(x=1)$" is not a sentence in the relevant language.



What you'll need to do is somehow define $1$ in the language of linear order alone; do you see a special property which $1$ has, in the structure $(0,1]$?




It's worth pointing out a subtlety of the notion of substructure: as long as our language has only relation symbols, then any subset of our structure is (or rather, is the underlying set of) a substructure, but this is no longer true if our language includes function symbols - e.g. ${$odd numbers$}$ is not (the underlying set of) a substructure of $(mathbb{N}; +)$, since it's not closed under $+$. In fact, even constant symbols (= zero-ary functions) kill this, since a subset not containing the thing named by a given constant symbol won't be a substructure.



It's also worth noting that you've in fact shown that $(0,1)$ is not even elementarily equivalent to $(0,1]$, and that this is actually a stronger property than not being an elementary substructure: e.g. $(0,{1over 2}]$ is elementarily equivalent (in fact, isomorphic) to $(0,1]$, but is not an elementary substructure of it (why not?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you misread their argument for the elementary substructure part.
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 21:02










  • $begingroup$
    @spaceisdarkgreen Oh derp, quite right! Fixed.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 21:03










  • $begingroup$
    I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
    $endgroup$
    – Leyla Alkan
    Dec 25 '18 at 6:54










  • $begingroup$
    @LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
    $endgroup$
    – Noah Schweber
    Dec 25 '18 at 19:05












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1 Answer
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active

oldest

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1 Answer
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active

oldest

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active

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3












$begingroup$

As spaceisdarkgreen points out, what you've done isn't correct: the language $L$ of pure linear order doesn't have a symbol for $1$, so "$exists x(x=1)$" is not a sentence in the relevant language.



What you'll need to do is somehow define $1$ in the language of linear order alone; do you see a special property which $1$ has, in the structure $(0,1]$?




It's worth pointing out a subtlety of the notion of substructure: as long as our language has only relation symbols, then any subset of our structure is (or rather, is the underlying set of) a substructure, but this is no longer true if our language includes function symbols - e.g. ${$odd numbers$}$ is not (the underlying set of) a substructure of $(mathbb{N}; +)$, since it's not closed under $+$. In fact, even constant symbols (= zero-ary functions) kill this, since a subset not containing the thing named by a given constant symbol won't be a substructure.



It's also worth noting that you've in fact shown that $(0,1)$ is not even elementarily equivalent to $(0,1]$, and that this is actually a stronger property than not being an elementary substructure: e.g. $(0,{1over 2}]$ is elementarily equivalent (in fact, isomorphic) to $(0,1]$, but is not an elementary substructure of it (why not?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you misread their argument for the elementary substructure part.
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 21:02










  • $begingroup$
    @spaceisdarkgreen Oh derp, quite right! Fixed.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 21:03










  • $begingroup$
    I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
    $endgroup$
    – Leyla Alkan
    Dec 25 '18 at 6:54










  • $begingroup$
    @LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
    $endgroup$
    – Noah Schweber
    Dec 25 '18 at 19:05
















3












$begingroup$

As spaceisdarkgreen points out, what you've done isn't correct: the language $L$ of pure linear order doesn't have a symbol for $1$, so "$exists x(x=1)$" is not a sentence in the relevant language.



What you'll need to do is somehow define $1$ in the language of linear order alone; do you see a special property which $1$ has, in the structure $(0,1]$?




It's worth pointing out a subtlety of the notion of substructure: as long as our language has only relation symbols, then any subset of our structure is (or rather, is the underlying set of) a substructure, but this is no longer true if our language includes function symbols - e.g. ${$odd numbers$}$ is not (the underlying set of) a substructure of $(mathbb{N}; +)$, since it's not closed under $+$. In fact, even constant symbols (= zero-ary functions) kill this, since a subset not containing the thing named by a given constant symbol won't be a substructure.



It's also worth noting that you've in fact shown that $(0,1)$ is not even elementarily equivalent to $(0,1]$, and that this is actually a stronger property than not being an elementary substructure: e.g. $(0,{1over 2}]$ is elementarily equivalent (in fact, isomorphic) to $(0,1]$, but is not an elementary substructure of it (why not?).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you misread their argument for the elementary substructure part.
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 21:02










  • $begingroup$
    @spaceisdarkgreen Oh derp, quite right! Fixed.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 21:03










  • $begingroup$
    I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
    $endgroup$
    – Leyla Alkan
    Dec 25 '18 at 6:54










  • $begingroup$
    @LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
    $endgroup$
    – Noah Schweber
    Dec 25 '18 at 19:05














3












3








3





$begingroup$

As spaceisdarkgreen points out, what you've done isn't correct: the language $L$ of pure linear order doesn't have a symbol for $1$, so "$exists x(x=1)$" is not a sentence in the relevant language.



What you'll need to do is somehow define $1$ in the language of linear order alone; do you see a special property which $1$ has, in the structure $(0,1]$?




It's worth pointing out a subtlety of the notion of substructure: as long as our language has only relation symbols, then any subset of our structure is (or rather, is the underlying set of) a substructure, but this is no longer true if our language includes function symbols - e.g. ${$odd numbers$}$ is not (the underlying set of) a substructure of $(mathbb{N}; +)$, since it's not closed under $+$. In fact, even constant symbols (= zero-ary functions) kill this, since a subset not containing the thing named by a given constant symbol won't be a substructure.



It's also worth noting that you've in fact shown that $(0,1)$ is not even elementarily equivalent to $(0,1]$, and that this is actually a stronger property than not being an elementary substructure: e.g. $(0,{1over 2}]$ is elementarily equivalent (in fact, isomorphic) to $(0,1]$, but is not an elementary substructure of it (why not?).






share|cite|improve this answer











$endgroup$



As spaceisdarkgreen points out, what you've done isn't correct: the language $L$ of pure linear order doesn't have a symbol for $1$, so "$exists x(x=1)$" is not a sentence in the relevant language.



What you'll need to do is somehow define $1$ in the language of linear order alone; do you see a special property which $1$ has, in the structure $(0,1]$?




It's worth pointing out a subtlety of the notion of substructure: as long as our language has only relation symbols, then any subset of our structure is (or rather, is the underlying set of) a substructure, but this is no longer true if our language includes function symbols - e.g. ${$odd numbers$}$ is not (the underlying set of) a substructure of $(mathbb{N}; +)$, since it's not closed under $+$. In fact, even constant symbols (= zero-ary functions) kill this, since a subset not containing the thing named by a given constant symbol won't be a substructure.



It's also worth noting that you've in fact shown that $(0,1)$ is not even elementarily equivalent to $(0,1]$, and that this is actually a stronger property than not being an elementary substructure: e.g. $(0,{1over 2}]$ is elementarily equivalent (in fact, isomorphic) to $(0,1]$, but is not an elementary substructure of it (why not?).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 21:04

























answered Dec 24 '18 at 20:36









Noah SchweberNoah Schweber

129k10155296




129k10155296












  • $begingroup$
    I think you misread their argument for the elementary substructure part.
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 21:02










  • $begingroup$
    @spaceisdarkgreen Oh derp, quite right! Fixed.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 21:03










  • $begingroup$
    I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
    $endgroup$
    – Leyla Alkan
    Dec 25 '18 at 6:54










  • $begingroup$
    @LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
    $endgroup$
    – Noah Schweber
    Dec 25 '18 at 19:05


















  • $begingroup$
    I think you misread their argument for the elementary substructure part.
    $endgroup$
    – spaceisdarkgreen
    Dec 24 '18 at 21:02










  • $begingroup$
    @spaceisdarkgreen Oh derp, quite right! Fixed.
    $endgroup$
    – Noah Schweber
    Dec 24 '18 at 21:03










  • $begingroup$
    I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
    $endgroup$
    – Leyla Alkan
    Dec 25 '18 at 6:54










  • $begingroup$
    @LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
    $endgroup$
    – Noah Schweber
    Dec 25 '18 at 19:05
















$begingroup$
I think you misread their argument for the elementary substructure part.
$endgroup$
– spaceisdarkgreen
Dec 24 '18 at 21:02




$begingroup$
I think you misread their argument for the elementary substructure part.
$endgroup$
– spaceisdarkgreen
Dec 24 '18 at 21:02












$begingroup$
@spaceisdarkgreen Oh derp, quite right! Fixed.
$endgroup$
– Noah Schweber
Dec 24 '18 at 21:03




$begingroup$
@spaceisdarkgreen Oh derp, quite right! Fixed.
$endgroup$
– Noah Schweber
Dec 24 '18 at 21:03












$begingroup$
I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
$endgroup$
– Leyla Alkan
Dec 25 '18 at 6:54




$begingroup$
I understood the part about substructure notion, but I can't come with an answer when you say "define 1 in the language of linear order alone" @NoahSchweber
$endgroup$
– Leyla Alkan
Dec 25 '18 at 6:54












$begingroup$
@LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:05




$begingroup$
@LeylaAlkan Do you see something about $1$, in the context of the linear order $(0,1]$, which makes it particularly "special" in any way (besides just having a fancy name)?
$endgroup$
– Noah Schweber
Dec 25 '18 at 19:05


















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