Show that $(0,1)$ is a substructure of $(0,1]$ in the language $L={<}$ but is not an elementary...
$begingroup$
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
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
$endgroup$
add a comment |
$begingroup$
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
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
$endgroup$
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
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
asked Dec 24 '18 at 20:29
Leyla AlkanLeyla Alkan
1,5401724
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
add a comment |
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
add a comment |
1 Answer
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$begingroup$
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?).
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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?).
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?).
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?).
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
$endgroup$
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?).
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
edited Dec 24 '18 at 21:04
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
answered Dec 24 '18 at 20:36
Noah SchweberNoah Schweber
129k10155296
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
$exists x (x=1)$ 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