$a in A$ is equivalent to $ { a } subset A $












0












$begingroup$


In Naive Set theory by Halmos, he states $a in A$ is equivalent to $ { a } subset A $. I am guessing that here he means logically equivalent. Intuitively, I see how they can be interchanged, but I am not sure how to prove it.



I attempted to set up a truth table. I tried breaking down $B subset A$ into the sentence $ forall x(x in B implies x in A) $.



I know that truth tables in predicate logic would be infinitely long, but I thought that in this special case, since we only have ${ a }$ I could employ a truth table with $B = { a }$ and $x = a$.



$$begin{array}{c|c|c|}
a in A & a in B & implies & a in A\ hline
bf{T} & T & bf{T} & T \ hline
bf{T} & F & bf{T} & T\ hline
bf{F} & T & bf{F} & F\ hline
F & F & T & F\ hline
end{array}$$



This truth table shows that these are not logically equivalent. What am I doing wrong?






elementary-set-theory predicate-logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you getting at something along the lines of $ { a, a, a,... } $
    $endgroup$
    – pmac
    Dec 14 '18 at 2:24










  • $begingroup$
    No. Since $a$ is the only element, $b=a$ necessarily. At least intuitively.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:37












  • $begingroup$
    So where in my attempt did I go wrong?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:39
















0












$begingroup$


In Naive Set theory by Halmos, he states $a in A$ is equivalent to $ { a } subset A $. I am guessing that here he means logically equivalent. Intuitively, I see how they can be interchanged, but I am not sure how to prove it.



I attempted to set up a truth table. I tried breaking down $B subset A$ into the sentence $ forall x(x in B implies x in A) $.



I know that truth tables in predicate logic would be infinitely long, but I thought that in this special case, since we only have ${ a }$ I could employ a truth table with $B = { a }$ and $x = a$.



$$begin{array}{c|c|c|}
a in A & a in B & implies & a in A\ hline
bf{T} & T & bf{T} & T \ hline
bf{T} & F & bf{T} & T\ hline
bf{F} & T & bf{F} & F\ hline
F & F & T & F\ hline
end{array}$$



This truth table shows that these are not logically equivalent. What am I doing wrong?






elementary-set-theory predicate-logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    Are you getting at something along the lines of $ { a, a, a,... } $
    $endgroup$
    – pmac
    Dec 14 '18 at 2:24










  • $begingroup$
    No. Since $a$ is the only element, $b=a$ necessarily. At least intuitively.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:37












  • $begingroup$
    So where in my attempt did I go wrong?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:39














0












0








0





$begingroup$


In Naive Set theory by Halmos, he states $a in A$ is equivalent to $ { a } subset A $. I am guessing that here he means logically equivalent. Intuitively, I see how they can be interchanged, but I am not sure how to prove it.



I attempted to set up a truth table. I tried breaking down $B subset A$ into the sentence $ forall x(x in B implies x in A) $.



I know that truth tables in predicate logic would be infinitely long, but I thought that in this special case, since we only have ${ a }$ I could employ a truth table with $B = { a }$ and $x = a$.



$$begin{array}{c|c|c|}
a in A & a in B & implies & a in A\ hline
bf{T} & T & bf{T} & T \ hline
bf{T} & F & bf{T} & T\ hline
bf{F} & T & bf{F} & F\ hline
F & F & T & F\ hline
end{array}$$



This truth table shows that these are not logically equivalent. What am I doing wrong?






elementary-set-theory predicate-logic







share|cite|improve this question









$endgroup$




In Naive Set theory by Halmos, he states $a in A$ is equivalent to $ { a } subset A $. I am guessing that here he means logically equivalent. Intuitively, I see how they can be interchanged, but I am not sure how to prove it.



I attempted to set up a truth table. I tried breaking down $B subset A$ into the sentence $ forall x(x in B implies x in A) $.



I know that truth tables in predicate logic would be infinitely long, but I thought that in this special case, since we only have ${ a }$ I could employ a truth table with $B = { a }$ and $x = a$.



$$begin{array}{c|c|c|}
a in A & a in B & implies & a in A\ hline
bf{T} & T & bf{T} & T \ hline
bf{T} & F & bf{T} & T\ hline
bf{F} & T & bf{F} & F\ hline
F & F & T & F\ hline
end{array}$$



This truth table shows that these are not logically equivalent. What am I doing wrong?






elementary-set-theory predicate-logic




elementary-set-theory predicate-logic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 2:11









pmacpmac

10116




10116












  • $begingroup$
    Are you getting at something along the lines of $ { a, a, a,... } $
    $endgroup$
    – pmac
    Dec 14 '18 at 2:24










  • $begingroup$
    No. Since $a$ is the only element, $b=a$ necessarily. At least intuitively.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:37












  • $begingroup$
    So where in my attempt did I go wrong?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:39


















  • $begingroup$
    Are you getting at something along the lines of $ { a, a, a,... } $
    $endgroup$
    – pmac
    Dec 14 '18 at 2:24










  • $begingroup$
    No. Since $a$ is the only element, $b=a$ necessarily. At least intuitively.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:37












  • $begingroup$
    So where in my attempt did I go wrong?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:39
















$begingroup$
Are you getting at something along the lines of $ { a, a, a,... } $
$endgroup$
– pmac
Dec 14 '18 at 2:24




$begingroup$
Are you getting at something along the lines of $ { a, a, a,... } $
$endgroup$
– pmac
Dec 14 '18 at 2:24












$begingroup$
No. Since $a$ is the only element, $b=a$ necessarily. At least intuitively.
$endgroup$
– pmac
Dec 14 '18 at 2:37






$begingroup$
No. Since $a$ is the only element, $b=a$ necessarily. At least intuitively.
$endgroup$
– pmac
Dec 14 '18 at 2:37














$begingroup$
So where in my attempt did I go wrong?
$endgroup$
– pmac
Dec 14 '18 at 2:39




$begingroup$
So where in my attempt did I go wrong?
$endgroup$
– pmac
Dec 14 '18 at 2:39










1 Answer
1






active

oldest

votes


















2












$begingroup$

You went wrong in assuming that this is a logical equivalence. It is not; set-theoretically they are equivalent, but logically they are not, as you yourself demonstrated.



Specifically, $a in { a }$ has to be true set-theoretically (and hence the last two rows of the truth-table do not apply), but it is not a logical truth.



Put a different way: once you assume some basic axioms regarding set-theory (in particular ones that define the notion of subset ... or what something like a singleton set means) you can derive ${ a } subset A$ from $a in A$, and vice versa. But without those axioms, you cannot. So that shows that these are not logically equivalent, but 'merely' equivalent in the context of set-theory.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53










  • $begingroup$
    Oh just saw your edit, but my questions still stands.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53








  • 1




    $begingroup$
    @pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
    $endgroup$
    – Bram28
    Dec 14 '18 at 2:56











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1 Answer
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1 Answer
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active

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active

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active

oldest

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2












$begingroup$

You went wrong in assuming that this is a logical equivalence. It is not; set-theoretically they are equivalent, but logically they are not, as you yourself demonstrated.



Specifically, $a in { a }$ has to be true set-theoretically (and hence the last two rows of the truth-table do not apply), but it is not a logical truth.



Put a different way: once you assume some basic axioms regarding set-theory (in particular ones that define the notion of subset ... or what something like a singleton set means) you can derive ${ a } subset A$ from $a in A$, and vice versa. But without those axioms, you cannot. So that shows that these are not logically equivalent, but 'merely' equivalent in the context of set-theory.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53










  • $begingroup$
    Oh just saw your edit, but my questions still stands.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53








  • 1




    $begingroup$
    @pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
    $endgroup$
    – Bram28
    Dec 14 '18 at 2:56
















2












$begingroup$

You went wrong in assuming that this is a logical equivalence. It is not; set-theoretically they are equivalent, but logically they are not, as you yourself demonstrated.



Specifically, $a in { a }$ has to be true set-theoretically (and hence the last two rows of the truth-table do not apply), but it is not a logical truth.



Put a different way: once you assume some basic axioms regarding set-theory (in particular ones that define the notion of subset ... or what something like a singleton set means) you can derive ${ a } subset A$ from $a in A$, and vice versa. But without those axioms, you cannot. So that shows that these are not logically equivalent, but 'merely' equivalent in the context of set-theory.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53










  • $begingroup$
    Oh just saw your edit, but my questions still stands.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53








  • 1




    $begingroup$
    @pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
    $endgroup$
    – Bram28
    Dec 14 '18 at 2:56














2












2








2





$begingroup$

You went wrong in assuming that this is a logical equivalence. It is not; set-theoretically they are equivalent, but logically they are not, as you yourself demonstrated.



Specifically, $a in { a }$ has to be true set-theoretically (and hence the last two rows of the truth-table do not apply), but it is not a logical truth.



Put a different way: once you assume some basic axioms regarding set-theory (in particular ones that define the notion of subset ... or what something like a singleton set means) you can derive ${ a } subset A$ from $a in A$, and vice versa. But without those axioms, you cannot. So that shows that these are not logically equivalent, but 'merely' equivalent in the context of set-theory.






share|cite|improve this answer











$endgroup$



You went wrong in assuming that this is a logical equivalence. It is not; set-theoretically they are equivalent, but logically they are not, as you yourself demonstrated.



Specifically, $a in { a }$ has to be true set-theoretically (and hence the last two rows of the truth-table do not apply), but it is not a logical truth.



Put a different way: once you assume some basic axioms regarding set-theory (in particular ones that define the notion of subset ... or what something like a singleton set means) you can derive ${ a } subset A$ from $a in A$, and vice versa. But without those axioms, you cannot. So that shows that these are not logically equivalent, but 'merely' equivalent in the context of set-theory.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 3:04

























answered Dec 14 '18 at 2:47









Bram28Bram28

63.2k44793




63.2k44793












  • $begingroup$
    But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53










  • $begingroup$
    Oh just saw your edit, but my questions still stands.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53








  • 1




    $begingroup$
    @pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
    $endgroup$
    – Bram28
    Dec 14 '18 at 2:56


















  • $begingroup$
    But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53










  • $begingroup$
    Oh just saw your edit, but my questions still stands.
    $endgroup$
    – pmac
    Dec 14 '18 at 2:53








  • 1




    $begingroup$
    @pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
    $endgroup$
    – Bram28
    Dec 14 '18 at 2:56
















$begingroup$
But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
$endgroup$
– pmac
Dec 14 '18 at 2:53




$begingroup$
But $ a in A $ is not a set and neither is $ { a } subset A $. These are sentences, right? So how can we talk about set-equivalent?
$endgroup$
– pmac
Dec 14 '18 at 2:53












$begingroup$
Oh just saw your edit, but my questions still stands.
$endgroup$
– pmac
Dec 14 '18 at 2:53






$begingroup$
Oh just saw your edit, but my questions still stands.
$endgroup$
– pmac
Dec 14 '18 at 2:53






1




1




$begingroup$
@pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
$endgroup$
– Bram28
Dec 14 '18 at 2:56




$begingroup$
@pmac Yes, they are statements. What I mean is that the statement $a in { a }$ has to be true in the context of set-theory, whereas from the perspective of pure logic one can set it to false.
$endgroup$
– Bram28
Dec 14 '18 at 2:56


















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