Mixing
Axiom of Foundation and transitive sets.
Assuming the Axiom of Foundation, show that every non-empty transtive set contains $0$ and show that every non-singleton transitive set contains $1$.
It's straightfoward to show the first part.
Suppose that $t$ is a nonempty transitive set.
By the Axiom of Foundation, there is $r in t$ such that $r cap t = emptyset$.
Now, as $r in t$ and $t$ is transitive, we have that $r subseteq t$ so that $emptyset subseteq r subseteq r cap t = emptyset$, so $emptyset = r in t$.
I'm stuck on the second part. If we suppose further that $t$ has at least two elements, then we may consider $q in t$ with $q neq emptyset$.
We can find $p in q$ with $q cap p = emptyset$, but I don't know where to go from there.
elementary-set-theory
asked Nov 20 '18 at 22:06
Sprinkle
39119
add a comment |
Assuming the Axiom of Foundation, show that every non-empty transtive set contains $0$ and show that every non-singleton transitive set contains $1$.
It's straightfoward to show the first part.
Suppose that $t$ is a nonempty transitive set.
By the Axiom of Foundation, there is $r in t$ such that $r cap t = emptyset$.
Now, as $r in t$ and $t$ is transitive, we have that $r subseteq t$ so that $emptyset subseteq r subseteq r cap t = emptyset$, so $emptyset = r in t$.
I'm stuck on the second part. If we suppose further that $t$ has at least two elements, then we may consider $q in t$ with $q neq emptyset$.
We can find $p in q$ with $q cap p = emptyset$, but I don't know where to go from there.
elementary-set-theory
asked Nov 20 '18 at 22:06
Sprinkle
39119
add a comment |
Assuming the Axiom of Foundation, show that every non-empty transtive set contains $0$ and show that every non-singleton transitive set contains $1$.
It's straightfoward to show the first part.
Suppose that $t$ is a nonempty transitive set.
By the Axiom of Foundation, there is $r in t$ such that $r cap t = emptyset$.
Now, as $r in t$ and $t$ is transitive, we have that $r subseteq t$ so that $emptyset subseteq r subseteq r cap t = emptyset$, so $emptyset = r in t$.
I'm stuck on the second part. If we suppose further that $t$ has at least two elements, then we may consider $q in t$ with $q neq emptyset$.
We can find $p in q$ with $q cap p = emptyset$, but I don't know where to go from there.
elementary-set-theory
asked Nov 20 '18 at 22:06
Sprinkle
39119
Assuming the Axiom of Foundation, show that every non-empty transtive set contains $0$ and show that every non-singleton transitive set contains $1$.
It's straightfoward to show the first part.
Suppose that $t$ is a nonempty transitive set.
By the Axiom of Foundation, there is $r in t$ such that $r cap t = emptyset$.
Now, as $r in t$ and $t$ is transitive, we have that $r subseteq t$ so that $emptyset subseteq r subseteq r cap t = emptyset$, so $emptyset = r in t$.
I'm stuck on the second part. If we suppose further that $t$ has at least two elements, then we may consider $q in t$ with $q neq emptyset$.
We can find $p in q$ with $q cap p = emptyset$, but I don't know where to go from there.
elementary-set-theory
elementary-set-theory
asked Nov 20 '18 at 22:06
Sprinkle
39119
asked Nov 20 '18 at 22:06
Sprinkle
39119
asked Nov 20 '18 at 22:06
Sprinkle
39119
asked Nov 20 '18 at 22:06
Sprinkle
39119
asked Nov 20 '18 at 22:06
Sprinkle
39119
39119
add a comment |
add a comment |
1 Answer
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Let t be a multipoint transitive set.
As discused $phi$ in t.
r = t - {$phi$} is not empty.
By regularity, exists a in r with empty a $cap$ r.
a in t; a subset t; if x in a, then x = $phi$.
As a is not empty, a = {$phi$}, 1 in t QED.
$phi$ is empty set.
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
add a comment |
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1 Answer
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1 Answer
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active
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votes
Let t be a multipoint transitive set.
As discused $phi$ in t.
r = t - {$phi$} is not empty.
By regularity, exists a in r with empty a $cap$ r.
a in t; a subset t; if x in a, then x = $phi$.
As a is not empty, a = {$phi$}, 1 in t QED.
$phi$ is empty set.
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
add a comment |
Let t be a multipoint transitive set.
As discused $phi$ in t.
r = t - {$phi$} is not empty.
By regularity, exists a in r with empty a $cap$ r.
a in t; a subset t; if x in a, then x = $phi$.
As a is not empty, a = {$phi$}, 1 in t QED.
$phi$ is empty set.
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
add a comment |
Let t be a multipoint transitive set.
As discused $phi$ in t.
r = t - {$phi$} is not empty.
By regularity, exists a in r with empty a $cap$ r.
a in t; a subset t; if x in a, then x = $phi$.
As a is not empty, a = {$phi$}, 1 in t QED.
$phi$ is empty set.
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
Let t be a multipoint transitive set.
As discused $phi$ in t.
r = t - {$phi$} is not empty.
By regularity, exists a in r with empty a $cap$ r.
a in t; a subset t; if x in a, then x = $phi$.
As a is not empty, a = {$phi$}, 1 in t QED.
$phi$ is empty set.
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
answered Nov 23 '18 at 5:41
William Elliot
7,3082620
7,3082620
add a comment |
add a comment |
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