Axiom of Foundation and transitive sets.
up vote
1
down vote
favorite
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 at 22:06
Sprinkle
39119
add a comment |
up vote
1
down vote
favorite
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 at 22:06
Sprinkle
39119
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 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 at 22:06
Sprinkle
39119
asked Nov 20 at 22:06
Sprinkle
39119
asked Nov 20 at 22:06
Sprinkle
39119
asked Nov 20 at 22:06
Sprinkle
39119
asked Nov 20 at 22:06
Sprinkle
39119
39119
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
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 at 5:41
William Elliot
6,9072518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 at 5:41
William Elliot
6,9072518
add a comment |
up vote
0
down vote
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 at 5:41
William Elliot
6,9072518
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 5:41
William Elliot
6,9072518
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 at 5:41
William Elliot
6,9072518
answered Nov 23 at 5:41
William Elliot
6,9072518
answered Nov 23 at 5:41
William Elliot
6,9072518
answered Nov 23 at 5:41
William Elliot
6,9072518
6,9072518
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006967%2faxiom-of-foundation-and-transitive-sets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
