Limit of a sequence related to non-well founded set theory [closed]
$begingroup$
Consider the following:
$$alpha_0 = {} $$
$$alpha_1 ={{}}$$
$$alpha_2 ={{{}}}$$
$$...$$
Clearly $alpha_i$ is considered a set for all $i in omega$. Then consider $ alpha=lim_{i rightarrow omega}alpha_i ={...{}...}$. Now $alpha$ is not allowed in most standard set theories, because $alpha={alpha}$, or in other words ${...{}...}= { {...{}...}}$. I think it is interesting that each $alpha_i$ is a set, but the limit $alpha$ is not -- or am I wrong and $alpha$ is in fact not the limit?
Another similar question is as follows, consider the infinite binary tree:
what do we say about the limit of the far left branch, is it not $alpha$ as we mentioned before? Or do we not include the limit of the far left branch in our set theory?
sequences-and-series logic set-theory foundations alternative-set-theories
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
$endgroup$
closed as unclear what you're asking by Eric Wofsey, Lord Shark the Unknown, Derek Elkins, Leucippus, choco_addicted Dec 17 '18 at 8:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the following:
$$alpha_0 = {} $$
$$alpha_1 ={{}}$$
$$alpha_2 ={{{}}}$$
$$...$$
Clearly $alpha_i$ is considered a set for all $i in omega$. Then consider $ alpha=lim_{i rightarrow omega}alpha_i ={...{}...}$. Now $alpha$ is not allowed in most standard set theories, because $alpha={alpha}$, or in other words ${...{}...}= { {...{}...}}$. I think it is interesting that each $alpha_i$ is a set, but the limit $alpha$ is not -- or am I wrong and $alpha$ is in fact not the limit?
Another similar question is as follows, consider the infinite binary tree:
what do we say about the limit of the far left branch, is it not $alpha$ as we mentioned before? Or do we not include the limit of the far left branch in our set theory?
sequences-and-series logic set-theory foundations alternative-set-theories
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
$endgroup$
closed as unclear what you're asking by Eric Wofsey, Lord Shark the Unknown, Derek Elkins, Leucippus, choco_addicted Dec 17 '18 at 8:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
"Limit" is not just a term you can magically use in whatever context you want. It is meaningless to talk about "limits" like this unless you define precisely what you mean.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 5:12
$begingroup$
maybe I am just going with the intuitive feel, that for each $alpha_i$ we write $i$ many parenthesis, so for the limit $alpha$ we write $omega$ many parenthesis. But what do you say about the infinite binary tree example? If we view each level as the powerset of the previous level, is not the limit of the left branch an element of $2^omega$, corresponding to $alpha$?
$endgroup$
– Philip White
Dec 17 '18 at 5:18
1
$begingroup$
It's not clear to me if the "limit" you're talking about is $$alpha={{{cdots cdots}}}tag1$$ or if it's $$alpha=cdots{{{}}}cdotstag2$$ Only the first one seems to satisfy $alpha={alpha}$.
$endgroup$
– bof
Dec 17 '18 at 5:19
$begingroup$
Using the standard definition of set-theoretic limit, the limit of your sequence is the empty set.
$endgroup$
– celtschk
Dec 17 '18 at 8:57
add a comment |
$begingroup$
Consider the following:
$$alpha_0 = {} $$
$$alpha_1 ={{}}$$
$$alpha_2 ={{{}}}$$
$$...$$
Clearly $alpha_i$ is considered a set for all $i in omega$. Then consider $ alpha=lim_{i rightarrow omega}alpha_i ={...{}...}$. Now $alpha$ is not allowed in most standard set theories, because $alpha={alpha}$, or in other words ${...{}...}= { {...{}...}}$. I think it is interesting that each $alpha_i$ is a set, but the limit $alpha$ is not -- or am I wrong and $alpha$ is in fact not the limit?
Another similar question is as follows, consider the infinite binary tree:
what do we say about the limit of the far left branch, is it not $alpha$ as we mentioned before? Or do we not include the limit of the far left branch in our set theory?
sequences-and-series logic set-theory foundations alternative-set-theories
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
$endgroup$
Consider the following:
$$alpha_0 = {} $$
$$alpha_1 ={{}}$$
$$alpha_2 ={{{}}}$$
$$...$$
Clearly $alpha_i$ is considered a set for all $i in omega$. Then consider $ alpha=lim_{i rightarrow omega}alpha_i ={...{}...}$. Now $alpha$ is not allowed in most standard set theories, because $alpha={alpha}$, or in other words ${...{}...}= { {...{}...}}$. I think it is interesting that each $alpha_i$ is a set, but the limit $alpha$ is not -- or am I wrong and $alpha$ is in fact not the limit?
Another similar question is as follows, consider the infinite binary tree:
what do we say about the limit of the far left branch, is it not $alpha$ as we mentioned before? Or do we not include the limit of the far left branch in our set theory?
sequences-and-series logic set-theory foundations alternative-set-theories
sequences-and-series logic set-theory foundations alternative-set-theories
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
asked Dec 17 '18 at 5:04
Philip WhitePhilip White
548417
548417
closed as unclear what you're asking by Eric Wofsey, Lord Shark the Unknown, Derek Elkins, Leucippus, choco_addicted Dec 17 '18 at 8:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Eric Wofsey, Lord Shark the Unknown, Derek Elkins, Leucippus, choco_addicted Dec 17 '18 at 8:53
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
"Limit" is not just a term you can magically use in whatever context you want. It is meaningless to talk about "limits" like this unless you define precisely what you mean.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 5:12
$begingroup$
maybe I am just going with the intuitive feel, that for each $alpha_i$ we write $i$ many parenthesis, so for the limit $alpha$ we write $omega$ many parenthesis. But what do you say about the infinite binary tree example? If we view each level as the powerset of the previous level, is not the limit of the left branch an element of $2^omega$, corresponding to $alpha$?
$endgroup$
– Philip White
Dec 17 '18 at 5:18
1
$begingroup$
It's not clear to me if the "limit" you're talking about is $$alpha={{{cdots cdots}}}tag1$$ or if it's $$alpha=cdots{{{}}}cdotstag2$$ Only the first one seems to satisfy $alpha={alpha}$.
$endgroup$
– bof
Dec 17 '18 at 5:19
$begingroup$
Using the standard definition of set-theoretic limit, the limit of your sequence is the empty set.
$endgroup$
– celtschk
Dec 17 '18 at 8:57
add a comment |
1
$begingroup$
"Limit" is not just a term you can magically use in whatever context you want. It is meaningless to talk about "limits" like this unless you define precisely what you mean.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 5:12
$begingroup$
maybe I am just going with the intuitive feel, that for each $alpha_i$ we write $i$ many parenthesis, so for the limit $alpha$ we write $omega$ many parenthesis. But what do you say about the infinite binary tree example? If we view each level as the powerset of the previous level, is not the limit of the left branch an element of $2^omega$, corresponding to $alpha$?
$endgroup$
– Philip White
Dec 17 '18 at 5:18
1
$begingroup$
It's not clear to me if the "limit" you're talking about is $$alpha={{{cdots cdots}}}tag1$$ or if it's $$alpha=cdots{{{}}}cdotstag2$$ Only the first one seems to satisfy $alpha={alpha}$.
$endgroup$
– bof
Dec 17 '18 at 5:19
$begingroup$
Using the standard definition of set-theoretic limit, the limit of your sequence is the empty set.
$endgroup$
– celtschk
Dec 17 '18 at 8:57
1
1
$begingroup$
"Limit" is not just a term you can magically use in whatever context you want. It is meaningless to talk about "limits" like this unless you define precisely what you mean.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 5:12
$begingroup$
"Limit" is not just a term you can magically use in whatever context you want. It is meaningless to talk about "limits" like this unless you define precisely what you mean.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 5:12
$begingroup$
maybe I am just going with the intuitive feel, that for each $alpha_i$ we write $i$ many parenthesis, so for the limit $alpha$ we write $omega$ many parenthesis. But what do you say about the infinite binary tree example? If we view each level as the powerset of the previous level, is not the limit of the left branch an element of $2^omega$, corresponding to $alpha$?
$endgroup$
– Philip White
Dec 17 '18 at 5:18
$begingroup$
maybe I am just going with the intuitive feel, that for each $alpha_i$ we write $i$ many parenthesis, so for the limit $alpha$ we write $omega$ many parenthesis. But what do you say about the infinite binary tree example? If we view each level as the powerset of the previous level, is not the limit of the left branch an element of $2^omega$, corresponding to $alpha$?
$endgroup$
– Philip White
Dec 17 '18 at 5:18
1
1
$begingroup$
It's not clear to me if the "limit" you're talking about is $$alpha={{{cdots cdots}}}tag1$$ or if it's $$alpha=cdots{{{}}}cdotstag2$$ Only the first one seems to satisfy $alpha={alpha}$.
$endgroup$
– bof
Dec 17 '18 at 5:19
$begingroup$
It's not clear to me if the "limit" you're talking about is $$alpha={{{cdots cdots}}}tag1$$ or if it's $$alpha=cdots{{{}}}cdotstag2$$ Only the first one seems to satisfy $alpha={alpha}$.
$endgroup$
– bof
Dec 17 '18 at 5:19
$begingroup$
Using the standard definition of set-theoretic limit, the limit of your sequence is the empty set.
$endgroup$
– celtschk
Dec 17 '18 at 8:57
$begingroup$
Using the standard definition of set-theoretic limit, the limit of your sequence is the empty set.
$endgroup$
– celtschk
Dec 17 '18 at 8:57
add a comment |
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$begingroup$
"Limit" is not just a term you can magically use in whatever context you want. It is meaningless to talk about "limits" like this unless you define precisely what you mean.
$endgroup$
– Eric Wofsey
Dec 17 '18 at 5:12
$begingroup$
maybe I am just going with the intuitive feel, that for each $alpha_i$ we write $i$ many parenthesis, so for the limit $alpha$ we write $omega$ many parenthesis. But what do you say about the infinite binary tree example? If we view each level as the powerset of the previous level, is not the limit of the left branch an element of $2^omega$, corresponding to $alpha$?
$endgroup$
– Philip White
Dec 17 '18 at 5:18
1
$begingroup$
It's not clear to me if the "limit" you're talking about is $$alpha={{{cdots cdots}}}tag1$$ or if it's $$alpha=cdots{{{}}}cdotstag2$$ Only the first one seems to satisfy $alpha={alpha}$.
$endgroup$
– bof
Dec 17 '18 at 5:19
$begingroup$
Using the standard definition of set-theoretic limit, the limit of your sequence is the empty set.
$endgroup$
– celtschk
Dec 17 '18 at 8:57