Mixing
On the definition of a infinite set by Tarski
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As far as I know, Tarski's definition of infinite set is:
A set $X$ is Tarski infinite, iff there exists a nonempty
subset $Ssubset P(X)$ such that for every $A in S$ there exists $B in S$ with $A subsetneq B$.
I think it is the same as the following definition.
A set $X$ is Tarski infinite, iff for every $A in (P(X)-{X})$ there exists $B in (P(X)-{X})$ with $A subsetneq B$.
And similarly for finite set,
A set $X$ is Tarski finite, iff exists $A in (P(X)-{X})$ s.t. there is no $B in (P(X)-{X})$ with $A subsetneq B$.
I think this is simpler. But is there a reason not to define it this way?
elementary-set-theory logic
asked Jan 30 at 14:38
amoogaeamoogae
728
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add a comment |
$begingroup$
As far as I know, Tarski's definition of infinite set is:
A set $X$ is Tarski infinite, iff there exists a nonempty
subset $Ssubset P(X)$ such that for every $A in S$ there exists $B in S$ with $A subsetneq B$.
I think it is the same as the following definition.
A set $X$ is Tarski infinite, iff for every $A in (P(X)-{X})$ there exists $B in (P(X)-{X})$ with $A subsetneq B$.
And similarly for finite set,
A set $X$ is Tarski finite, iff exists $A in (P(X)-{X})$ s.t. there is no $B in (P(X)-{X})$ with $A subsetneq B$.
I think this is simpler. But is there a reason not to define it this way?
elementary-set-theory logic
asked Jan 30 at 14:38
amoogaeamoogae
728
$endgroup$
4
$begingroup$
$mathbb Nsetminus{0}∈P(mathbb N)setminus{mathbb N}$
$endgroup$
– Holo
Jan 30 at 14:46
add a comment |
$begingroup$
As far as I know, Tarski's definition of infinite set is:
A set $X$ is Tarski infinite, iff there exists a nonempty
subset $Ssubset P(X)$ such that for every $A in S$ there exists $B in S$ with $A subsetneq B$.
I think it is the same as the following definition.
A set $X$ is Tarski infinite, iff for every $A in (P(X)-{X})$ there exists $B in (P(X)-{X})$ with $A subsetneq B$.
And similarly for finite set,
A set $X$ is Tarski finite, iff exists $A in (P(X)-{X})$ s.t. there is no $B in (P(X)-{X})$ with $A subsetneq B$.
I think this is simpler. But is there a reason not to define it this way?
elementary-set-theory logic
asked Jan 30 at 14:38
amoogaeamoogae
728
$endgroup$
As far as I know, Tarski's definition of infinite set is:
A set $X$ is Tarski infinite, iff there exists a nonempty
subset $Ssubset P(X)$ such that for every $A in S$ there exists $B in S$ with $A subsetneq B$.
I think it is the same as the following definition.
A set $X$ is Tarski infinite, iff for every $A in (P(X)-{X})$ there exists $B in (P(X)-{X})$ with $A subsetneq B$.
And similarly for finite set,
A set $X$ is Tarski finite, iff exists $A in (P(X)-{X})$ s.t. there is no $B in (P(X)-{X})$ with $A subsetneq B$.
I think this is simpler. But is there a reason not to define it this way?
elementary-set-theory logic
elementary-set-theory logic
asked Jan 30 at 14:38
amoogaeamoogae
728
asked Jan 30 at 14:38
amoogaeamoogae
728
edited Jan 30 at 14:53
amoogae
asked Jan 30 at 14:38
amoogaeamoogae
728
asked Jan 30 at 14:38
amoogaeamoogae
728
asked Jan 30 at 14:38
amoogaeamoogae
728
728
4
$begingroup$
$mathbb Nsetminus{0}∈P(mathbb N)setminus{mathbb N}$
$endgroup$
– Holo
Jan 30 at 14:46
add a comment |
4
$begingroup$
$mathbb Nsetminus{0}∈P(mathbb N)setminus{mathbb N}$
$endgroup$
– Holo
Jan 30 at 14:46
4
4
$begingroup$
$mathbb Nsetminus{0}∈P(mathbb N)setminus{mathbb N}$
$endgroup$
– Holo
Jan 30 at 14:46
$begingroup$
$mathbb Nsetminus{0}∈P(mathbb N)setminus{mathbb N}$
$endgroup$
– Holo
Jan 30 at 14:46
add a comment |
1 Answer
1
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oldest
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$begingroup$
The definition that you quote says that the set $X$ is Tarski-infinite if there is some strictly increasing infinite chain $S$ of subsets of $X$. Your proposed equivalent definition implies that every proper subset $A$ of $X$ can be extended to a larger proper subset of $X$; but that doesn't work, because an infinite $X$ will have subsets that don't belong to any strictly increasing infinite chain : if you take $X = Bbb{N}$ and $A = X setminus {0}$, then there are no subsets of $X$ strictly between $A$ and $X$, but $X$ is infinite. In that example, the set $S = {{0}, {0, 1}, {0, 1, 2}, ldots}$ gives a witness to the infiniteness of $X$ according to Tarski's definition.
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
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$begingroup$
The definition that you quote says that the set $X$ is Tarski-infinite if there is some strictly increasing infinite chain $S$ of subsets of $X$. Your proposed equivalent definition implies that every proper subset $A$ of $X$ can be extended to a larger proper subset of $X$; but that doesn't work, because an infinite $X$ will have subsets that don't belong to any strictly increasing infinite chain : if you take $X = Bbb{N}$ and $A = X setminus {0}$, then there are no subsets of $X$ strictly between $A$ and $X$, but $X$ is infinite. In that example, the set $S = {{0}, {0, 1}, {0, 1, 2}, ldots}$ gives a witness to the infiniteness of $X$ according to Tarski's definition.
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
$endgroup$
add a comment |
$begingroup$
The definition that you quote says that the set $X$ is Tarski-infinite if there is some strictly increasing infinite chain $S$ of subsets of $X$. Your proposed equivalent definition implies that every proper subset $A$ of $X$ can be extended to a larger proper subset of $X$; but that doesn't work, because an infinite $X$ will have subsets that don't belong to any strictly increasing infinite chain : if you take $X = Bbb{N}$ and $A = X setminus {0}$, then there are no subsets of $X$ strictly between $A$ and $X$, but $X$ is infinite. In that example, the set $S = {{0}, {0, 1}, {0, 1, 2}, ldots}$ gives a witness to the infiniteness of $X$ according to Tarski's definition.
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
$endgroup$
add a comment |
$begingroup$
The definition that you quote says that the set $X$ is Tarski-infinite if there is some strictly increasing infinite chain $S$ of subsets of $X$. Your proposed equivalent definition implies that every proper subset $A$ of $X$ can be extended to a larger proper subset of $X$; but that doesn't work, because an infinite $X$ will have subsets that don't belong to any strictly increasing infinite chain : if you take $X = Bbb{N}$ and $A = X setminus {0}$, then there are no subsets of $X$ strictly between $A$ and $X$, but $X$ is infinite. In that example, the set $S = {{0}, {0, 1}, {0, 1, 2}, ldots}$ gives a witness to the infiniteness of $X$ according to Tarski's definition.
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
$endgroup$
The definition that you quote says that the set $X$ is Tarski-infinite if there is some strictly increasing infinite chain $S$ of subsets of $X$. Your proposed equivalent definition implies that every proper subset $A$ of $X$ can be extended to a larger proper subset of $X$; but that doesn't work, because an infinite $X$ will have subsets that don't belong to any strictly increasing infinite chain : if you take $X = Bbb{N}$ and $A = X setminus {0}$, then there are no subsets of $X$ strictly between $A$ and $X$, but $X$ is infinite. In that example, the set $S = {{0}, {0, 1}, {0, 1, 2}, ldots}$ gives a witness to the infiniteness of $X$ according to Tarski's definition.
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
answered Jan 31 at 22:01
Rob ArthanRob Arthan
29.6k42967
29.6k42967
add a comment |
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4
$begingroup$
$mathbb Nsetminus{0}∈P(mathbb N)setminus{mathbb N}$
$endgroup$
– Holo
Jan 30 at 14:46