Infinite chain ordered by inclusion
$begingroup$
Does there exists an infinite chain in $P(mathbb{N})$ ordered by inclusion?
I think that no because we have always add at least one element but its not a proof so I'm not sure.
elementary-set-theory
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 16:36
avan1235avan1235
3297
$endgroup$
add a comment |
$begingroup$
Does there exists an infinite chain in $P(mathbb{N})$ ordered by inclusion?
I think that no because we have always add at least one element but its not a proof so I'm not sure.
elementary-set-theory
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 16:36
avan1235avan1235
3297
$endgroup$
$begingroup$
What is $P(Bbb N)$?
$endgroup$
– Clayton
Jan 24 at 16:38
4
$begingroup$
There is both an increasing and a decreasing (countably)infinite chain in $mathbb N$ by inclusion. The add-one each time gives an increasing chain, and "remove-one" from $mathbb N$ each time gives a decreasing chain. (I think $P(mathbb N)$ is the power set of $mathbb N$). As for uncountably infinite, I think you should think about this.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:40
add a comment |
$begingroup$
Does there exists an infinite chain in $P(mathbb{N})$ ordered by inclusion?
I think that no because we have always add at least one element but its not a proof so I'm not sure.
elementary-set-theory
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 16:36
avan1235avan1235
3297
$endgroup$
Does there exists an infinite chain in $P(mathbb{N})$ ordered by inclusion?
I think that no because we have always add at least one element but its not a proof so I'm not sure.
elementary-set-theory
elementary-set-theory
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 16:36
avan1235avan1235
3297
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
asked Jan 24 at 16:36
avan1235avan1235
3297
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
edited Jan 24 at 17:27
Andrés E. Caicedo
65.6k8160250
65.6k8160250
asked Jan 24 at 16:36
avan1235avan1235
3297
asked Jan 24 at 16:36
avan1235avan1235
3297
asked Jan 24 at 16:36
avan1235avan1235
3297
3297
$begingroup$
What is $P(Bbb N)$?
$endgroup$
– Clayton
Jan 24 at 16:38
4
$begingroup$
There is both an increasing and a decreasing (countably)infinite chain in $mathbb N$ by inclusion. The add-one each time gives an increasing chain, and "remove-one" from $mathbb N$ each time gives a decreasing chain. (I think $P(mathbb N)$ is the power set of $mathbb N$). As for uncountably infinite, I think you should think about this.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:40
add a comment |
$begingroup$
What is $P(Bbb N)$?
$endgroup$
– Clayton
Jan 24 at 16:38
4
$begingroup$
There is both an increasing and a decreasing (countably)infinite chain in $mathbb N$ by inclusion. The add-one each time gives an increasing chain, and "remove-one" from $mathbb N$ each time gives a decreasing chain. (I think $P(mathbb N)$ is the power set of $mathbb N$). As for uncountably infinite, I think you should think about this.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:40
$begingroup$
What is $P(Bbb N)$?
$endgroup$
– Clayton
Jan 24 at 16:38
$begingroup$
What is $P(Bbb N)$?
$endgroup$
– Clayton
Jan 24 at 16:38
4
4
$begingroup$
There is both an increasing and a decreasing (countably)infinite chain in $mathbb N$ by inclusion. The add-one each time gives an increasing chain, and "remove-one" from $mathbb N$ each time gives a decreasing chain. (I think $P(mathbb N)$ is the power set of $mathbb N$). As for uncountably infinite, I think you should think about this.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:40
$begingroup$
There is both an increasing and a decreasing (countably)infinite chain in $mathbb N$ by inclusion. The add-one each time gives an increasing chain, and "remove-one" from $mathbb N$ each time gives a decreasing chain. (I think $P(mathbb N)$ is the power set of $mathbb N$). As for uncountably infinite, I think you should think about this.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:40
add a comment |
1 Answer
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$begingroup$
Here's an example of a (countably) infinite chain by inclusion: Let $Z_d = {n | n$ is divisible by $d}$.
$Z_2 supset Z_4 supset Z_8 supset ldots supset Z_{2^k} supset ldots$
answered Jan 24 at 17:47
DubsDubs
55926
$endgroup$
add a comment |
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$begingroup$
Here's an example of a (countably) infinite chain by inclusion: Let $Z_d = {n | n$ is divisible by $d}$.
$Z_2 supset Z_4 supset Z_8 supset ldots supset Z_{2^k} supset ldots$
answered Jan 24 at 17:47
DubsDubs
55926
$endgroup$
add a comment |
$begingroup$
Here's an example of a (countably) infinite chain by inclusion: Let $Z_d = {n | n$ is divisible by $d}$.
$Z_2 supset Z_4 supset Z_8 supset ldots supset Z_{2^k} supset ldots$
answered Jan 24 at 17:47
DubsDubs
55926
$endgroup$
add a comment |
$begingroup$
Here's an example of a (countably) infinite chain by inclusion: Let $Z_d = {n | n$ is divisible by $d}$.
$Z_2 supset Z_4 supset Z_8 supset ldots supset Z_{2^k} supset ldots$
answered Jan 24 at 17:47
DubsDubs
55926
$endgroup$
Here's an example of a (countably) infinite chain by inclusion: Let $Z_d = {n | n$ is divisible by $d}$.
$Z_2 supset Z_4 supset Z_8 supset ldots supset Z_{2^k} supset ldots$
answered Jan 24 at 17:47
DubsDubs
55926
answered Jan 24 at 17:47
DubsDubs
55926
answered Jan 24 at 17:47
DubsDubs
55926
answered Jan 24 at 17:47
DubsDubs
55926
55926
add a comment |
add a comment |
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$begingroup$
What is $P(Bbb N)$?
$endgroup$
– Clayton
Jan 24 at 16:38
4
$begingroup$
There is both an increasing and a decreasing (countably)infinite chain in $mathbb N$ by inclusion. The add-one each time gives an increasing chain, and "remove-one" from $mathbb N$ each time gives a decreasing chain. (I think $P(mathbb N)$ is the power set of $mathbb N$). As for uncountably infinite, I think you should think about this.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:40