Mixing
If $Asim B$(both dedekind infinite), is it then that $Asim Bcup {x}$
1
$begingroup$
If the symbol $Asim B $signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?:$$A sim B Rightarrow A sim Bcup{x}$$
elementary-set-theory cardinals
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
asked Jan 6 at 13:18
user13910user13910
61
$endgroup$
add a comment |
1
$begingroup$
If the symbol $Asim B $signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?:$$A sim B Rightarrow A sim Bcup{x}$$
elementary-set-theory cardinals
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
asked Jan 6 at 13:18
user13910user13910
61
$endgroup$
$begingroup$
Which of the definitions of Dedekind-infinite are you using?
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself.
$endgroup$
– user13910
Jan 6 at 13:47
add a comment |
1
1
1
$begingroup$
If the symbol $Asim B $signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?:$$A sim B Rightarrow A sim Bcup{x}$$
elementary-set-theory cardinals
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
asked Jan 6 at 13:18
user13910user13910
61
$endgroup$
If the symbol $Asim B $signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?:$$A sim B Rightarrow A sim Bcup{x}$$
elementary-set-theory cardinals
elementary-set-theory cardinals
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
asked Jan 6 at 13:18
user13910user13910
61
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
asked Jan 6 at 13:18
user13910user13910
61
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
edited Jan 6 at 13:35
Andrés E. Caicedo
65.2k8158247
65.2k8158247
asked Jan 6 at 13:18
user13910user13910
61
asked Jan 6 at 13:18
user13910user13910
61
asked Jan 6 at 13:18
user13910user13910
61
61
$begingroup$
Which of the definitions of Dedekind-infinite are you using?
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself.
$endgroup$
– user13910
Jan 6 at 13:47
add a comment |
$begingroup$
Which of the definitions of Dedekind-infinite are you using?
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself.
$endgroup$
– user13910
Jan 6 at 13:47
$begingroup$
Which of the definitions of Dedekind-infinite are you using?
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
Which of the definitions of Dedekind-infinite are you using?
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself.
$endgroup$
– user13910
Jan 6 at 13:47
$begingroup$
If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself.
$endgroup$
– user13910
Jan 6 at 13:47
add a comment |
2 Answers
2
active
oldest
votes
0
$begingroup$
Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
1
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
1
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
add a comment |
0
$begingroup$
Let $f : A to B$ be a bijection, and fix an injection $g : mathbb{N} to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n in mathbb{N}$.
Define $f' : A to B cup { x }$ by letting
$$f'(a) = begin{cases} f(a) & text{if } a notin mathrm{im}(g) \ x & text{if } a=a_0 \ f(a_{n-1}) & text{if } a=a_n text{ for some } n>0 end{cases}$$
You need to prove that $f'$ is a bijection.
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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oldest
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0
$begingroup$
Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
1
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
1
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
add a comment |
0
$begingroup$
Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
1
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
1
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
add a comment |
0
0
0
$begingroup$
Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
$endgroup$
Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
answered Jan 6 at 13:43
Lucas HenriqueLucas Henrique
1,024414
1,024414
1
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
1
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
add a comment |
1
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
1
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
1
1
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
The axiom of choice has no business being involved here.
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
It could not, but I can. And it works.
$endgroup$
– Lucas Henrique
Jan 6 at 13:45
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
$begingroup$
@AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713
$endgroup$
– Stupid Questions Inc
Jan 7 at 6:28
1
1
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed.
$endgroup$
– Asaf Karagila♦
Jan 7 at 7:52
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
$begingroup$
@AsafKaragila thank you $mathsf{AC}$ master
$endgroup$
– Stupid Questions Inc
Jan 7 at 9:21
add a comment |
0
$begingroup$
Let $f : A to B$ be a bijection, and fix an injection $g : mathbb{N} to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n in mathbb{N}$.
Define $f' : A to B cup { x }$ by letting
$$f'(a) = begin{cases} f(a) & text{if } a notin mathrm{im}(g) \ x & text{if } a=a_0 \ f(a_{n-1}) & text{if } a=a_n text{ for some } n>0 end{cases}$$
You need to prove that $f'$ is a bijection.
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
add a comment |
0
$begingroup$
Let $f : A to B$ be a bijection, and fix an injection $g : mathbb{N} to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n in mathbb{N}$.
Define $f' : A to B cup { x }$ by letting
$$f'(a) = begin{cases} f(a) & text{if } a notin mathrm{im}(g) \ x & text{if } a=a_0 \ f(a_{n-1}) & text{if } a=a_n text{ for some } n>0 end{cases}$$
You need to prove that $f'$ is a bijection.
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
add a comment |
0
0
0
$begingroup$
Let $f : A to B$ be a bijection, and fix an injection $g : mathbb{N} to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n in mathbb{N}$.
Define $f' : A to B cup { x }$ by letting
$$f'(a) = begin{cases} f(a) & text{if } a notin mathrm{im}(g) \ x & text{if } a=a_0 \ f(a_{n-1}) & text{if } a=a_n text{ for some } n>0 end{cases}$$
You need to prove that $f'$ is a bijection.
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
Let $f : A to B$ be a bijection, and fix an injection $g : mathbb{N} to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n in mathbb{N}$.
Define $f' : A to B cup { x }$ by letting
$$f'(a) = begin{cases} f(a) & text{if } a notin mathrm{im}(g) \ x & text{if } a=a_0 \ f(a_{n-1}) & text{if } a=a_n text{ for some } n>0 end{cases}$$
You need to prove that $f'$ is a bijection.
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
edited Jan 6 at 13:51
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
answered Jan 6 at 13:32
Clive NewsteadClive Newstead
51.2k474135
51.2k474135
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
add a comment |
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do).
$endgroup$
– user13910
Jan 6 at 13:44
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
@user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A to B$ into a bijection $A to B cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $mathbb{N} to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer.
$endgroup$
– Clive Newstead
Jan 6 at 13:51
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
$begingroup$
P.S. I've changed the notation in the answer slightly to make it more readable.
$endgroup$
– Clive Newstead
Jan 6 at 13:53
add a comment |
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$begingroup$
Which of the definitions of Dedekind-infinite are you using?
$endgroup$
– Asaf Karagila♦
Jan 6 at 13:44
$begingroup$
If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself.
$endgroup$
– user13910
Jan 6 at 13:47