Size of infinte sets cardinality Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Let $A$ be an infinite set and B a countable set, show that $vert A cup B vert = vert A vert$Comparing the cardinality of setsShow that open segment $(a,b)$, close segment $[a,b]$ have the same cardinality as $mathbbR$Trouble understanding cardinalityIs $2^aleph_0=c$?Prove that all of the following sets have the same cardinalitycardinality of $mathbb R$ is the same as the cardinality of $mathbb R^2$Cardinality of infinite setsSets of equal cardinalityIs this a superior measure of the size of sets, to cardinality?Cardinality of the unit square and union of sets of size $c$ are equal
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Size of infinte sets cardinality
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Let $A$ be an infinite set and B a countable set, show that $vert A cup B vert = vert A vert$Comparing the cardinality of setsShow that open segment $(a,b)$, close segment $[a,b]$ have the same cardinality as $mathbbR$Trouble understanding cardinalityIs $2^aleph_0=c$?Prove that all of the following sets have the same cardinalitycardinality of $mathbb R$ is the same as the cardinality of $mathbb R^2$Cardinality of infinite setsSets of equal cardinalityIs this a superior measure of the size of sets, to cardinality?Cardinality of the unit square and union of sets of size $c$ are equal
$begingroup$
The question is as follows:
Prove that if R is uncountable and T is a countable subset of R, then the cardinality of RT is the same as the cardinality of R.
What i have:
I know that R is uncountable so it has a countable subset (this is a theorem of uncountable sets). Let T be this subset so T has the same cardinality as the set of natural numbers(by the definition of T being countable). My intution is telling me that we will have to use the cantor bernstein theorem to prove they have same cardinalities. So for that the first thing i did showed was that |RT| <= |R| (pretty clear as its RT, |R| means cardinality of R).i got a bit confused while trying to show that |R| <= |RT|. Maybe we can show this by defining a bijection f from R --> RT, such that f(x) = x when x is in RT, but i dont know what to do if x is in R and not in RT, if i can define that function then i can conclude |R| <= |RT|, then use the cantor-bernstein theorem and then im done. Or maybe im doing this all wrong i cant think of any other way Any help is much appreciated!!
number-theory elementary-set-theory cardinals
asked Mar 26 at 7:35
JohmJohm
162
$endgroup$
|
show 1 more comment
$begingroup$
The question is as follows:
Prove that if R is uncountable and T is a countable subset of R, then the cardinality of RT is the same as the cardinality of R.
What i have:
I know that R is uncountable so it has a countable subset (this is a theorem of uncountable sets). Let T be this subset so T has the same cardinality as the set of natural numbers(by the definition of T being countable). My intution is telling me that we will have to use the cantor bernstein theorem to prove they have same cardinalities. So for that the first thing i did showed was that |RT| <= |R| (pretty clear as its RT, |R| means cardinality of R).i got a bit confused while trying to show that |R| <= |RT|. Maybe we can show this by defining a bijection f from R --> RT, such that f(x) = x when x is in RT, but i dont know what to do if x is in R and not in RT, if i can define that function then i can conclude |R| <= |RT|, then use the cantor-bernstein theorem and then im done. Or maybe im doing this all wrong i cant think of any other way Any help is much appreciated!!
number-theory elementary-set-theory cardinals
asked Mar 26 at 7:35
JohmJohm
162
$endgroup$
1
$begingroup$
What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $mathbb N$", others use "countable" to mean "has at most the cardinality of $mathbb N$". Are finite sets "countable" for you?
$endgroup$
– bof
Mar 26 at 7:40
$begingroup$
same cardinality as N, yes finite sets are countable
$endgroup$
– Johm
Mar 26 at 7:41
1
$begingroup$
You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $mathbb N$.)
$endgroup$
– bof
Mar 26 at 7:44
$begingroup$
yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said
$endgroup$
– Johm
Mar 26 at 7:47
1
$begingroup$
Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set?
$endgroup$
– bof
Mar 26 at 7:49
|
show 1 more comment
$begingroup$
The question is as follows:
Prove that if R is uncountable and T is a countable subset of R, then the cardinality of RT is the same as the cardinality of R.
What i have:
I know that R is uncountable so it has a countable subset (this is a theorem of uncountable sets). Let T be this subset so T has the same cardinality as the set of natural numbers(by the definition of T being countable). My intution is telling me that we will have to use the cantor bernstein theorem to prove they have same cardinalities. So for that the first thing i did showed was that |RT| <= |R| (pretty clear as its RT, |R| means cardinality of R).i got a bit confused while trying to show that |R| <= |RT|. Maybe we can show this by defining a bijection f from R --> RT, such that f(x) = x when x is in RT, but i dont know what to do if x is in R and not in RT, if i can define that function then i can conclude |R| <= |RT|, then use the cantor-bernstein theorem and then im done. Or maybe im doing this all wrong i cant think of any other way Any help is much appreciated!!
number-theory elementary-set-theory cardinals
asked Mar 26 at 7:35
JohmJohm
162
$endgroup$
The question is as follows:
Prove that if R is uncountable and T is a countable subset of R, then the cardinality of RT is the same as the cardinality of R.
What i have:
I know that R is uncountable so it has a countable subset (this is a theorem of uncountable sets). Let T be this subset so T has the same cardinality as the set of natural numbers(by the definition of T being countable). My intution is telling me that we will have to use the cantor bernstein theorem to prove they have same cardinalities. So for that the first thing i did showed was that |RT| <= |R| (pretty clear as its RT, |R| means cardinality of R).i got a bit confused while trying to show that |R| <= |RT|. Maybe we can show this by defining a bijection f from R --> RT, such that f(x) = x when x is in RT, but i dont know what to do if x is in R and not in RT, if i can define that function then i can conclude |R| <= |RT|, then use the cantor-bernstein theorem and then im done. Or maybe im doing this all wrong i cant think of any other way Any help is much appreciated!!
number-theory elementary-set-theory cardinals
number-theory elementary-set-theory cardinals
asked Mar 26 at 7:35
JohmJohm
162
asked Mar 26 at 7:35
JohmJohm
162
asked Mar 26 at 7:35
JohmJohm
162
asked Mar 26 at 7:35
JohmJohm
162
asked Mar 26 at 7:35
JohmJohm
162
162
1
$begingroup$
What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $mathbb N$", others use "countable" to mean "has at most the cardinality of $mathbb N$". Are finite sets "countable" for you?
$endgroup$
– bof
Mar 26 at 7:40
$begingroup$
same cardinality as N, yes finite sets are countable
$endgroup$
– Johm
Mar 26 at 7:41
1
$begingroup$
You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $mathbb N$.)
$endgroup$
– bof
Mar 26 at 7:44
$begingroup$
yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said
$endgroup$
– Johm
Mar 26 at 7:47
1
$begingroup$
Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set?
$endgroup$
– bof
Mar 26 at 7:49
|
show 1 more comment
1
$begingroup$
What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $mathbb N$", others use "countable" to mean "has at most the cardinality of $mathbb N$". Are finite sets "countable" for you?
$endgroup$
– bof
Mar 26 at 7:40
$begingroup$
same cardinality as N, yes finite sets are countable
$endgroup$
– Johm
Mar 26 at 7:41
1
$begingroup$
You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $mathbb N$.)
$endgroup$
– bof
Mar 26 at 7:44
$begingroup$
yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said
$endgroup$
– Johm
Mar 26 at 7:47
1
$begingroup$
Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set?
$endgroup$
– bof
Mar 26 at 7:49
1
1
$begingroup$
What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $mathbb N$", others use "countable" to mean "has at most the cardinality of $mathbb N$". Are finite sets "countable" for you?
$endgroup$
– bof
Mar 26 at 7:40
$begingroup$
What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $mathbb N$", others use "countable" to mean "has at most the cardinality of $mathbb N$". Are finite sets "countable" for you?
$endgroup$
– bof
Mar 26 at 7:40
$begingroup$
same cardinality as N, yes finite sets are countable
$endgroup$
– Johm
Mar 26 at 7:41
$begingroup$
same cardinality as N, yes finite sets are countable
$endgroup$
– Johm
Mar 26 at 7:41
1
1
$begingroup$
You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $mathbb N$.)
$endgroup$
– bof
Mar 26 at 7:44
$begingroup$
You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $mathbb N$.)
$endgroup$
– bof
Mar 26 at 7:44
$begingroup$
yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said
$endgroup$
– Johm
Mar 26 at 7:47
$begingroup$
yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said
$endgroup$
– Johm
Mar 26 at 7:47
1
1
$begingroup$
Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set?
$endgroup$
– bof
Mar 26 at 7:49
$begingroup$
Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set?
$endgroup$
– bof
Mar 26 at 7:49
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
I think you can find a quite nice bijection between $R$ and $Rsetminus T$, actually. I would go at it like so:
- Prove that $Rsetminus T$ is uncountable
- Conclude that there must exist some countable $T_1subseteq Rsetminus T$, with $|T_1|=|T|$.
- From there, we can find some countable $T_2subseteq Rsetminus (Tcup T_1)$, and continue from there to find countable $T_1,T_2,T_3,dots T_n,dots$, such that for each $n$, we have $T_nsubseteq Rsetminus(Tcup T_1cupcdotscup T_n-1)$
- Now, we can map the values of $xin T$ to elements in $T_1$, elements in $T_1$ to $T_2$, and so on. All values not in any of $T_i$ get mapped to themselves.
answered Mar 26 at 7:46
5xum5xum
92.7k394162
$endgroup$
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
add a comment |
$begingroup$
Let $T$ be a countable subset of an uncountable set $R$.
Since $R$ is uncountable and $T$ is countable, it follows that $Rsetminus T$ is uncountable. Therefore, there is a countably infinite set $Ssubseteq Rsetminus T$. Then $Scup T$ is also a countably infinite set. Therefore there is a bijection
$$f:Sto Scup T.$$
Of course there is a trivial bijection
$$g:Rsetminus(Scup T)to Rsetminus(Scup T).$$
The union of these two bijections is a bijection
$$fcup g:Rsetminus Tto R.$$
answered Mar 26 at 8:01
bofbof
52.6k559121
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you can find a quite nice bijection between $R$ and $Rsetminus T$, actually. I would go at it like so:
- Prove that $Rsetminus T$ is uncountable
- Conclude that there must exist some countable $T_1subseteq Rsetminus T$, with $|T_1|=|T|$.
- From there, we can find some countable $T_2subseteq Rsetminus (Tcup T_1)$, and continue from there to find countable $T_1,T_2,T_3,dots T_n,dots$, such that for each $n$, we have $T_nsubseteq Rsetminus(Tcup T_1cupcdotscup T_n-1)$
- Now, we can map the values of $xin T$ to elements in $T_1$, elements in $T_1$ to $T_2$, and so on. All values not in any of $T_i$ get mapped to themselves.
answered Mar 26 at 7:46
5xum5xum
92.7k394162
$endgroup$
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
add a comment |
$begingroup$
I think you can find a quite nice bijection between $R$ and $Rsetminus T$, actually. I would go at it like so:
- Prove that $Rsetminus T$ is uncountable
- Conclude that there must exist some countable $T_1subseteq Rsetminus T$, with $|T_1|=|T|$.
- From there, we can find some countable $T_2subseteq Rsetminus (Tcup T_1)$, and continue from there to find countable $T_1,T_2,T_3,dots T_n,dots$, such that for each $n$, we have $T_nsubseteq Rsetminus(Tcup T_1cupcdotscup T_n-1)$
- Now, we can map the values of $xin T$ to elements in $T_1$, elements in $T_1$ to $T_2$, and so on. All values not in any of $T_i$ get mapped to themselves.
answered Mar 26 at 7:46
5xum5xum
92.7k394162
$endgroup$
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
add a comment |
$begingroup$
I think you can find a quite nice bijection between $R$ and $Rsetminus T$, actually. I would go at it like so:
- Prove that $Rsetminus T$ is uncountable
- Conclude that there must exist some countable $T_1subseteq Rsetminus T$, with $|T_1|=|T|$.
- From there, we can find some countable $T_2subseteq Rsetminus (Tcup T_1)$, and continue from there to find countable $T_1,T_2,T_3,dots T_n,dots$, such that for each $n$, we have $T_nsubseteq Rsetminus(Tcup T_1cupcdotscup T_n-1)$
- Now, we can map the values of $xin T$ to elements in $T_1$, elements in $T_1$ to $T_2$, and so on. All values not in any of $T_i$ get mapped to themselves.
answered Mar 26 at 7:46
5xum5xum
92.7k394162
$endgroup$
I think you can find a quite nice bijection between $R$ and $Rsetminus T$, actually. I would go at it like so:
- Prove that $Rsetminus T$ is uncountable
- Conclude that there must exist some countable $T_1subseteq Rsetminus T$, with $|T_1|=|T|$.
- From there, we can find some countable $T_2subseteq Rsetminus (Tcup T_1)$, and continue from there to find countable $T_1,T_2,T_3,dots T_n,dots$, such that for each $n$, we have $T_nsubseteq Rsetminus(Tcup T_1cupcdotscup T_n-1)$
- Now, we can map the values of $xin T$ to elements in $T_1$, elements in $T_1$ to $T_2$, and so on. All values not in any of $T_i$ get mapped to themselves.
answered Mar 26 at 7:46
5xum5xum
92.7k394162
answered Mar 26 at 7:46
5xum5xum
92.7k394162
answered Mar 26 at 7:46
5xum5xum
92.7k394162
answered Mar 26 at 7:46
5xum5xum
92.7k394162
92.7k394162
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
add a comment |
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
Omg thank you so much!! i get it now i didnt even think about the countable subsets of T since T is countable itself. Appreciate it!!
$endgroup$
– Johm
Mar 26 at 7:52
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
@Johm I am not talking about the countable subsets of $T$. I am talking about the countable subsets of $Rsetminus T$.
$endgroup$
– 5xum
Mar 26 at 7:53
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
$begingroup$
I was literally misreading the entire thing God. But i understand it
$endgroup$
– Johm
Mar 26 at 8:01
add a comment |
$begingroup$
Let $T$ be a countable subset of an uncountable set $R$.
Since $R$ is uncountable and $T$ is countable, it follows that $Rsetminus T$ is uncountable. Therefore, there is a countably infinite set $Ssubseteq Rsetminus T$. Then $Scup T$ is also a countably infinite set. Therefore there is a bijection
$$f:Sto Scup T.$$
Of course there is a trivial bijection
$$g:Rsetminus(Scup T)to Rsetminus(Scup T).$$
The union of these two bijections is a bijection
$$fcup g:Rsetminus Tto R.$$
answered Mar 26 at 8:01
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Let $T$ be a countable subset of an uncountable set $R$.
Since $R$ is uncountable and $T$ is countable, it follows that $Rsetminus T$ is uncountable. Therefore, there is a countably infinite set $Ssubseteq Rsetminus T$. Then $Scup T$ is also a countably infinite set. Therefore there is a bijection
$$f:Sto Scup T.$$
Of course there is a trivial bijection
$$g:Rsetminus(Scup T)to Rsetminus(Scup T).$$
The union of these two bijections is a bijection
$$fcup g:Rsetminus Tto R.$$
answered Mar 26 at 8:01
bofbof
52.6k559121
$endgroup$
add a comment |
$begingroup$
Let $T$ be a countable subset of an uncountable set $R$.
Since $R$ is uncountable and $T$ is countable, it follows that $Rsetminus T$ is uncountable. Therefore, there is a countably infinite set $Ssubseteq Rsetminus T$. Then $Scup T$ is also a countably infinite set. Therefore there is a bijection
$$f:Sto Scup T.$$
Of course there is a trivial bijection
$$g:Rsetminus(Scup T)to Rsetminus(Scup T).$$
The union of these two bijections is a bijection
$$fcup g:Rsetminus Tto R.$$
answered Mar 26 at 8:01
bofbof
52.6k559121
$endgroup$
Let $T$ be a countable subset of an uncountable set $R$.
Since $R$ is uncountable and $T$ is countable, it follows that $Rsetminus T$ is uncountable. Therefore, there is a countably infinite set $Ssubseteq Rsetminus T$. Then $Scup T$ is also a countably infinite set. Therefore there is a bijection
$$f:Sto Scup T.$$
Of course there is a trivial bijection
$$g:Rsetminus(Scup T)to Rsetminus(Scup T).$$
The union of these two bijections is a bijection
$$fcup g:Rsetminus Tto R.$$
answered Mar 26 at 8:01
bofbof
52.6k559121
answered Mar 26 at 8:01
bofbof
52.6k559121
answered Mar 26 at 8:01
bofbof
52.6k559121
answered Mar 26 at 8:01
bofbof
52.6k559121
52.6k559121
add a comment |
add a comment |
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1
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What does "countable" mean in this question? Some authors use "countable" to mean "has the same cardinality as $mathbb N$", others use "countable" to mean "has at most the cardinality of $mathbb N$". Are finite sets "countable" for you?
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– bof
Mar 26 at 7:40
$begingroup$
same cardinality as N, yes finite sets are countable
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– Johm
Mar 26 at 7:41
1
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You just gave two contradictory answers to my question: (1) countable sets have the same cardinality as $mathbb N$ and (2) finite sets are countable. (A finite sets does not have the same cardinality as $mathbb N$.)
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– bof
Mar 26 at 7:44
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yes that is true, what i meant to say was that in this question countable means has the SAME cardinality as N, and finite sets are countable but they do not have same cardinality as N which is true like you said
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– Johm
Mar 26 at 7:47
1
$begingroup$
Your replies continue to contradict themselves. When you say that $T$ is a countable subset of $N$, is it safe to assume that $T$ is infinite? Or could $T$ be a finite set?
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– bof
Mar 26 at 7:49