How to prove the cardinality of a closed interval is equal to the continuum?
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Say we have a closed interval $[a,b]$. How can I prove $|[a,b]|$ equal to $c$, where $c$ is the continuum.
I already proved $|(a,b)| = |(c,d)|$. But I don't know whether it is useful.
elementary-set-theory
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
asked Nov 26 at 10:15
user8314628
1767
add a comment |
up vote
0
down vote
favorite
Say we have a closed interval $[a,b]$. How can I prove $|[a,b]|$ equal to $c$, where $c$ is the continuum.
I already proved $|(a,b)| = |(c,d)|$. But I don't know whether it is useful.
elementary-set-theory
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
asked Nov 26 at 10:15
user8314628
1767
Show cardinality of $[a,b]=(a,b)cup {a,b}$ and $mathbb{R}$ are same.
– Yadati Kiran
Nov 26 at 10:16
Yea, I think that is what $c$ means. But I don't know how to prove it.
– user8314628
Nov 26 at 10:17
Find a bijection between the two sets.
– Yadati Kiran
Nov 26 at 10:18
Oh I get it. $[a,b]$ belongs to $R$, so I can apply the Schroder-Bernstein theorem! $|(a,b)|leq R$. For any $(c,d)$ in $R$, $|(c,d)|leq (a,b)$.
– user8314628
Nov 26 at 10:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Say we have a closed interval $[a,b]$. How can I prove $|[a,b]|$ equal to $c$, where $c$ is the continuum.
I already proved $|(a,b)| = |(c,d)|$. But I don't know whether it is useful.
elementary-set-theory
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
asked Nov 26 at 10:15
user8314628
1767
Say we have a closed interval $[a,b]$. How can I prove $|[a,b]|$ equal to $c$, where $c$ is the continuum.
I already proved $|(a,b)| = |(c,d)|$. But I don't know whether it is useful.
elementary-set-theory
elementary-set-theory
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
asked Nov 26 at 10:15
user8314628
1767
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
asked Nov 26 at 10:15
user8314628
1767
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
edited Nov 26 at 12:31
Andrés E. Caicedo
64.5k8157245
64.5k8157245
asked Nov 26 at 10:15
user8314628
1767
asked Nov 26 at 10:15
user8314628
1767
asked Nov 26 at 10:15
user8314628
1767
1767
Show cardinality of $[a,b]=(a,b)cup {a,b}$ and $mathbb{R}$ are same.
– Yadati Kiran
Nov 26 at 10:16
Yea, I think that is what $c$ means. But I don't know how to prove it.
– user8314628
Nov 26 at 10:17
Find a bijection between the two sets.
– Yadati Kiran
Nov 26 at 10:18
Oh I get it. $[a,b]$ belongs to $R$, so I can apply the Schroder-Bernstein theorem! $|(a,b)|leq R$. For any $(c,d)$ in $R$, $|(c,d)|leq (a,b)$.
– user8314628
Nov 26 at 10:20
add a comment |
Show cardinality of $[a,b]=(a,b)cup {a,b}$ and $mathbb{R}$ are same.
– Yadati Kiran
Nov 26 at 10:16
Yea, I think that is what $c$ means. But I don't know how to prove it.
– user8314628
Nov 26 at 10:17
Find a bijection between the two sets.
– Yadati Kiran
Nov 26 at 10:18
Oh I get it. $[a,b]$ belongs to $R$, so I can apply the Schroder-Bernstein theorem! $|(a,b)|leq R$. For any $(c,d)$ in $R$, $|(c,d)|leq (a,b)$.
– user8314628
Nov 26 at 10:20
Show cardinality of $[a,b]=(a,b)cup {a,b}$ and $mathbb{R}$ are same.
– Yadati Kiran
Nov 26 at 10:16
Show cardinality of $[a,b]=(a,b)cup {a,b}$ and $mathbb{R}$ are same.
– Yadati Kiran
Nov 26 at 10:16
Yea, I think that is what $c$ means. But I don't know how to prove it.
– user8314628
Nov 26 at 10:17
Yea, I think that is what $c$ means. But I don't know how to prove it.
– user8314628
Nov 26 at 10:17
Find a bijection between the two sets.
– Yadati Kiran
Nov 26 at 10:18
Find a bijection between the two sets.
– Yadati Kiran
Nov 26 at 10:18
Oh I get it. $[a,b]$ belongs to $R$, so I can apply the Schroder-Bernstein theorem! $|(a,b)|leq R$. For any $(c,d)$ in $R$, $|(c,d)|leq (a,b)$.
– user8314628
Nov 26 at 10:20
Oh I get it. $[a,b]$ belongs to $R$, so I can apply the Schroder-Bernstein theorem! $|(a,b)|leq R$. For any $(c,d)$ in $R$, $|(c,d)|leq (a,b)$.
– user8314628
Nov 26 at 10:20
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
You can also use the fact that: "If $P$ is a non empty perfect set in $mathbb{R}^k$ then $P$ is uncountable". Show $[a,b]$ is a perfect set.
You may also define a mapping as $f:mathbb{R}to[0,1]:::f(x)=begin{cases}dfrac{1}{pi}left(dfrac{pi}{2}+arctan(x)right) &xin(0,1)\x &xin{0,1}end{cases}$
answered Nov 26 at 10:23
Yadati Kiran
1,253417
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
add a comment |
up vote
0
down vote
Hint: You only need to find an injection $mathbb R to [a,b]$. Some such injection can be achieved by modifying $tan^{-1}$. (It's also possible to construct one by hand that is piecewise linear.)
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You can also use the fact that: "If $P$ is a non empty perfect set in $mathbb{R}^k$ then $P$ is uncountable". Show $[a,b]$ is a perfect set.
You may also define a mapping as $f:mathbb{R}to[0,1]:::f(x)=begin{cases}dfrac{1}{pi}left(dfrac{pi}{2}+arctan(x)right) &xin(0,1)\x &xin{0,1}end{cases}$
answered Nov 26 at 10:23
Yadati Kiran
1,253417
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
add a comment |
up vote
0
down vote
accepted
You can also use the fact that: "If $P$ is a non empty perfect set in $mathbb{R}^k$ then $P$ is uncountable". Show $[a,b]$ is a perfect set.
You may also define a mapping as $f:mathbb{R}to[0,1]:::f(x)=begin{cases}dfrac{1}{pi}left(dfrac{pi}{2}+arctan(x)right) &xin(0,1)\x &xin{0,1}end{cases}$
answered Nov 26 at 10:23
Yadati Kiran
1,253417
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You can also use the fact that: "If $P$ is a non empty perfect set in $mathbb{R}^k$ then $P$ is uncountable". Show $[a,b]$ is a perfect set.
You may also define a mapping as $f:mathbb{R}to[0,1]:::f(x)=begin{cases}dfrac{1}{pi}left(dfrac{pi}{2}+arctan(x)right) &xin(0,1)\x &xin{0,1}end{cases}$
answered Nov 26 at 10:23
Yadati Kiran
1,253417
You can also use the fact that: "If $P$ is a non empty perfect set in $mathbb{R}^k$ then $P$ is uncountable". Show $[a,b]$ is a perfect set.
You may also define a mapping as $f:mathbb{R}to[0,1]:::f(x)=begin{cases}dfrac{1}{pi}left(dfrac{pi}{2}+arctan(x)right) &xin(0,1)\x &xin{0,1}end{cases}$
answered Nov 26 at 10:23
Yadati Kiran
1,253417
edited Nov 26 at 10:32
answered Nov 26 at 10:23
Yadati Kiran
1,253417
answered Nov 26 at 10:23
Yadati Kiran
1,253417
answered Nov 26 at 10:23
Yadati Kiran
1,253417
1,253417
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
add a comment |
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
What you wrote in the first paragraph is not enough. You do not simply want to show that intervals are uncountable, as there are many different uncountable cardinalities.
– Andrés E. Caicedo
Nov 26 at 13:07
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
@AndrésE.Caicedo: Yes I agree.
– Yadati Kiran
Nov 26 at 13:29
add a comment |
up vote
0
down vote
Hint: You only need to find an injection $mathbb R to [a,b]$. Some such injection can be achieved by modifying $tan^{-1}$. (It's also possible to construct one by hand that is piecewise linear.)
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
add a comment |
up vote
0
down vote
Hint: You only need to find an injection $mathbb R to [a,b]$. Some such injection can be achieved by modifying $tan^{-1}$. (It's also possible to construct one by hand that is piecewise linear.)
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: You only need to find an injection $mathbb R to [a,b]$. Some such injection can be achieved by modifying $tan^{-1}$. (It's also possible to construct one by hand that is piecewise linear.)
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
Hint: You only need to find an injection $mathbb R to [a,b]$. Some such injection can be achieved by modifying $tan^{-1}$. (It's also possible to construct one by hand that is piecewise linear.)
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
answered Nov 26 at 10:20
Stefan Mesken
14.4k32046
14.4k32046
add a comment |
add a comment |
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Show cardinality of $[a,b]=(a,b)cup {a,b}$ and $mathbb{R}$ are same.
– Yadati Kiran
Nov 26 at 10:16
Yea, I think that is what $c$ means. But I don't know how to prove it.
– user8314628
Nov 26 at 10:17
Find a bijection between the two sets.
– Yadati Kiran
Nov 26 at 10:18
Oh I get it. $[a,b]$ belongs to $R$, so I can apply the Schroder-Bernstein theorem! $|(a,b)|leq R$. For any $(c,d)$ in $R$, $|(c,d)|leq (a,b)$.
– user8314628
Nov 26 at 10:20