How to show $2^{ℵ_0} leq mathfrak c$ [duplicate]
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Easiest way to prove that $2^{aleph_0} = c$
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I want to show $2^{ℵ_0}=mathfrak c$.
I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:
Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have
$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$
But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?
elementary-set-theory cardinals
edited Nov 26 at 13:14
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asked Nov 26 at 12:49
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marked as duplicate by Lord_Farin, amWhy elementary-set-theory
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Dec 11 at 22:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
6
down vote
favorite
This question already has an answer here:
Easiest way to prove that $2^{aleph_0} = c$
3 answers
I want to show $2^{ℵ_0}=mathfrak c$.
I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:
Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have
$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$
But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?
elementary-set-theory cardinals
edited Nov 26 at 13:14
user126154
5,323716
asked Nov 26 at 12:49
user8314628
1817
marked as duplicate by Lord_Farin, amWhy elementary-set-theory
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Dec 11 at 22:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Look at the Cantor set.
– Andrés E. Caicedo
Nov 26 at 13:04
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
This question already has an answer here:
Easiest way to prove that $2^{aleph_0} = c$
3 answers
I want to show $2^{ℵ_0}=mathfrak c$.
I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:
Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have
$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$
But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?
elementary-set-theory cardinals
edited Nov 26 at 13:14
user126154
5,323716
asked Nov 26 at 12:49
user8314628
1817
This question already has an answer here:
Easiest way to prove that $2^{aleph_0} = c$
3 answers
I want to show $2^{ℵ_0}=mathfrak c$.
I already showed $mathfrak c leq 2^{ℵ_0}$ as follows:
Each real number is constructed from an integer part and a decimal fraction. The decimal fraction is countable and has $ℵ_0$ digits. So we have
$mathfrak c leq ℵ_0 * 10^{ℵ_0} leq 2^{ℵ_0} * (2^4)^{ℵ_0} = 2^{ℵ_0}$ since $ℵ_0 + 4ℵ_0=ℵ_0$
But how can I prove the other way $2^{ℵ_0} leq mathfrak c$?
This question already has an answer here:
Easiest way to prove that $2^{aleph_0} = c$
3 answers
elementary-set-theory cardinals
elementary-set-theory cardinals
edited Nov 26 at 13:14
user126154
5,323716
asked Nov 26 at 12:49
user8314628
1817
edited Nov 26 at 13:14
user126154
5,323716
asked Nov 26 at 12:49
user8314628
1817
edited Nov 26 at 13:14
user126154
5,323716
edited Nov 26 at 13:14
user126154
5,323716
edited Nov 26 at 13:14
user126154
5,323716
5,323716
asked Nov 26 at 12:49
user8314628
1817
asked Nov 26 at 12:49
user8314628
1817
asked Nov 26 at 12:49
user8314628
1817
1817
marked as duplicate by Lord_Farin, amWhy elementary-set-theory
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Dec 11 at 22:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord_Farin, amWhy elementary-set-theory
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Dec 11 at 22:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Look at the Cantor set.
– Andrés E. Caicedo
Nov 26 at 13:04
add a comment |
1
Look at the Cantor set.
– Andrés E. Caicedo
Nov 26 at 13:04
1
1
Look at the Cantor set.
– Andrés E. Caicedo
Nov 26 at 13:04
Look at the Cantor set.
– Andrés E. Caicedo
Nov 26 at 13:04
add a comment |
2 Answers
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$2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.
answered Nov 26 at 13:09
user126154
5,323716
add a comment |
up vote
1
down vote
You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$
Then this function is injective. So $$
text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
leq text{Cardinal}(mathbb{R}) $$ So $2^{ℵ_0} leq mathfrak c$.
answered Nov 26 at 13:23
Dadrahm
4599
2
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
1
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
|
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.
answered Nov 26 at 13:09
user126154
5,323716
add a comment |
up vote
1
down vote
$2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.
answered Nov 26 at 13:09
user126154
5,323716
add a comment |
up vote
1
down vote
up vote
1
down vote
$2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.
answered Nov 26 at 13:09
user126154
5,323716
$2^{aleph_0}$ is the cardinality of all reals (belonging to $(0,1)$ if you prefer) that you can write by using only $0,1$. Those numbers clearly form a subset of $mathbb R$ which must therefore have cardinality at least $2^{aleph_0}$.
answered Nov 26 at 13:09
user126154
5,323716
answered Nov 26 at 13:09
user126154
5,323716
answered Nov 26 at 13:09
user126154
5,323716
answered Nov 26 at 13:09
user126154
5,323716
5,323716
add a comment |
add a comment |
up vote
1
down vote
You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$
Then this function is injective. So $$
text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
leq text{Cardinal}(mathbb{R}) $$ So $2^{ℵ_0} leq mathfrak c$.
answered Nov 26 at 13:23
Dadrahm
4599
2
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
1
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
|
show 4 more comments
up vote
1
down vote
You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$
Then this function is injective. So $$
text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
leq text{Cardinal}(mathbb{R}) $$ So $2^{ℵ_0} leq mathfrak c$.
answered Nov 26 at 13:23
Dadrahm
4599
2
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
1
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
|
show 4 more comments
up vote
1
down vote
up vote
1
down vote
You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$
Then this function is injective. So $$
text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
leq text{Cardinal}(mathbb{R}) $$ So $2^{ℵ_0} leq mathfrak c$.
answered Nov 26 at 13:23
Dadrahm
4599
You can define a function $F$ from the set ${ (x_n) | nin mathbb{N}, x_nin { 0,1} }$ to $mathbb {R}$ such that $$F[(x_1,x_2,x_3,ldots )]=0 . x_1x_2x_3ldots $$
Then this function is injective. So $$
text{Cardinal} { (x_n) | nin mathbb{N}, x_nin { 0,1} }
leq text{Cardinal}(mathbb{R}) $$ So $2^{ℵ_0} leq mathfrak c$.
answered Nov 26 at 13:23
Dadrahm
4599
edited Nov 26 at 14:16
answered Nov 26 at 13:23
Dadrahm
4599
answered Nov 26 at 13:23
Dadrahm
4599
answered Nov 26 at 13:23
Dadrahm
4599
4599
2
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
1
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
|
show 4 more comments
2
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
1
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
2
2
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
It’s not 1-1 because $F(0,1,1,dots)=F(1,0,0,0dots)$, if I’m not mistaken. But I think the idea is accurate.
– Brahadeesh
Nov 26 at 13:28
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
No. It's One to One and your saying is false%%
– Dadrahm
Nov 26 at 13:32
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
Well, I'll let you know that the downvote is mine.
– Brahadeesh
Nov 26 at 13:40
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
But why! was my solution false?
– Dadrahm
Nov 26 at 13:50
1
1
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
@GitGud For the same reason that 0.999...=1, assuming that $0.x_1 x_2 x_3 dots$ is in binary.
– Brahadeesh
Nov 27 at 15:36
|
show 4 more comments
1
Look at the Cantor set.
– Andrés E. Caicedo
Nov 26 at 13:04