Why is ${f,|,fcolon Atomathbb N}$ not uncountable?
$begingroup$
Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.
I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.
Please help me where is I am wrong.
elementary-set-theory cardinals
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
$endgroup$
add a comment |
$begingroup$
Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.
I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.
Please help me where is I am wrong.
elementary-set-theory cardinals
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
$endgroup$
$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18
$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19
$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47
$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02
add a comment |
$begingroup$
Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.
I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.
Please help me where is I am wrong.
elementary-set-theory cardinals
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
$endgroup$
Let $S={,f,|,fcolon Atomathbb N}$, where $A={1,2}$.
I thought cardinality of $S$ is $2^{|mathbb N|}=aleph_0$. But my friend told that my answer was wrong.
Please help me where is I am wrong.
elementary-set-theory cardinals
elementary-set-theory cardinals
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
edited Dec 8 '18 at 12:56
Asaf Karagila♦
306k33438769
306k33438769
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
asked Dec 8 '18 at 11:07
MathLoverMathLover
54010
54010
$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18
$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19
$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47
$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02
add a comment |
$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18
$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19
$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47
$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02
$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18
$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18
$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19
$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19
$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47
$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47
$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02
$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02
add a comment |
1 Answer
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$begingroup$
The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.
In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.
In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
$endgroup$
add a comment |
$begingroup$
The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.
In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
$endgroup$
add a comment |
$begingroup$
The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.
In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
$endgroup$
The cardinality of the set of functions from a set $A$ to a set $B$ is $|B|^{|A|}$.
In particular in your case the size of $S$ is $|mathbb{N}|^2$ (not $2^{|mathbb{N}|}$) which is the size of $mathbb{N}timesmathbb{N}$ (hence countable).
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
edited Dec 8 '18 at 13:35
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
answered Dec 8 '18 at 11:14
YankoYanko
7,7901830
7,7901830
add a comment |
add a comment |
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$begingroup$
Why do you think the cardinality of $S$ would be $2^{|mathbb N|}$?
$endgroup$
– Henrik
Dec 8 '18 at 11:18
$begingroup$
Because for any function to defined every point has 2 choices so there are 2^|N|.I thought
$endgroup$
– MathLover
Dec 8 '18 at 11:19
$begingroup$
Note that the set $S = { f$ a function$ : f:mathbb{N}to{1,2}}$ is uncountable. To see why, think about $mathcal{P}(mathbb{N})$ and $f^{-1}(1)$ for $fin S$.
$endgroup$
– Santana Afton
Dec 8 '18 at 12:47
$begingroup$
You could just list down all the functions in your set and see what the answer is.
$endgroup$
– Andrés E. Caicedo
Dec 8 '18 at 14:02