Countability of set of all functions with domain $mathbb{Z}$ and codomain $mathbb{R}$
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I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:
elementary-set-theory
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
$endgroup$
add a comment |
$begingroup$
I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:
elementary-set-theory
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
$endgroup$
$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25
add a comment |
$begingroup$
I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:
elementary-set-theory
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
$endgroup$
I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:
elementary-set-theory
elementary-set-theory
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
asked Nov 29 '18 at 23:22
darylnakdarylnak
167111
167111
$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25
add a comment |
$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25
$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25
$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).
Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
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1
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
1
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
add a comment |
$begingroup$
Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.
Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
$$
|mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
$$
More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).
Actually, also the converse inequality holds, in your case:
$$
|mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
2^{aleph_0}=|mathbb{R}|
$$
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
$endgroup$
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
$begingroup$
Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).
Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
$endgroup$
1
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
1
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
add a comment |
$begingroup$
Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).
Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
$endgroup$
1
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
1
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
add a comment |
$begingroup$
Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).
Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
$endgroup$
Because $mathbb{N}$ and $mathbb{Z}$ have the same cardinality, the space of functions from $mathbb{Z}$ to $mathbb{R}$ has the same size as the space of functions from $mathbb{N}$ to $mathbb{R}$ (this is the space of all real-valued-sequences).
Cantor's diagonality argument gives us that the space of all ${0,1}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
answered Nov 29 '18 at 23:29
Alberto TakaseAlberto Takase
2,083418
2,083418
1
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
1
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
add a comment |
1
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
1
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
1
1
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
$begingroup$
You might not even need to go this far. The set of all constant functions $mathbb{Z} to mathbb{R}$ is already uncountable.
$endgroup$
– platty
Nov 29 '18 at 23:31
1
1
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
$begingroup$
I agree. It is because $mathbb{R}$ is uncountable. This is typically proven via Cantor's diagonality argument. Both have merits because of the common lemma with Cantor's diagonality argument.
$endgroup$
– Alberto Takase
Nov 29 '18 at 23:35
add a comment |
$begingroup$
Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.
Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
$$
|mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
$$
More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).
Actually, also the converse inequality holds, in your case:
$$
|mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
2^{aleph_0}=|mathbb{R}|
$$
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.
Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
$$
|mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
$$
More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).
Actually, also the converse inequality holds, in your case:
$$
|mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
2^{aleph_0}=|mathbb{R}|
$$
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.
Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
$$
|mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
$$
More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).
Actually, also the converse inequality holds, in your case:
$$
|mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
2^{aleph_0}=|mathbb{R}|
$$
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
$endgroup$
Your proof would be complete if you actually show that the set has cardinality $ge|mathbb{R}|$. This is essentially obvious: for every $rinmathbb{R}$, define $bar{r}colonmathbb{Z}tomathbb{R}$ to be the constant function $bar{r}(x)=r$.
Then $rmapstobar{r}$ is an injective map $mathbb{R}to mathbb{R}^{mathbb{Z}}$, proving that
$$
|mathbb{R}|le|mathbb{R}^{mathbb{Z}}|
$$
More generally, $|X|le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Yto X$).
Actually, also the converse inequality holds, in your case:
$$
|mathbb{R}^{mathbb{Z}}|=(2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=
2^{aleph_0}=|mathbb{R}|
$$
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
answered Nov 29 '18 at 23:32
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
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$begingroup$
A set is infinite if its cardinality is equal to $|Z^+|$, still countable but infinite.
$endgroup$
– gd1035
Nov 29 '18 at 23:25