What is $mathbb R^omega$?
$begingroup$
I have seen $mathbb R^omega$ mentioned in my topology texts but cannot find where $omega$ is defined. Could someone please tell me what it means in comparison to $mathbb R^n$?
general-topology notation
edited Jan 2 at 12:09
Wojowu
19.4k23274
asked Apr 16 '14 at 14:57
McTMcT
241314
$endgroup$
add a comment |
$begingroup$
I have seen $mathbb R^omega$ mentioned in my topology texts but cannot find where $omega$ is defined. Could someone please tell me what it means in comparison to $mathbb R^n$?
general-topology notation
edited Jan 2 at 12:09
Wojowu
19.4k23274
asked Apr 16 '14 at 14:57
McTMcT
241314
$endgroup$
6
$begingroup$
$omega$ is the smallest infinite ordinal. When you're not considering the ordinal aspects, $mathbb{N}$ is a commonly used name. $mathbb{R}^omega$ is the space of all real sequences, the product of countably many copies of $mathbb{R}$.
$endgroup$
– Daniel Fischer
Apr 16 '14 at 15:05
$begingroup$
See also math.stackexchange.com/questions/616651/….
$endgroup$
– lhf
Apr 16 '14 at 15:09
add a comment |
$begingroup$
I have seen $mathbb R^omega$ mentioned in my topology texts but cannot find where $omega$ is defined. Could someone please tell me what it means in comparison to $mathbb R^n$?
general-topology notation
edited Jan 2 at 12:09
Wojowu
19.4k23274
asked Apr 16 '14 at 14:57
McTMcT
241314
$endgroup$
I have seen $mathbb R^omega$ mentioned in my topology texts but cannot find where $omega$ is defined. Could someone please tell me what it means in comparison to $mathbb R^n$?
general-topology notation
general-topology notation
edited Jan 2 at 12:09
Wojowu
19.4k23274
asked Apr 16 '14 at 14:57
McTMcT
241314
edited Jan 2 at 12:09
Wojowu
19.4k23274
asked Apr 16 '14 at 14:57
McTMcT
241314
edited Jan 2 at 12:09
Wojowu
19.4k23274
edited Jan 2 at 12:09
Wojowu
19.4k23274
edited Jan 2 at 12:09
Wojowu
19.4k23274
19.4k23274
asked Apr 16 '14 at 14:57
McTMcT
241314
asked Apr 16 '14 at 14:57
McTMcT
241314
asked Apr 16 '14 at 14:57
McTMcT
241314
241314
6
$begingroup$
$omega$ is the smallest infinite ordinal. When you're not considering the ordinal aspects, $mathbb{N}$ is a commonly used name. $mathbb{R}^omega$ is the space of all real sequences, the product of countably many copies of $mathbb{R}$.
$endgroup$
– Daniel Fischer
Apr 16 '14 at 15:05
$begingroup$
See also math.stackexchange.com/questions/616651/….
$endgroup$
– lhf
Apr 16 '14 at 15:09
add a comment |
6
$begingroup$
$omega$ is the smallest infinite ordinal. When you're not considering the ordinal aspects, $mathbb{N}$ is a commonly used name. $mathbb{R}^omega$ is the space of all real sequences, the product of countably many copies of $mathbb{R}$.
$endgroup$
– Daniel Fischer
Apr 16 '14 at 15:05
$begingroup$
See also math.stackexchange.com/questions/616651/….
$endgroup$
– lhf
Apr 16 '14 at 15:09
6
6
$begingroup$
$omega$ is the smallest infinite ordinal. When you're not considering the ordinal aspects, $mathbb{N}$ is a commonly used name. $mathbb{R}^omega$ is the space of all real sequences, the product of countably many copies of $mathbb{R}$.
$endgroup$
– Daniel Fischer
Apr 16 '14 at 15:05
$begingroup$
$omega$ is the smallest infinite ordinal. When you're not considering the ordinal aspects, $mathbb{N}$ is a commonly used name. $mathbb{R}^omega$ is the space of all real sequences, the product of countably many copies of $mathbb{R}$.
$endgroup$
– Daniel Fischer
Apr 16 '14 at 15:05
$begingroup$
See also math.stackexchange.com/questions/616651/….
$endgroup$
– lhf
Apr 16 '14 at 15:09
$begingroup$
See also math.stackexchange.com/questions/616651/….
$endgroup$
– lhf
Apr 16 '14 at 15:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
tl; dr;
$$X^omega = Xtimes Xtimescdots$$
is an infinite countable product of $X$, i.e. the set of all countable sequences over $X$.
Longer explanation: so $omega$ is an ordinal number defined as the ordinal number of naturals $mathbb{N}$. If we use the Von Neumann definition of ordinals then for an ordinal number $k$ we have a very straight forward generalization of finite Cartesian product:
$$X^k={f:kto X | ftext{ is a function}}$$
and so you can think about $X^k$ as the set of all ordered "sequences" with values in $X$. For finite $n$ this coincides with classical finite product because $n$ as a Von Neumann ordinal is recursively defined as ${0,1,2,ldots,n-1}$ with $0=emptyset$. For $omega$ (which in Von Neumann case is simply $mathbb{N}$) this coincides with the usual notion of sequences. For uncountable ordinals this probably goes beyond any intuition. Nevertheless typically by $X^omega$ people understand the countable infinite product $Xtimes Xtimescdots$.
Note that there are alternative (and arguably simplier) definitions of an infinite Cartesian product, e.g. the set of all choice functions.
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
$endgroup$
add a comment |
$begingroup$
The cardinality of $Bbb N$, which is usually written $aleph_0$ (the smallest infinite cardinal), is sometimes/often also written $omega$. In other words
$$|Bbb N| = aleph_0 = omega$$
Hence $Bbb R^{omega} = Bbb R times Bbb R times Bbb R times dots$ is the cartesian product of countably infinitely many $Bbb R$.
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
$endgroup$
1
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
1
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
1
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
1
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
tl; dr;
$$X^omega = Xtimes Xtimescdots$$
is an infinite countable product of $X$, i.e. the set of all countable sequences over $X$.
Longer explanation: so $omega$ is an ordinal number defined as the ordinal number of naturals $mathbb{N}$. If we use the Von Neumann definition of ordinals then for an ordinal number $k$ we have a very straight forward generalization of finite Cartesian product:
$$X^k={f:kto X | ftext{ is a function}}$$
and so you can think about $X^k$ as the set of all ordered "sequences" with values in $X$. For finite $n$ this coincides with classical finite product because $n$ as a Von Neumann ordinal is recursively defined as ${0,1,2,ldots,n-1}$ with $0=emptyset$. For $omega$ (which in Von Neumann case is simply $mathbb{N}$) this coincides with the usual notion of sequences. For uncountable ordinals this probably goes beyond any intuition. Nevertheless typically by $X^omega$ people understand the countable infinite product $Xtimes Xtimescdots$.
Note that there are alternative (and arguably simplier) definitions of an infinite Cartesian product, e.g. the set of all choice functions.
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
$endgroup$
add a comment |
$begingroup$
tl; dr;
$$X^omega = Xtimes Xtimescdots$$
is an infinite countable product of $X$, i.e. the set of all countable sequences over $X$.
Longer explanation: so $omega$ is an ordinal number defined as the ordinal number of naturals $mathbb{N}$. If we use the Von Neumann definition of ordinals then for an ordinal number $k$ we have a very straight forward generalization of finite Cartesian product:
$$X^k={f:kto X | ftext{ is a function}}$$
and so you can think about $X^k$ as the set of all ordered "sequences" with values in $X$. For finite $n$ this coincides with classical finite product because $n$ as a Von Neumann ordinal is recursively defined as ${0,1,2,ldots,n-1}$ with $0=emptyset$. For $omega$ (which in Von Neumann case is simply $mathbb{N}$) this coincides with the usual notion of sequences. For uncountable ordinals this probably goes beyond any intuition. Nevertheless typically by $X^omega$ people understand the countable infinite product $Xtimes Xtimescdots$.
Note that there are alternative (and arguably simplier) definitions of an infinite Cartesian product, e.g. the set of all choice functions.
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
$endgroup$
add a comment |
$begingroup$
tl; dr;
$$X^omega = Xtimes Xtimescdots$$
is an infinite countable product of $X$, i.e. the set of all countable sequences over $X$.
Longer explanation: so $omega$ is an ordinal number defined as the ordinal number of naturals $mathbb{N}$. If we use the Von Neumann definition of ordinals then for an ordinal number $k$ we have a very straight forward generalization of finite Cartesian product:
$$X^k={f:kto X | ftext{ is a function}}$$
and so you can think about $X^k$ as the set of all ordered "sequences" with values in $X$. For finite $n$ this coincides with classical finite product because $n$ as a Von Neumann ordinal is recursively defined as ${0,1,2,ldots,n-1}$ with $0=emptyset$. For $omega$ (which in Von Neumann case is simply $mathbb{N}$) this coincides with the usual notion of sequences. For uncountable ordinals this probably goes beyond any intuition. Nevertheless typically by $X^omega$ people understand the countable infinite product $Xtimes Xtimescdots$.
Note that there are alternative (and arguably simplier) definitions of an infinite Cartesian product, e.g. the set of all choice functions.
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
$endgroup$
tl; dr;
$$X^omega = Xtimes Xtimescdots$$
is an infinite countable product of $X$, i.e. the set of all countable sequences over $X$.
Longer explanation: so $omega$ is an ordinal number defined as the ordinal number of naturals $mathbb{N}$. If we use the Von Neumann definition of ordinals then for an ordinal number $k$ we have a very straight forward generalization of finite Cartesian product:
$$X^k={f:kto X | ftext{ is a function}}$$
and so you can think about $X^k$ as the set of all ordered "sequences" with values in $X$. For finite $n$ this coincides with classical finite product because $n$ as a Von Neumann ordinal is recursively defined as ${0,1,2,ldots,n-1}$ with $0=emptyset$. For $omega$ (which in Von Neumann case is simply $mathbb{N}$) this coincides with the usual notion of sequences. For uncountable ordinals this probably goes beyond any intuition. Nevertheless typically by $X^omega$ people understand the countable infinite product $Xtimes Xtimescdots$.
Note that there are alternative (and arguably simplier) definitions of an infinite Cartesian product, e.g. the set of all choice functions.
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
edited Jan 2 at 12:16
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
answered Jan 2 at 11:36
freakishfreakish
13.1k1630
13.1k1630
add a comment |
add a comment |
$begingroup$
The cardinality of $Bbb N$, which is usually written $aleph_0$ (the smallest infinite cardinal), is sometimes/often also written $omega$. In other words
$$|Bbb N| = aleph_0 = omega$$
Hence $Bbb R^{omega} = Bbb R times Bbb R times Bbb R times dots$ is the cartesian product of countably infinitely many $Bbb R$.
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
$endgroup$
1
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
1
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
1
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
1
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
|
show 1 more comment
$begingroup$
The cardinality of $Bbb N$, which is usually written $aleph_0$ (the smallest infinite cardinal), is sometimes/often also written $omega$. In other words
$$|Bbb N| = aleph_0 = omega$$
Hence $Bbb R^{omega} = Bbb R times Bbb R times Bbb R times dots$ is the cartesian product of countably infinitely many $Bbb R$.
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
$endgroup$
1
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
1
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
1
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
1
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
|
show 1 more comment
$begingroup$
The cardinality of $Bbb N$, which is usually written $aleph_0$ (the smallest infinite cardinal), is sometimes/often also written $omega$. In other words
$$|Bbb N| = aleph_0 = omega$$
Hence $Bbb R^{omega} = Bbb R times Bbb R times Bbb R times dots$ is the cartesian product of countably infinitely many $Bbb R$.
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
$endgroup$
The cardinality of $Bbb N$, which is usually written $aleph_0$ (the smallest infinite cardinal), is sometimes/often also written $omega$. In other words
$$|Bbb N| = aleph_0 = omega$$
Hence $Bbb R^{omega} = Bbb R times Bbb R times Bbb R times dots$ is the cartesian product of countably infinitely many $Bbb R$.
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
answered Apr 16 '14 at 15:10
naslundxnaslundx
7,99952941
7,99952941
1
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
1
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
1
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
1
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
|
show 1 more comment
1
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
1
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
1
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
1
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
1
1
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
$begingroup$
$omega$ is not a cardinal number. It's an ordinal number. And $aleph_0=omega$ is simply wrong.
$endgroup$
– freakish
Jan 2 at 11:24
1
1
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
$begingroup$
That is not true
$endgroup$
– José Alejandro Aburto Araneda
Jan 2 at 11:27
1
1
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@freakish In set theory, cardinal numbers are often identified with corresponding initial ordinals. This is not "simply wrong", but at best it's misleading
$endgroup$
– Wojowu
Jan 2 at 12:11
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
$begingroup$
@Wojowu These are apples and oranges. Perhaps there are people who identify them (how?) but they are simply wrong. It's like saying that there are people identifying groups with their underlying sets. No, this is wrong. Ordinals have much richer structure.
$endgroup$
– freakish
Jan 2 at 12:13
1
1
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
$begingroup$
@freakish Let me say more - some people define cardinal numbers as initial ordinals - for instance Halmos or Jech. This is different from what is done with groups and underlying sets.
$endgroup$
– Wojowu
Jan 2 at 12:21
|
show 1 more comment
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6
$begingroup$
$omega$ is the smallest infinite ordinal. When you're not considering the ordinal aspects, $mathbb{N}$ is a commonly used name. $mathbb{R}^omega$ is the space of all real sequences, the product of countably many copies of $mathbb{R}$.
$endgroup$
– Daniel Fischer
Apr 16 '14 at 15:05
$begingroup$
See also math.stackexchange.com/questions/616651/….
$endgroup$
– lhf
Apr 16 '14 at 15:09