Mixing
Easy question, but what is a subset of $mathbb R^n$?
$begingroup$
I recognize $mathbb R^n$ to be every single possible vector in $mathbb R^n$, but a subset is any of those points right? So technically a subset of $mathbb R^2$ can be just {[0, 1]}, right?
Well then it's pretty obvious that any subspace of $mathbb R^n$ will be a subset of $mathbb R^n$, but how come we have to state at the start of any "proving it's a subspace" proof by stating that it is a subset of $mathbb R^n$?
Also am I correct in thinking that a subspace is just a span of a particular "dimension in $mathbb R^n$"? So a subspace is either a line, plane, hyperplane, etc.?
linear-algebra
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
asked Feb 2 at 22:20
mingming
4606
$endgroup$
add a comment |
$begingroup$
I recognize $mathbb R^n$ to be every single possible vector in $mathbb R^n$, but a subset is any of those points right? So technically a subset of $mathbb R^2$ can be just {[0, 1]}, right?
Well then it's pretty obvious that any subspace of $mathbb R^n$ will be a subset of $mathbb R^n$, but how come we have to state at the start of any "proving it's a subspace" proof by stating that it is a subset of $mathbb R^n$?
Also am I correct in thinking that a subspace is just a span of a particular "dimension in $mathbb R^n$"? So a subspace is either a line, plane, hyperplane, etc.?
linear-algebra
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
asked Feb 2 at 22:20
mingming
4606
$endgroup$
1
$begingroup$
The subset has to have the same kind of elements. So ${(0,y) in mathbb{R}^2 : 0leq y leq 1}$ is a subset of $mathbb{R}^2$, but the unit interval $[0,1]$ is not, because all points in the unit interval have only one dimension. For example $1/2 notin mathbb{R}^2$. But $(0,1/2) in mathbb{R}^2$. ANd ${(0,1/2)} subseteq mathbb{R}^2$. I am assuming by $[0,1]$ you mean the set of all points $x in mathbb{R}$ such taht $0leq x leq 1$.
$endgroup$
– Michael
Feb 2 at 22:28
1
$begingroup$
Usually it is a requirement that if U is a subspace then it must be a subset. But in general this isn't much of anything to show. Sometimes people who haven't seen linear algebra before will think that a two dimensional subspace of $mathbb{R}^{3}$ is a subspace of $mathbb{R}^{2}$ (because it's two-dimensional...) but at some point this part is barely worth mentioning.
$endgroup$
– Morgan Rodgers
Feb 2 at 22:39
$begingroup$
@Michael by [0, 1], I mean the vector [0, 1] being the only vector in the "subset". If that is what the definition of subset means, just any set in $R^n$. So the whole subset, the set { [0, 1] }, is a set that contains just one vector, [0, 1]. So you're saying this wouldn't be considered a subset in $R^n$ because.. well I'm a little confused why not. What do you mean by saying the "unit interval [0, 1]"? Why can't I just say $y = 0$ instead of $0 leq y leq 1$
$endgroup$
– ming
Feb 2 at 23:28
$begingroup$
@ming You can find the definition of 'subset' here en.m.wikipedia.org/wiki/Subset
$endgroup$
– AnyAD
Feb 3 at 5:03
$begingroup$
@ming : A subset of $mathbb{R}^n$ is any set of points in $mathbb{R}^n$ (the subset can have just one point, or 12, or it can be the whole of $mathbb{R}^n$ itself, it can also have no points and that is called the empty set). The comment history above already makes it clear that what I thought you meant by $[0,1]$ was the interval of real numbers between $0$ and $1$, but that is not what you really meant.
$endgroup$
– Michael
Feb 3 at 6:20
add a comment |
$begingroup$
I recognize $mathbb R^n$ to be every single possible vector in $mathbb R^n$, but a subset is any of those points right? So technically a subset of $mathbb R^2$ can be just {[0, 1]}, right?
Well then it's pretty obvious that any subspace of $mathbb R^n$ will be a subset of $mathbb R^n$, but how come we have to state at the start of any "proving it's a subspace" proof by stating that it is a subset of $mathbb R^n$?
Also am I correct in thinking that a subspace is just a span of a particular "dimension in $mathbb R^n$"? So a subspace is either a line, plane, hyperplane, etc.?
linear-algebra
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
asked Feb 2 at 22:20
mingming
4606
$endgroup$
I recognize $mathbb R^n$ to be every single possible vector in $mathbb R^n$, but a subset is any of those points right? So technically a subset of $mathbb R^2$ can be just {[0, 1]}, right?
Well then it's pretty obvious that any subspace of $mathbb R^n$ will be a subset of $mathbb R^n$, but how come we have to state at the start of any "proving it's a subspace" proof by stating that it is a subset of $mathbb R^n$?
Also am I correct in thinking that a subspace is just a span of a particular "dimension in $mathbb R^n$"? So a subspace is either a line, plane, hyperplane, etc.?
linear-algebra
linear-algebra
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
asked Feb 2 at 22:20
mingming
4606
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
asked Feb 2 at 22:20
mingming
4606
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
edited Feb 3 at 0:13
J. W. Tanner
4,7871420
4,7871420
asked Feb 2 at 22:20
mingming
4606
asked Feb 2 at 22:20
mingming
4606
asked Feb 2 at 22:20
mingming
4606
4606
1
$begingroup$
The subset has to have the same kind of elements. So ${(0,y) in mathbb{R}^2 : 0leq y leq 1}$ is a subset of $mathbb{R}^2$, but the unit interval $[0,1]$ is not, because all points in the unit interval have only one dimension. For example $1/2 notin mathbb{R}^2$. But $(0,1/2) in mathbb{R}^2$. ANd ${(0,1/2)} subseteq mathbb{R}^2$. I am assuming by $[0,1]$ you mean the set of all points $x in mathbb{R}$ such taht $0leq x leq 1$.
$endgroup$
– Michael
Feb 2 at 22:28
1
$begingroup$
Usually it is a requirement that if U is a subspace then it must be a subset. But in general this isn't much of anything to show. Sometimes people who haven't seen linear algebra before will think that a two dimensional subspace of $mathbb{R}^{3}$ is a subspace of $mathbb{R}^{2}$ (because it's two-dimensional...) but at some point this part is barely worth mentioning.
$endgroup$
– Morgan Rodgers
Feb 2 at 22:39
$begingroup$
@Michael by [0, 1], I mean the vector [0, 1] being the only vector in the "subset". If that is what the definition of subset means, just any set in $R^n$. So the whole subset, the set { [0, 1] }, is a set that contains just one vector, [0, 1]. So you're saying this wouldn't be considered a subset in $R^n$ because.. well I'm a little confused why not. What do you mean by saying the "unit interval [0, 1]"? Why can't I just say $y = 0$ instead of $0 leq y leq 1$
$endgroup$
– ming
Feb 2 at 23:28
$begingroup$
@ming You can find the definition of 'subset' here en.m.wikipedia.org/wiki/Subset
$endgroup$
– AnyAD
Feb 3 at 5:03
$begingroup$
@ming : A subset of $mathbb{R}^n$ is any set of points in $mathbb{R}^n$ (the subset can have just one point, or 12, or it can be the whole of $mathbb{R}^n$ itself, it can also have no points and that is called the empty set). The comment history above already makes it clear that what I thought you meant by $[0,1]$ was the interval of real numbers between $0$ and $1$, but that is not what you really meant.
$endgroup$
– Michael
Feb 3 at 6:20
add a comment |
1
$begingroup$
The subset has to have the same kind of elements. So ${(0,y) in mathbb{R}^2 : 0leq y leq 1}$ is a subset of $mathbb{R}^2$, but the unit interval $[0,1]$ is not, because all points in the unit interval have only one dimension. For example $1/2 notin mathbb{R}^2$. But $(0,1/2) in mathbb{R}^2$. ANd ${(0,1/2)} subseteq mathbb{R}^2$. I am assuming by $[0,1]$ you mean the set of all points $x in mathbb{R}$ such taht $0leq x leq 1$.
$endgroup$
– Michael
Feb 2 at 22:28
1
$begingroup$
Usually it is a requirement that if U is a subspace then it must be a subset. But in general this isn't much of anything to show. Sometimes people who haven't seen linear algebra before will think that a two dimensional subspace of $mathbb{R}^{3}$ is a subspace of $mathbb{R}^{2}$ (because it's two-dimensional...) but at some point this part is barely worth mentioning.
$endgroup$
– Morgan Rodgers
Feb 2 at 22:39
$begingroup$
@Michael by [0, 1], I mean the vector [0, 1] being the only vector in the "subset". If that is what the definition of subset means, just any set in $R^n$. So the whole subset, the set { [0, 1] }, is a set that contains just one vector, [0, 1]. So you're saying this wouldn't be considered a subset in $R^n$ because.. well I'm a little confused why not. What do you mean by saying the "unit interval [0, 1]"? Why can't I just say $y = 0$ instead of $0 leq y leq 1$
$endgroup$
– ming
Feb 2 at 23:28
$begingroup$
@ming You can find the definition of 'subset' here en.m.wikipedia.org/wiki/Subset
$endgroup$
– AnyAD
Feb 3 at 5:03
$begingroup$
@ming : A subset of $mathbb{R}^n$ is any set of points in $mathbb{R}^n$ (the subset can have just one point, or 12, or it can be the whole of $mathbb{R}^n$ itself, it can also have no points and that is called the empty set). The comment history above already makes it clear that what I thought you meant by $[0,1]$ was the interval of real numbers between $0$ and $1$, but that is not what you really meant.
$endgroup$
– Michael
Feb 3 at 6:20
1
1
$begingroup$
The subset has to have the same kind of elements. So ${(0,y) in mathbb{R}^2 : 0leq y leq 1}$ is a subset of $mathbb{R}^2$, but the unit interval $[0,1]$ is not, because all points in the unit interval have only one dimension. For example $1/2 notin mathbb{R}^2$. But $(0,1/2) in mathbb{R}^2$. ANd ${(0,1/2)} subseteq mathbb{R}^2$. I am assuming by $[0,1]$ you mean the set of all points $x in mathbb{R}$ such taht $0leq x leq 1$.
$endgroup$
– Michael
Feb 2 at 22:28
$begingroup$
The subset has to have the same kind of elements. So ${(0,y) in mathbb{R}^2 : 0leq y leq 1}$ is a subset of $mathbb{R}^2$, but the unit interval $[0,1]$ is not, because all points in the unit interval have only one dimension. For example $1/2 notin mathbb{R}^2$. But $(0,1/2) in mathbb{R}^2$. ANd ${(0,1/2)} subseteq mathbb{R}^2$. I am assuming by $[0,1]$ you mean the set of all points $x in mathbb{R}$ such taht $0leq x leq 1$.
$endgroup$
– Michael
Feb 2 at 22:28
1
1
$begingroup$
Usually it is a requirement that if U is a subspace then it must be a subset. But in general this isn't much of anything to show. Sometimes people who haven't seen linear algebra before will think that a two dimensional subspace of $mathbb{R}^{3}$ is a subspace of $mathbb{R}^{2}$ (because it's two-dimensional...) but at some point this part is barely worth mentioning.
$endgroup$
– Morgan Rodgers
Feb 2 at 22:39
$begingroup$
Usually it is a requirement that if U is a subspace then it must be a subset. But in general this isn't much of anything to show. Sometimes people who haven't seen linear algebra before will think that a two dimensional subspace of $mathbb{R}^{3}$ is a subspace of $mathbb{R}^{2}$ (because it's two-dimensional...) but at some point this part is barely worth mentioning.
$endgroup$
– Morgan Rodgers
Feb 2 at 22:39
$begingroup$
@Michael by [0, 1], I mean the vector [0, 1] being the only vector in the "subset". If that is what the definition of subset means, just any set in $R^n$. So the whole subset, the set { [0, 1] }, is a set that contains just one vector, [0, 1]. So you're saying this wouldn't be considered a subset in $R^n$ because.. well I'm a little confused why not. What do you mean by saying the "unit interval [0, 1]"? Why can't I just say $y = 0$ instead of $0 leq y leq 1$
$endgroup$
– ming
Feb 2 at 23:28
$begingroup$
@Michael by [0, 1], I mean the vector [0, 1] being the only vector in the "subset". If that is what the definition of subset means, just any set in $R^n$. So the whole subset, the set { [0, 1] }, is a set that contains just one vector, [0, 1]. So you're saying this wouldn't be considered a subset in $R^n$ because.. well I'm a little confused why not. What do you mean by saying the "unit interval [0, 1]"? Why can't I just say $y = 0$ instead of $0 leq y leq 1$
$endgroup$
– ming
Feb 2 at 23:28
$begingroup$
@ming You can find the definition of 'subset' here en.m.wikipedia.org/wiki/Subset
$endgroup$
– AnyAD
Feb 3 at 5:03
$begingroup$
@ming You can find the definition of 'subset' here en.m.wikipedia.org/wiki/Subset
$endgroup$
– AnyAD
Feb 3 at 5:03
$begingroup$
@ming : A subset of $mathbb{R}^n$ is any set of points in $mathbb{R}^n$ (the subset can have just one point, or 12, or it can be the whole of $mathbb{R}^n$ itself, it can also have no points and that is called the empty set). The comment history above already makes it clear that what I thought you meant by $[0,1]$ was the interval of real numbers between $0$ and $1$, but that is not what you really meant.
$endgroup$
– Michael
Feb 3 at 6:20
$begingroup$
@ming : A subset of $mathbb{R}^n$ is any set of points in $mathbb{R}^n$ (the subset can have just one point, or 12, or it can be the whole of $mathbb{R}^n$ itself, it can also have no points and that is called the empty set). The comment history above already makes it clear that what I thought you meant by $[0,1]$ was the interval of real numbers between $0$ and $1$, but that is not what you really meant.
$endgroup$
– Michael
Feb 3 at 6:20
add a comment |
1 Answer
1
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$begingroup$
Starting with the set of $n$-tuples, and defining the operations of vector addition and scalar multiplication on the set (in the 'usual' way) of these $n$-tuples, we get the vector space $mathbb{R}^n$.
Any 'collection' of elements from this set is by definition a subset of $mathbb{R}^n$. It is also a subspace of $mathbb{R}^n$ iff it satisfies certain other conditions (is a vector space when considered on its own).
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
$endgroup$
add a comment |
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$begingroup$
Starting with the set of $n$-tuples, and defining the operations of vector addition and scalar multiplication on the set (in the 'usual' way) of these $n$-tuples, we get the vector space $mathbb{R}^n$.
Any 'collection' of elements from this set is by definition a subset of $mathbb{R}^n$. It is also a subspace of $mathbb{R}^n$ iff it satisfies certain other conditions (is a vector space when considered on its own).
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
$endgroup$
add a comment |
$begingroup$
Starting with the set of $n$-tuples, and defining the operations of vector addition and scalar multiplication on the set (in the 'usual' way) of these $n$-tuples, we get the vector space $mathbb{R}^n$.
Any 'collection' of elements from this set is by definition a subset of $mathbb{R}^n$. It is also a subspace of $mathbb{R}^n$ iff it satisfies certain other conditions (is a vector space when considered on its own).
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
$endgroup$
add a comment |
$begingroup$
Starting with the set of $n$-tuples, and defining the operations of vector addition and scalar multiplication on the set (in the 'usual' way) of these $n$-tuples, we get the vector space $mathbb{R}^n$.
Any 'collection' of elements from this set is by definition a subset of $mathbb{R}^n$. It is also a subspace of $mathbb{R}^n$ iff it satisfies certain other conditions (is a vector space when considered on its own).
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
$endgroup$
Starting with the set of $n$-tuples, and defining the operations of vector addition and scalar multiplication on the set (in the 'usual' way) of these $n$-tuples, we get the vector space $mathbb{R}^n$.
Any 'collection' of elements from this set is by definition a subset of $mathbb{R}^n$. It is also a subspace of $mathbb{R}^n$ iff it satisfies certain other conditions (is a vector space when considered on its own).
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
answered Feb 2 at 22:31
AnyADAnyAD
2,113812
2,113812
add a comment |
add a comment |
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$begingroup$
The subset has to have the same kind of elements. So ${(0,y) in mathbb{R}^2 : 0leq y leq 1}$ is a subset of $mathbb{R}^2$, but the unit interval $[0,1]$ is not, because all points in the unit interval have only one dimension. For example $1/2 notin mathbb{R}^2$. But $(0,1/2) in mathbb{R}^2$. ANd ${(0,1/2)} subseteq mathbb{R}^2$. I am assuming by $[0,1]$ you mean the set of all points $x in mathbb{R}$ such taht $0leq x leq 1$.
$endgroup$
– Michael
Feb 2 at 22:28
1
$begingroup$
Usually it is a requirement that if U is a subspace then it must be a subset. But in general this isn't much of anything to show. Sometimes people who haven't seen linear algebra before will think that a two dimensional subspace of $mathbb{R}^{3}$ is a subspace of $mathbb{R}^{2}$ (because it's two-dimensional...) but at some point this part is barely worth mentioning.
$endgroup$
– Morgan Rodgers
Feb 2 at 22:39
$begingroup$
@Michael by [0, 1], I mean the vector [0, 1] being the only vector in the "subset". If that is what the definition of subset means, just any set in $R^n$. So the whole subset, the set { [0, 1] }, is a set that contains just one vector, [0, 1]. So you're saying this wouldn't be considered a subset in $R^n$ because.. well I'm a little confused why not. What do you mean by saying the "unit interval [0, 1]"? Why can't I just say $y = 0$ instead of $0 leq y leq 1$
$endgroup$
– ming
Feb 2 at 23:28
$begingroup$
@ming You can find the definition of 'subset' here en.m.wikipedia.org/wiki/Subset
$endgroup$
– AnyAD
Feb 3 at 5:03
$begingroup$
@ming : A subset of $mathbb{R}^n$ is any set of points in $mathbb{R}^n$ (the subset can have just one point, or 12, or it can be the whole of $mathbb{R}^n$ itself, it can also have no points and that is called the empty set). The comment history above already makes it clear that what I thought you meant by $[0,1]$ was the interval of real numbers between $0$ and $1$, but that is not what you really meant.
$endgroup$
– Michael
Feb 3 at 6:20