Can Someone Explain the Set ${𝑥 ∈ ℝ^𝑁 ∣ d(x,0) = 1}$?
$begingroup$
I'm trying to understand this set. I understand most of the individual parts such as $∈$ means "an element of", $ℝ$ is all real numbers, and $d(x,0) = 1$ is the distance from $x$ to $0$, as in absolute value.
I don't understand what $x$ means in this situation though. Is it considered a variable? Is it multiple values?
I also don't understand what the actual elements of the set are. Is the value of $x$ and element in the set? Also, are there any other numbers with an absolute value of $1$ other than $1$ and $-1$? If not, are those the only possible values for $x$?
In this situation, $N$ is a number of dimensions, but I don't understand how that relates to real numbers, nor what real numbers to any power would be.
Thanks.
linear-algebra
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
asked Jan 9 at 3:29
PriyaPriya
41
$endgroup$
add a comment |
$begingroup$
I'm trying to understand this set. I understand most of the individual parts such as $∈$ means "an element of", $ℝ$ is all real numbers, and $d(x,0) = 1$ is the distance from $x$ to $0$, as in absolute value.
I don't understand what $x$ means in this situation though. Is it considered a variable? Is it multiple values?
I also don't understand what the actual elements of the set are. Is the value of $x$ and element in the set? Also, are there any other numbers with an absolute value of $1$ other than $1$ and $-1$? If not, are those the only possible values for $x$?
In this situation, $N$ is a number of dimensions, but I don't understand how that relates to real numbers, nor what real numbers to any power would be.
Thanks.
linear-algebra
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
asked Jan 9 at 3:29
PriyaPriya
41
$endgroup$
$begingroup$
What are you working with here? you have defined a subset of $R^n$ that satisfies some condition $d(x,0) = 1$... what is 'distance' here? how is this distance defined?
$endgroup$
– DavidG
Jan 12 at 9:09
add a comment |
$begingroup$
I'm trying to understand this set. I understand most of the individual parts such as $∈$ means "an element of", $ℝ$ is all real numbers, and $d(x,0) = 1$ is the distance from $x$ to $0$, as in absolute value.
I don't understand what $x$ means in this situation though. Is it considered a variable? Is it multiple values?
I also don't understand what the actual elements of the set are. Is the value of $x$ and element in the set? Also, are there any other numbers with an absolute value of $1$ other than $1$ and $-1$? If not, are those the only possible values for $x$?
In this situation, $N$ is a number of dimensions, but I don't understand how that relates to real numbers, nor what real numbers to any power would be.
Thanks.
linear-algebra
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
asked Jan 9 at 3:29
PriyaPriya
41
$endgroup$
I'm trying to understand this set. I understand most of the individual parts such as $∈$ means "an element of", $ℝ$ is all real numbers, and $d(x,0) = 1$ is the distance from $x$ to $0$, as in absolute value.
I don't understand what $x$ means in this situation though. Is it considered a variable? Is it multiple values?
I also don't understand what the actual elements of the set are. Is the value of $x$ and element in the set? Also, are there any other numbers with an absolute value of $1$ other than $1$ and $-1$? If not, are those the only possible values for $x$?
In this situation, $N$ is a number of dimensions, but I don't understand how that relates to real numbers, nor what real numbers to any power would be.
Thanks.
linear-algebra
linear-algebra
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
asked Jan 9 at 3:29
PriyaPriya
41
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
asked Jan 9 at 3:29
PriyaPriya
41
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
edited Jan 9 at 8:07
Asaf Karagila♦
305k33435766
305k33435766
asked Jan 9 at 3:29
PriyaPriya
41
asked Jan 9 at 3:29
PriyaPriya
41
asked Jan 9 at 3:29
PriyaPriya
41
41
$begingroup$
What are you working with here? you have defined a subset of $R^n$ that satisfies some condition $d(x,0) = 1$... what is 'distance' here? how is this distance defined?
$endgroup$
– DavidG
Jan 12 at 9:09
add a comment |
$begingroup$
What are you working with here? you have defined a subset of $R^n$ that satisfies some condition $d(x,0) = 1$... what is 'distance' here? how is this distance defined?
$endgroup$
– DavidG
Jan 12 at 9:09
$begingroup$
What are you working with here? you have defined a subset of $R^n$ that satisfies some condition $d(x,0) = 1$... what is 'distance' here? how is this distance defined?
$endgroup$
– DavidG
Jan 12 at 9:09
$begingroup$
What are you working with here? you have defined a subset of $R^n$ that satisfies some condition $d(x,0) = 1$... what is 'distance' here? how is this distance defined?
$endgroup$
– DavidG
Jan 12 at 9:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$x$ is just a name. Translated into English, the set
$${x in Bbb R^n | d(x,0) = 1 }$$
would read "all elements of the set $Bbb R^n$, such that the distance between that element and $0$ is $1$." (Presumably $0$ here denotes the "origin," so to speak, if you were considering graphing in $Bbb R^n$.) This is essentially the $n$-dimensional unit sphere - for example, in $Bbb R^2$, this set denotes all points on the circle of radius $1$.
The name $x$ is just a shorthand, it doesn't mean anything other than giving elements of $Bbb R^n$ some sort of name in the setbuilder notation above.
$Bbb R^n$, to answer your other question, is the $n$-dimensional analogoue of the real number line. For example, $n=2$ gives you the $xy$-plane; $n=3$ gives you typical $3D$ Cartesian space; and so on.
If you're at all familiar with the notion of Cartesian product, $Bbb R^n$ can also be expressed as
$$mathbb{R}^n = underbrace{mathbb{R} times mathbb{R} times mathbb{R} times ... times mathbb{R} times mathbb{R}}_{text{n times}}$$
and thus that allows us to express elements of $Bbb R^n$ by $n$-tuples. For example, for $n=2$, we would have our typical $(x,y)$ ordered pairs from the $xy$-plane. Generally then, for the $n$-dimensional case, you could express elements of $Bbb R^n$ by ordered $n$-tuples $(x_1, x_2, x_3, ..., x_n)$, where $x_k$ is a real number for all $k$.
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
$endgroup$
1
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
add a comment |
$begingroup$
The notation $d(x,0)=1$ seems to indicate that this is a problem in a general metric space. This is not necessarily the absolute value, as there can be more abstract metrics. This set is the unit ball in $mathbb{R}^N$ with respect to this metric.
answered Jan 9 at 5:20
Tyler6Tyler6
838413
$endgroup$
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
add a comment |
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2 Answers
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2 Answers
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active
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oldest
votes
$begingroup$
$x$ is just a name. Translated into English, the set
$${x in Bbb R^n | d(x,0) = 1 }$$
would read "all elements of the set $Bbb R^n$, such that the distance between that element and $0$ is $1$." (Presumably $0$ here denotes the "origin," so to speak, if you were considering graphing in $Bbb R^n$.) This is essentially the $n$-dimensional unit sphere - for example, in $Bbb R^2$, this set denotes all points on the circle of radius $1$.
The name $x$ is just a shorthand, it doesn't mean anything other than giving elements of $Bbb R^n$ some sort of name in the setbuilder notation above.
$Bbb R^n$, to answer your other question, is the $n$-dimensional analogoue of the real number line. For example, $n=2$ gives you the $xy$-plane; $n=3$ gives you typical $3D$ Cartesian space; and so on.
If you're at all familiar with the notion of Cartesian product, $Bbb R^n$ can also be expressed as
$$mathbb{R}^n = underbrace{mathbb{R} times mathbb{R} times mathbb{R} times ... times mathbb{R} times mathbb{R}}_{text{n times}}$$
and thus that allows us to express elements of $Bbb R^n$ by $n$-tuples. For example, for $n=2$, we would have our typical $(x,y)$ ordered pairs from the $xy$-plane. Generally then, for the $n$-dimensional case, you could express elements of $Bbb R^n$ by ordered $n$-tuples $(x_1, x_2, x_3, ..., x_n)$, where $x_k$ is a real number for all $k$.
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
$endgroup$
1
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
add a comment |
$begingroup$
$x$ is just a name. Translated into English, the set
$${x in Bbb R^n | d(x,0) = 1 }$$
would read "all elements of the set $Bbb R^n$, such that the distance between that element and $0$ is $1$." (Presumably $0$ here denotes the "origin," so to speak, if you were considering graphing in $Bbb R^n$.) This is essentially the $n$-dimensional unit sphere - for example, in $Bbb R^2$, this set denotes all points on the circle of radius $1$.
The name $x$ is just a shorthand, it doesn't mean anything other than giving elements of $Bbb R^n$ some sort of name in the setbuilder notation above.
$Bbb R^n$, to answer your other question, is the $n$-dimensional analogoue of the real number line. For example, $n=2$ gives you the $xy$-plane; $n=3$ gives you typical $3D$ Cartesian space; and so on.
If you're at all familiar with the notion of Cartesian product, $Bbb R^n$ can also be expressed as
$$mathbb{R}^n = underbrace{mathbb{R} times mathbb{R} times mathbb{R} times ... times mathbb{R} times mathbb{R}}_{text{n times}}$$
and thus that allows us to express elements of $Bbb R^n$ by $n$-tuples. For example, for $n=2$, we would have our typical $(x,y)$ ordered pairs from the $xy$-plane. Generally then, for the $n$-dimensional case, you could express elements of $Bbb R^n$ by ordered $n$-tuples $(x_1, x_2, x_3, ..., x_n)$, where $x_k$ is a real number for all $k$.
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
$endgroup$
1
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
add a comment |
$begingroup$
$x$ is just a name. Translated into English, the set
$${x in Bbb R^n | d(x,0) = 1 }$$
would read "all elements of the set $Bbb R^n$, such that the distance between that element and $0$ is $1$." (Presumably $0$ here denotes the "origin," so to speak, if you were considering graphing in $Bbb R^n$.) This is essentially the $n$-dimensional unit sphere - for example, in $Bbb R^2$, this set denotes all points on the circle of radius $1$.
The name $x$ is just a shorthand, it doesn't mean anything other than giving elements of $Bbb R^n$ some sort of name in the setbuilder notation above.
$Bbb R^n$, to answer your other question, is the $n$-dimensional analogoue of the real number line. For example, $n=2$ gives you the $xy$-plane; $n=3$ gives you typical $3D$ Cartesian space; and so on.
If you're at all familiar with the notion of Cartesian product, $Bbb R^n$ can also be expressed as
$$mathbb{R}^n = underbrace{mathbb{R} times mathbb{R} times mathbb{R} times ... times mathbb{R} times mathbb{R}}_{text{n times}}$$
and thus that allows us to express elements of $Bbb R^n$ by $n$-tuples. For example, for $n=2$, we would have our typical $(x,y)$ ordered pairs from the $xy$-plane. Generally then, for the $n$-dimensional case, you could express elements of $Bbb R^n$ by ordered $n$-tuples $(x_1, x_2, x_3, ..., x_n)$, where $x_k$ is a real number for all $k$.
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
$endgroup$
$x$ is just a name. Translated into English, the set
$${x in Bbb R^n | d(x,0) = 1 }$$
would read "all elements of the set $Bbb R^n$, such that the distance between that element and $0$ is $1$." (Presumably $0$ here denotes the "origin," so to speak, if you were considering graphing in $Bbb R^n$.) This is essentially the $n$-dimensional unit sphere - for example, in $Bbb R^2$, this set denotes all points on the circle of radius $1$.
The name $x$ is just a shorthand, it doesn't mean anything other than giving elements of $Bbb R^n$ some sort of name in the setbuilder notation above.
$Bbb R^n$, to answer your other question, is the $n$-dimensional analogoue of the real number line. For example, $n=2$ gives you the $xy$-plane; $n=3$ gives you typical $3D$ Cartesian space; and so on.
If you're at all familiar with the notion of Cartesian product, $Bbb R^n$ can also be expressed as
$$mathbb{R}^n = underbrace{mathbb{R} times mathbb{R} times mathbb{R} times ... times mathbb{R} times mathbb{R}}_{text{n times}}$$
and thus that allows us to express elements of $Bbb R^n$ by $n$-tuples. For example, for $n=2$, we would have our typical $(x,y)$ ordered pairs from the $xy$-plane. Generally then, for the $n$-dimensional case, you could express elements of $Bbb R^n$ by ordered $n$-tuples $(x_1, x_2, x_3, ..., x_n)$, where $x_k$ is a real number for all $k$.
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
edited Jan 9 at 7:36
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
answered Jan 9 at 3:33
Eevee TrainerEevee Trainer
6,62311237
6,62311237
1
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
add a comment |
1
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
1
1
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Might be worth mentioning that $Bbb R^n$ is the cartesian product $$underbrace{Bbb RtimesBbb Rtimes...timesBbb R}_{text{n times}}$$ consisting of $n-$tuples with real entries.
$endgroup$
– Shubham Johri
Jan 9 at 7:29
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Thanks for catching that, no idea how I overlooked elaborating on it. ^_^
$endgroup$
– Eevee Trainer
Jan 9 at 7:30
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
$begingroup$
Why does $d(x,0)$ refer to the unit sphere? I can't see a definition for $d(x,0)$ and more so if the set is indeed a metric space, inner produce space, normed space. It seems you have inferred it's in reference to the Euclidean distance which is defined via an induced norm (via the Euclidean Inner Product Space). If that turns out to be correct, your solution is spot on. If not, this could potentially be misleading.
$endgroup$
– DavidG
Jan 12 at 9:08
add a comment |
$begingroup$
The notation $d(x,0)=1$ seems to indicate that this is a problem in a general metric space. This is not necessarily the absolute value, as there can be more abstract metrics. This set is the unit ball in $mathbb{R}^N$ with respect to this metric.
answered Jan 9 at 5:20
Tyler6Tyler6
838413
$endgroup$
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
add a comment |
$begingroup$
The notation $d(x,0)=1$ seems to indicate that this is a problem in a general metric space. This is not necessarily the absolute value, as there can be more abstract metrics. This set is the unit ball in $mathbb{R}^N$ with respect to this metric.
answered Jan 9 at 5:20
Tyler6Tyler6
838413
$endgroup$
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
add a comment |
$begingroup$
The notation $d(x,0)=1$ seems to indicate that this is a problem in a general metric space. This is not necessarily the absolute value, as there can be more abstract metrics. This set is the unit ball in $mathbb{R}^N$ with respect to this metric.
answered Jan 9 at 5:20
Tyler6Tyler6
838413
$endgroup$
The notation $d(x,0)=1$ seems to indicate that this is a problem in a general metric space. This is not necessarily the absolute value, as there can be more abstract metrics. This set is the unit ball in $mathbb{R}^N$ with respect to this metric.
answered Jan 9 at 5:20
Tyler6Tyler6
838413
answered Jan 9 at 5:20
Tyler6Tyler6
838413
answered Jan 9 at 5:20
Tyler6Tyler6
838413
answered Jan 9 at 5:20
Tyler6Tyler6
838413
838413
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
add a comment |
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
$begingroup$
There is a lot of unknowns in the question as presented. I'm assuming that it's a distance formed from an induced norm of an inner product space, but we don't know for sure (unless I've missed something of course).
$endgroup$
– DavidG
Jan 12 at 9:10
add a comment |
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$begingroup$
What are you working with here? you have defined a subset of $R^n$ that satisfies some condition $d(x,0) = 1$... what is 'distance' here? how is this distance defined?
$endgroup$
– DavidG
Jan 12 at 9:09